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372 Vectors<br />
5<br />
Vectors<br />
5.1 VECTORS<br />
A <strong>vector</strong> is a quantity having both magnitude and direction such as force, velocity<br />
acceleration, displacement etc.<br />
5.2 ADDITION OF VECTORS<br />
Let a and b be two given <strong>vector</strong>s<br />
OA<br />
= a and AB<br />
= b then <strong>vector</strong> OB<br />
<br />
is called the<br />
sum of a and b .<br />
Symbolically<br />
— — <br />
OA AB = OB — <br />
<br />
a b = OB — <br />
5.3 RECTANGULAR RESOLUTION OF A VECTOR<br />
Let OX, OY, OZ be the three rectangular axes. Let ^i , ^j , ^k be three unit <strong>vector</strong>s and<br />
parallel to three axes.<br />
If OP<br />
= n and the co-ordinates of P be (x, y, z) Z<br />
OA<br />
= x ^i , OB<br />
= y ^j and OC<br />
<br />
= z ^k<br />
(x, y, z)<br />
C<br />
P<br />
<br />
<br />
OP<br />
<br />
= OF FP<br />
OP<br />
<br />
= ( OA AF)<br />
FP<br />
OP<br />
<br />
= OA OB OC<br />
<br />
r<br />
= x ^i + y ^j + z ^k<br />
OP 2 = OF 2 + FP 2<br />
= (OA 2 + AF 2 ) + FP 2 = OA 2 + OB 2 + OC 2 = x 2 + y 2 + z 2<br />
2 2 2<br />
OP = x y z<br />
<br />
2 2 2<br />
| r | = x y z<br />
5.4 UNIT VECTOR<br />
Let a <strong>vector</strong> be x ^i + y ^j + z ^k .<br />
Unit <strong>vector</strong> =<br />
^ ^ ^<br />
xi yj<br />
zk<br />
2 2 2<br />
x y z<br />
372<br />
X<br />
O<br />
<br />
xi<br />
A<br />
O<br />
<br />
k<br />
– r<br />
a + b<br />
a<br />
y j<br />
<br />
j<br />
zk <br />
A<br />
<br />
xi<br />
F<br />
(x, y)<br />
B<br />
b<br />
B<br />
Y
Vectors 373<br />
Example 1. If a and b be two unit <strong>vector</strong>s and be the angle between them, then find<br />
the value of such that a + b is a unit <strong>vector</strong>. (Nagpur, University, Winter 2001)<br />
Solution.Let<br />
<br />
OA = a be a unit <strong>vector</strong> and<br />
be the angle between a and b .<br />
If<br />
<br />
OB = c = a + b is also a unit <strong>vector</strong> then, we have<br />
<br />
| OA | = 1<br />
<br />
| OB | = 1<br />
<br />
| OB | = 1<br />
OAB is an equilateral triangle.<br />
Hence each angle of OAB is 3<br />
5.5 POSITION VECTOR OF A POINT<br />
<br />
AB = b is another unit <strong>vector</strong> and <br />
Ans.<br />
The position <strong>vector</strong> of a point A with respect to origin O is the <strong>vector</strong> OA which is<br />
used to specify the position of A w.r.t. O.<br />
—<br />
To find AB if the position <strong>vector</strong>s of the point A and point B are given.<br />
If the position <strong>vector</strong>s of A and B are a and b . Let the origin be O.<br />
Then<br />
<br />
<br />
= <br />
OA<br />
<br />
a,<br />
OB b<br />
<br />
OA AB = OB<br />
AB<br />
<br />
= OB OA<br />
AB<br />
= b <br />
<br />
O<br />
a<br />
AB<br />
<br />
= Position <strong>vector</strong> of B – Position <strong>vector</strong> of A<br />
Example 2. If A and B are (3, 4, 5) and (6, 8, 9), find AB<br />
Solution.<br />
AB<br />
<br />
.<br />
= Position <strong>vector</strong> of B – Position <strong>vector</strong> of A<br />
= (6 iˆ 8 ˆj 9 kˆ) (3ˆi 4ˆj 5 kˆ)<br />
= 3iˆ<br />
4 ˆj 4 kˆ<br />
Ans.<br />
5.6 RATIO FORMULA<br />
To find the position <strong>vector</strong> of the point which divides the line joining two given<br />
points.<br />
Let A and B be two points and a point C divides AB in the ratio of m : n.<br />
Let O be the origin, then<br />
OA<br />
= <br />
<br />
A<br />
m C n B<br />
a , and OB b, OC ? (a)<br />
(b)<br />
<br />
<br />
OC<br />
= OA AC<br />
<br />
m<br />
m <br />
= OA AB AC<br />
AB <br />
m n m n <br />
m <br />
<br />
= a .( b a ) ( AB b a)<br />
m n<br />
O<br />
a<br />
A<br />
<br />
c<br />
a + b = c<br />
<br />
B (b)<br />
O<br />
b<br />
B<br />
A (a)
374 Vectors<br />
OC<br />
=<br />
<br />
m n<br />
Cor. If m = n = 1, then C will be the mid-point, and<br />
<br />
mb na<br />
OC<br />
a b<br />
=<br />
2<br />
5.7 PRODUCT OF TWO VECTORS<br />
<br />
<br />
The product of two <strong>vector</strong>s results in two different ways, the one is a number and<br />
the other is <strong>vector</strong>. So, there are two types of product of two <strong>vector</strong>s, namely scalar<br />
product and <strong>vector</strong> product. They are written as <br />
a .<br />
<br />
b and a <br />
<br />
b .<br />
5.8 SCALAR, OR DOT PRODUCT<br />
The scalar, or dot product of two <strong>vector</strong>s a and b is defined to be <br />
a<br />
scalar where is the angle between a and b .<br />
Symbolically, <br />
a .<br />
<br />
b = a b cos <br />
<br />
b cos i.e.,<br />
Due to a dot between a and <br />
b this product is also called dot product.<br />
The scalar product is commutative<br />
To Prove.<br />
Proof.<br />
<br />
<br />
<br />
<br />
a . b = b . a<br />
<br />
<br />
b . a = b a<br />
cos ( )<br />
b<br />
B<br />
= <br />
a b cos <br />
0<br />
A<br />
a<br />
= a .<br />
<br />
b Proved.<br />
Geometrical interpretation. The scalar product of two <strong>vector</strong>s is the product of one<br />
<strong>vector</strong> and the length of the projection of the other in the direction of the first.<br />
Let<br />
then<br />
<br />
=<br />
OA<br />
<br />
<br />
<br />
<br />
a and OB b<br />
a . b = (OA) . (OB) cos <br />
ON<br />
= OA . OB . OB<br />
= OA . ON<br />
= (Length of a ) (projection of b along a )<br />
5.9 USEFUL RESULTS<br />
^<br />
^<br />
i . i = (1) (1) cos 0° = 1 Similarly, ^ j .<br />
^<br />
j = 1,<br />
^<br />
^<br />
^<br />
^<br />
k . k = 1<br />
i . j = (1) (1) cos 90° = 0 Similarly, ^ j . k ^<br />
= 0, k . i = 0<br />
Note. If the dot product of two <strong>vector</strong>s is zero then <strong>vector</strong>s are prependicular to each other.<br />
5.10 WORK DONE AS A SCALAR PRODUCT<br />
If a constant force F acting on a particle displaces it from A to B then,<br />
Work done = (component of F along AB). Displacement<br />
= F cos . AB<br />
= <br />
F . AB<br />
Work done = Force . Displacement<br />
0<br />
A<br />
<br />
<br />
^<br />
b<br />
^<br />
a<br />
N<br />
B<br />
F<br />
A<br />
B
Vectors 375<br />
5.11 VECTOR PRODUCT OR CROSS PRODUCT<br />
1.<br />
<br />
The <strong>vector</strong>, or cross product of two <strong>vector</strong>s a<br />
and b is defined to be a <strong>vector</strong> such that<br />
(i) Its magnitude is<br />
<br />
a<br />
<br />
b<br />
sin , where is the<br />
<br />
<br />
<br />
b<br />
angle between a and b .<br />
(ii) Its direction is perpendicular to both <strong>vector</strong>s<br />
<br />
a and b<br />
<br />
.<br />
(iii) It forms with a right handed system.<br />
Let ^<br />
be a unit <strong>vector</strong> perpendicular to both the <strong>vector</strong>s a and b .<br />
<br />
<br />
a b =<br />
2. Useful results<br />
<br />
a<br />
<br />
b<br />
<br />
sin .<br />
<br />
Since ^i , ^j , ^k are three mutually perpendicular unit <strong>vector</strong>s, then<br />
^<br />
^<br />
^ ^ ^ ^<br />
i i = j j k k 0<br />
^<br />
^<br />
i j =<br />
^ ^ ^<br />
j i k<br />
^ ^ ^ ^<br />
j i i j<br />
^ ^ ^<br />
ĵ k ˆ = k j <br />
^ ^ ^ ^<br />
i and k j j k<br />
^ ^ ^<br />
kˆ<br />
i ˆ<br />
^ ^ ^ ^<br />
= i k j i k k i<br />
5.12 VECTOR PRODUCT EXPRESSED AS A DETERMINANT<br />
If<br />
<br />
^ ^ ^<br />
a = a i a j a k<br />
1 2 3<br />
^ ^ ^<br />
b = b1 i b2 j b3<br />
k<br />
<br />
a b =<br />
=<br />
^ ^ ^ ^ ^ ^<br />
1 2 3 1 2 3<br />
( a i a j a k) ( b i b j b k)<br />
^ ^ ^ ^ ^ ^ ^ ^ ^ ^<br />
1 1( ) 1 2( ) 1 3( ) 2 1( ) 2 2 ( )<br />
^ ^ ^ ^ ^ ^ ^ ^<br />
ab 2 3 j k ab 3 1k i ab 3 2 k j ab 3 3 k k<br />
ab i i ab i j ab i k ab j i ab j j<br />
^ ^ ^ ^ ^ ^<br />
1 2 1 3 2 1 2 3 3 1 3 2<br />
= ab k a b j ab k a b i ab j a b i<br />
=<br />
=<br />
^ ^ ^<br />
2 3 3 2 1 3 3 1 1 2 2 1<br />
^ ^ ^<br />
( ab ab) i ( a b a b ) j ( ab ab)<br />
k<br />
i j k<br />
a a a<br />
1 2 3<br />
b b b<br />
1 2 3<br />
5.13 AREA OF PARALLELOGRAM<br />
( ) ( ) ( ) ( )<br />
Example 3. Find the area of a parallelogram whose adjacent sides are i – 2j + 3 k and<br />
2i + j – 4k.<br />
^ ^ ^<br />
i j k<br />
Solution. Vector area of gm = 1 2 3<br />
2 1 4<br />
<br />
<br />
a
376 Vectors<br />
^ ^ ^<br />
^ ^ ^<br />
= (8 3) i ( 4 6) j (1 4) k = 5i 10 j 5k<br />
2 2 2<br />
Area of parallelogram = (5) (10) (5) = 5 6 Ans.<br />
5.14 MOMENT OF A FORCE<br />
O<br />
Let a force F ( PQ ) act at a point P.<br />
Moment of <br />
F about O<br />
= Product of force F and perpendicular<br />
distance (ON. ^<br />
)<br />
<br />
= (PQ) (ON)( ^<br />
) = (PQ) (OP) sin (^<br />
) =<br />
<br />
M r F<br />
5.15 ANGULAR VELOCITY<br />
<br />
<br />
OP PQ<br />
Let a rigid body be rotating about the axis OA with the angular<br />
velocity which is a <strong>vector</strong> and its magnitude is radians per second<br />
and its direction is parallel to the axis of rotation OA.<br />
Let P be any point on the body such that OP = r and<br />
AOP = and AP OA. Let the velocity of P be V.<br />
Let be a unit <strong>vector</strong> perpendicular to and r .<br />
<br />
r = ( r sin ) ^ = ( AP) = (Speed of P) ^<br />
r<br />
N P F<br />
A<br />
Axis<br />
<br />
B<br />
<br />
V<br />
r<br />
Q<br />
P<br />
= Velocity of P to and r<br />
<br />
Hence V = <br />
r<br />
5.16 SCALAR TRIPLE PRODUCT<br />
Let a, b,<br />
<br />
c be three <strong>vector</strong>s then their dot product is written as a .( b <br />
c)or[ <br />
a b c ].<br />
If<br />
<br />
<br />
a =<br />
^ <br />
<br />
^ ^ ^ ^ ^ ^ ^ <br />
^<br />
1 2 3 , 1 2 3 , and 1 2 3<br />
a i a j a k b b i b j b k c c i c j c k<br />
^ ^ ^ ^ ^ ^ ^ ^ ^<br />
1 2 3 1 2 3 1 2 3<br />
a .( b c ) = ( a i a j a k).[( b i b j b k) ( c i c j c k)]<br />
=<br />
^ ^ ^ ^ ^ ^<br />
1 2 3 2 3 3 2 3 1 1 3 1 2<br />
2 1<br />
( a i a j a k).[( bc bc ) i ( b c b c ) j ( bc bc) k]<br />
= a 1<br />
(b 2<br />
c 3<br />
– b 3<br />
c 2<br />
) + a 2<br />
(b 3<br />
c 1<br />
– b 1<br />
c 3<br />
) + a 3<br />
(b 1<br />
c 2<br />
– b 2<br />
c 1<br />
)<br />
=<br />
a a a<br />
1 2 3<br />
b b b<br />
1 2 3<br />
c c c<br />
1 2 3<br />
Similarly, b .( c <br />
a)and c .( a <br />
<br />
b ) have the same value.<br />
<br />
a .( b c ) = b .( c <br />
<br />
a ) = c .( a <br />
<br />
b )<br />
The value of the product depends upon the cyclic order of the <strong>vector</strong>, but is<br />
independent of the position of the dot and cross. These may be interchanged.<br />
The value of the product changes if the order is non-cyclic.<br />
Note.<br />
<br />
a ( b . c) and ( a . b)<br />
c are meaningless.<br />
O
Vectors 377<br />
5.17 GEOMETRICAL INTERPRETATION<br />
The scalar triple product a .( b <br />
<br />
c ) represents the volume of the parallelopiped<br />
having a , b , c as its co-terminous edges.<br />
<br />
a .( b c ) = a .Area of gm OBDC <br />
= Area of gm OBDC × perpendicular distance<br />
A<br />
between the parallel faces OBDC and AEFG. a<br />
–<br />
= Volume of the parallelopiped<br />
Note. (1) If a .( b <br />
<br />
c ) = 0, then a, b,<br />
<br />
c are<br />
coplanar.<br />
1 <br />
(2) Volume of tetrahedron ( )<br />
6 a b c .<br />
Example 4. Find the volume of parallelopiped if<br />
^<br />
^ ^ ^ ^ ^ ^ ^ ^ ^<br />
a 3 i 7 j 5k, b 3i 7j 3k,<br />
and c 7 i 5 j 3k<br />
are the three co-terminous edges of the parallelopiped.<br />
Solution.<br />
Volume = a .( b <br />
<br />
c )<br />
3 7 5<br />
=<br />
3 7 3<br />
7 5 3<br />
= 108 – 210 – 170 = – 272<br />
Volume = 272 cube units.<br />
= – 3 (–21 – 15) – 7 (9 + 21) + 5 (15 – 49)<br />
Example 5. Show that the volume of the tetrahedron having<br />
<br />
Ans.<br />
A B, B C,<br />
C A as<br />
concurrent edges is twice the volume of the tetrahendron having A , B , C <br />
as concurrent edges.<br />
1 <br />
Solution. Volume of tetrahendron = ( ) .[( ) ( )]<br />
6 A B B C C A<br />
1 <br />
= ( ) .[ ]<br />
6 A B B C B A C C C A <br />
[ C C 0]<br />
1 <br />
= ( ) .( )<br />
6 A B B C B A C A<br />
1 <br />
= [ .( ) .( ) .( ) .( ) .( ) .( )]<br />
6 A B C A B A A C A B B C B B A B C A<br />
1 1 <br />
= [ A.( BC) B.( CA)] A.( B<br />
C)<br />
6 3<br />
<br />
= 2 1 [ ]<br />
6 ABC<br />
= 2 Volume of tetrahedron having A , B , C <br />
, as concurrent edges. Proved.<br />
EXERCISE 5.1<br />
1. Find the volume of the parallelopiped with adjacent sides.<br />
<br />
OA = 3 i j, OB j 2 k, and OC i 5 j 4k<br />
extending from the origin of co-ordinates O. Ans. 20<br />
2. Find the volume of the tetrahedron whose vertices are the points A (2, –1, –3), B (4, 1, 3)<br />
C (3, 2, –1) and D (1, 4, 2).<br />
O<br />
n^<br />
–<br />
b<br />
G<br />
B<br />
–<br />
c<br />
C<br />
E<br />
Ans.<br />
F<br />
D<br />
1<br />
7 3
378 Vectors<br />
3. Choose y in order that the <strong>vector</strong>s<br />
^ ^ ^ ^ ^<br />
<br />
a 7 i yj kˆ<br />
, b 3 i 2 j k,<br />
^ ^ ^<br />
c 5 i 3 j k are linearly dependent. Ans. y = 4<br />
4. Prove that<br />
<br />
[ a b, b c, c a] 2[ a b c]<br />
5.18 COPLANARITY QUESTIONS<br />
Example 6. Find the volume of tetrahedron having vertices<br />
^ ^ ^<br />
^ ^ ^<br />
^ ^ ^ ^ ^<br />
( j k), ( 4i 5j qk), ( 3i 9j<br />
4k ) and 4( i j k)<br />
.<br />
Also find the value of q for which these four points are coplanar.<br />
(Nagpur University, Summer 2004, 2003, 2002)<br />
<br />
^ ^ ^ ^ ^ ^ ^ ^ ^ ^ ^<br />
Solution. Let A = j k, B4 i 5 jqk, C 3 i 9 j4 k, D 4( i j k)<br />
AB = <br />
^ ^ ^ ^ ^ ^ ^<br />
B A 4 i 5 j qk ( jkˆ<br />
) 4i 6 j( q<br />
1) k<br />
AC = <br />
^ ^ ^ ^ ^ ^ ^<br />
C A (3 i 9ˆj 4 k) ( jk) 3 i 10 j5<br />
k<br />
AD = ^ ^ ^ ^ ^ ^ ^ ^<br />
D A 4( i jk) ( jk) 4i 5 j<br />
5k<br />
Volume of the tetrahedron = 1 [ AB AC AD]<br />
6<br />
4 6 q 1<br />
1<br />
= 3 10 5 = 1 {4(50 25) 6(15 20) ( q 1)(15 40)}<br />
6<br />
6<br />
4 5 5<br />
= 1 {100 210 55 ( q 1)} = 1 ( 110 55 55 q)<br />
6<br />
6<br />
= 1 ( 5555 q) 55 ( q1)<br />
6 6<br />
If four points A, B, C and D are coplanar, then ( AB AC AD ) = 0<br />
i.e., Volume of the tetrahedron = 0<br />
55<br />
<br />
( q 1) = 0 q = 1 Ans.<br />
6<br />
Example 7. If four points whose position <strong>vector</strong>s are a, b, c,<br />
d are coplanar, show that<br />
<br />
[ a b c] [ a d b] [ a d c] [ d b c ] (Nagpur University, Summer 2005)<br />
Solution. Let A, B, C, D be four points whose position <strong>vector</strong>s are a, b, c,<br />
<br />
d .<br />
<br />
AD = d a, BD d b and CD d <br />
<br />
c<br />
<br />
If AD, BD,<br />
CD are coplanar, then<br />
<br />
<br />
AD .( BD CD ) = 0<br />
<br />
( d a) .[( d b) ( d c )] = 0<br />
<br />
<br />
<br />
( d a) .[ d d d c b d b c ] = 0<br />
<br />
( d a) .[ d c b d b c ] = 0<br />
<br />
d .( d c) d .( b d) d .( b c) a .( d c) a .( b a) a .( b c ) = 0<br />
<br />
0 0 [ dbc] [ ddc] [ dbd] [ abc ] = 0<br />
<br />
<br />
[ abc ] [ abd] [ adc] [ dbc]<br />
Proved.
Vectors 379<br />
EXERCISE 5.2<br />
1. Determine such that<br />
^ ^ ^ ^ ^ ^ ^ ^<br />
a i j k, b 2i 4 k, and c i j 3 k are coplanar. Ans. = 5/3<br />
2. Show that the four points<br />
^ ^ ^ ^ ^ ^ ^ ^ ^ ^ ^ ^<br />
6 i 3 j 2 k,3i 2 j 4 k,5i 7 j 3 k and 13 i 17 j k are coplanar.<br />
3. Find the constant a such that the <strong>vector</strong>s<br />
^ ^ ^ ^ ^ ^ ^ ^ ^<br />
2 i j k, i 2 j 3 k,and 3 i a j 5 k are coplanar. Ans. – 4<br />
4. Prove that four points<br />
^ ^ ^ ^ ^ ^ ^ ^ ^ ^ ^<br />
4 i 5 j k, ( j k), 3 i 9 j 4 k,4( i j k)<br />
are coplanar.<br />
<br />
5. If the <strong>vector</strong>s a, b and c are coplanar, show that<br />
<br />
a b c<br />
<br />
a . a a . b a . c<br />
<br />
b . a b . b b . c<br />
5.19 VECTOR PRODUCT OF THREE VECTORS (A.M.I.E.T.E., Summer, 2004, 2000)<br />
= 0<br />
Let a , b and c be three <strong>vector</strong>s then their <strong>vector</strong> product is written as a ( b × <br />
c ).<br />
Let<br />
<br />
^ ^ ^<br />
a = a i a j a k<br />
1 2 3 ,<br />
^ ^ ^<br />
b = b1 i b2 j b3 k,<br />
^ ^ ^<br />
c = c1 i c2 j c3<br />
k<br />
^ ^ ^ ^ ^ ^ ^ ^ ^<br />
( a1 i a2 j a3 k) ( b1 i b2 j b3 k) ( c1 i c2 j c3<br />
k)<br />
^ ^ ^ ^ ^ ^<br />
= a1 i a2 j a3 k bc 2 3 bc 3 2 i b3c1 b1c3 j bc 1 2<br />
bc 2 1 k<br />
a ( b c ) = <br />
=<br />
( ) [( ) ( ) ( ) ]<br />
^<br />
2 1 2 2 1 3 3 1 1 3 3 2 3 3 2 1 1 2<br />
2 1<br />
[ a ( b c b c ) a ( bc bc )] i [ a ( bc bc ) a ( b c b c )] j<br />
[ a ( b c b c ) a ( bc bc) k]<br />
1 3 1 1 3 2 2 3 3 2<br />
^ ^ ^ ^ ^ ^<br />
= ac 1 1 ac 2 2 ac 3 3 b1 i b2 j b3 k a1 b1 ab 2 2 ab 3 3 c1 i c2 j c3<br />
k<br />
= <br />
( )( ) ( )( )<br />
( a . c) b ( a . b) c .<br />
Ans.<br />
Example 8. Prove that :<br />
<br />
a ( b c) b ( c a) c ( a b ) 0 (Nagpur University, Winter 2008)<br />
Solution. Here, we have<br />
<br />
a ( b c) b ( c a) c ( a b )<br />
= <br />
[( a . c) b ( a . b) c] [( b . a) c ( b . c) a] [( c . b) b ( c . a) b]<br />
= <br />
[( b . a) c ( a . b) c] [( c . b) a ( b . c) a] [( a . c) b ( c . a) b]<br />
<br />
= [( a . b) c ( a . b) c] [( b . c) a ( b . c) a] [( c . a) b ( c . a) b]<br />
= 0 + 0 + 0 = 0 Proved.<br />
Example 9. Prove that :<br />
^ ^ ^ ^ ^ ^ <br />
i ( a i) j ( a j) k ( a k) 2 a (Nagpur University, Winter 2003)<br />
^ ^ ^<br />
Solution. Let a = a i a j a k<br />
1 2 3<br />
^<br />
^
380 Vectors<br />
Now, L.H.S. =<br />
^ ^ ^ ^ ^ ^<br />
i ( a i) j ( a j) k ( a k)<br />
<br />
<br />
= ^ i ( a ^ ^ ^ ^ ^ ^ ^ ^ ^<br />
1 i a2 j a3 k) i j ( a1 i a2 j a3<br />
k)<br />
j<br />
=<br />
<br />
<br />
^ ^ ^ ^ ^<br />
k ( a1 i a2 j a3<br />
k)<br />
k<br />
^ ^ ^ ^ ^ ^ ^ ^ ^ ^ ^ ^ ^ ^ <br />
i a1( i i) a2( j i) a3( k i) j a1( i j) a2( j j) a3<br />
( k<br />
j)<br />
<br />
^ ^ ^ ^ ^ ^ ^ <br />
k a1( i k) a2( j k) a3( k k)<br />
<br />
0 0 0<br />
<br />
<br />
= ^ i a ^ ^ ^ ^ ^ ^ ^ ^<br />
2 k a3 j j a1 k a3 i k a1 j a2<br />
i<br />
=<br />
=<br />
=<br />
^ ^ ^ ^ ^ ^ ^ ^ ^ ^ ^ ^<br />
2 3 1 3 1 2<br />
a ( i k) a ( i j) a ( j k) a ( j i) a ( k j) a ( k i)<br />
^ ^ ^ ^ ^ ^<br />
2 3 1 3 1 2<br />
^ ^ ^<br />
1 2 3<br />
a j a k a i a k a i a j = 2a i 2a j 2a k<br />
^ ^ ^<br />
1 2 3<br />
<br />
2( a i a j a k) 2 a Proved.<br />
Example 10. Show that for any scalar , the <strong>vector</strong>s <br />
x,<br />
<br />
y given by<br />
<br />
<br />
q ( a b) (1 p ) p ( a b )<br />
x a , y a <br />
satisfy the equations<br />
2 2<br />
a<br />
q<br />
a<br />
<br />
px qy aand x y b . (Nagpur University, Winter 2004)<br />
Solution. The given equations are<br />
<br />
<br />
px qy = a ...(1)<br />
<br />
<br />
x y = b ...(2)<br />
Multiplying equation (1) <strong>vector</strong>ially by x , we get<br />
<br />
x ( px qy ) = x <br />
<br />
a<br />
<br />
p ( x x) q ( x y ) = x <br />
<br />
a<br />
<br />
q ( x y ) = x <br />
a,<br />
as x <br />
x 0<br />
<br />
x a = qb ,<br />
[From (2) x <br />
y <br />
<br />
b ] ...(3)<br />
<br />
<br />
Multiplying (3) <strong>vector</strong>ially by a , we have<br />
<br />
a ( x a ) = a q <br />
b<br />
<br />
( a . a) x ( a . x)<br />
a =<br />
2<br />
<br />
a x ( a . x)<br />
a =<br />
<br />
<br />
q ( a b )<br />
<br />
<br />
q ( a b ) <br />
2<br />
<br />
a x =<br />
<br />
( a . x) a q ( a b)<br />
x = <br />
2 2<br />
a a<br />
<br />
( a . x) a q ( a b)
Vectors 381<br />
<br />
x =<br />
a <br />
<br />
<br />
q ( a b )<br />
Substituting the value of x in (1), we get<br />
<br />
<br />
<br />
q ( a b ) <br />
qy = a p a <br />
2 <br />
a <br />
<br />
<br />
<br />
(1 p) a p ( a b )<br />
y = <br />
2<br />
q a<br />
EXERCISE 5.3<br />
1. Show that a ( b a) ( a b)<br />
<br />
a<br />
2. Write the correct answer<br />
(a)<br />
<br />
A B C lies in the plane of<br />
( )<br />
(i)<br />
<br />
A and B (ii) B and<br />
<br />
C (iii)<br />
(b) The value of a .( b c) ( a + b <br />
<br />
c ) is<br />
(i) Zero<br />
(ii)<br />
<br />
[ a , b , c] [ b , c , a ] (iii)<br />
5.20 SCALAR PRODUCT OF FOUR VECTORS<br />
Prove the identity<br />
<br />
( a b) .( c d ) = ( a . c) ( b . d) ( a . d) ( b . c)<br />
Proof.<br />
<br />
( a b) .( c d ) =<br />
= <br />
<br />
<br />
( a b) . r<br />
a<br />
2<br />
where =<br />
<br />
<br />
q ( a b ) <br />
p a qy<br />
2<br />
= a<br />
a <br />
<br />
<br />
<br />
2<br />
<br />
a . x<br />
a<br />
Ans.<br />
<br />
C and A Ans. (ii)<br />
<br />
[ a , b , c ] (iv) None of these<br />
Ans. (ii)<br />
a .( b r)<br />
dot and cross can be interchanged. Put c <br />
d <br />
<br />
r<br />
= <br />
<br />
<br />
a .[ b ( c d )] = a .[( b . d) c ( b . c) <br />
d ]<br />
= <br />
=<br />
( a . c) ( b . d) ( a . d) ( b . c)<br />
<br />
a . c a . d<br />
<br />
b . c b . d<br />
EXERCISE 5.4<br />
Proved.<br />
1. If a 2i 3 j k, b i 2j 4 k, c i j k,find ( a b) .( a <br />
c).<br />
Ans. –74<br />
2. Prove that<br />
<br />
2<br />
<br />
( a b) .( a c) a ( b . c) ( a . b)( a . c ).<br />
5.21 VECTOR PRODUCT OF FOUR VECTORS<br />
Let a, b,<br />
<br />
c and d be four <strong>vector</strong>s then their <strong>vector</strong> product is written as<br />
Now,<br />
<br />
( a b) ( c d)<br />
<br />
( a b) ( c d ) = r ( c <br />
<br />
d )<br />
[Put a <br />
b <br />
<br />
r ]<br />
= <br />
( r . d) c ( r . c)<br />
d
382 Vectors<br />
<br />
=<br />
<br />
[( a b). d] c [( a b) . c]<br />
d<br />
= <br />
[ a b d] c [ a b c]<br />
d<br />
( a b) ( c d)<br />
lies in the plane of c and d . ...(1)<br />
<br />
Again, ( a b) ( c d ) = ( a <br />
b)<br />
<br />
<br />
s [Put = c <br />
d <br />
<br />
s ]<br />
<br />
<br />
= s ( a b ) = ( s . b) a ( s . a)<br />
b<br />
=<br />
<br />
[( c d). b] a [( c d) . a]<br />
b =<br />
<br />
[( b c d] a [ a c d]<br />
b<br />
<br />
( a b) ( c d)<br />
lies in the plane of a and b . ...(2)<br />
<br />
Geometrical interpretation : From (1) and (2) we conclude that ( a b) ( c d ) is<br />
a <strong>vector</strong> parallel to the line of intersection of the plane containing a , <br />
b and plane<br />
containing c , d .<br />
Example 11. Show that<br />
<br />
( B C) ( A D) ( C A) ( B D) ( A B) ( C D) 2 ( A BC)<br />
D<br />
Solution. L.H.S. =<br />
=<br />
=<br />
=<br />
=<br />
<br />
( B C) ( A D) ( C A) ( B D) ( A B) ( C D)<br />
<br />
[( BCD) A ( B CA) D] [( C AD) B ( C AB) D] [( B CD) A ( A CD) B]<br />
<br />
( BCD) A ( B CD) A ( C AD) B ( A CD) B ( B CA) D ( C AB)<br />
D<br />
<br />
( ACD) B ( ACD) B ( AB C) D ( ABC)<br />
D<br />
<br />
<br />
2( AB C)<br />
D = R.H.S.<br />
Show that:<br />
<br />
1. ( b c) ( c a) c ( a b c)<br />
when<br />
2.<br />
3.<br />
4.<br />
5.<br />
6.<br />
7.<br />
8.<br />
EXERCISE 5.5<br />
<br />
2<br />
[ b c, c a, a b)] [ a v c]<br />
<br />
d [ a { b ( c d)}] [( b . d)[ a . ( c d)]<br />
<br />
2<br />
<br />
a [ a [ a ( a b)] a ( b a)<br />
<br />
[( a b) ( a c)] . d ( a . d) [ a b c]<br />
2 2 <br />
2<br />
2 ^ ^ ^<br />
2a a i a j a k<br />
^ ^ ^ ^ ^ ^<br />
a b [( i a). b] i [( j a) . b] j [( k a). b]<br />
k<br />
<br />
( a b c ) stands for scalar triple product.<br />
<br />
p [( a q) ( b r)] q [( a r) ( b p)] r [( a p) ( b q )] 0<br />
Proved.
Vectors 383<br />
5.22 VECTOR FUNCTION<br />
If <strong>vector</strong> r is a function of a scalar variable t, then we write<br />
<br />
<br />
r = r()<br />
t<br />
If a particle is moving along a curved path then the position <strong>vector</strong> <br />
r of the particle is a<br />
function of t. If the component of f (t) along x-axis, y-axis, z-axis are f 1<br />
(t), f 2<br />
(t), f 3<br />
(t) respectively.<br />
Then,<br />
— <br />
f()<br />
t<br />
5.23 DIFFERENTIATION OF VECTORS<br />
<br />
= f1() t i f2() t j f3()<br />
t k<br />
Let O be the origin and P be the position of a moving particle at time t.<br />
—<br />
Let<br />
OP = <br />
r<br />
Let Q be the position of the particle at the time t + t and<br />
the position <strong>vector</strong> of Q is OQ — = r r<br />
<br />
—<br />
<br />
<br />
PQ = — <br />
OQ OP<br />
— <br />
<br />
<br />
= ( r r)<br />
r r<br />
r<br />
is a <strong>vector</strong>. As t 0, Q tends to P and the chord<br />
t<br />
becomes the tangent at P.<br />
<br />
<br />
dr r<br />
We define<br />
=<br />
lim<br />
t<br />
0<br />
dt t , then<br />
<br />
dr<br />
dt<br />
<br />
dr<br />
dt<br />
Similarly,<br />
<br />
is a <strong>vector</strong> in the direction of the tangent at P.<br />
is also called the differential coefficient of r <br />
d<br />
2<br />
<br />
dt<br />
2<br />
r<br />
is the second order derivative of r .<br />
r + r<br />
O<br />
with respect to ‘t’.<br />
dr<br />
d r<br />
gives the velocity of the particle at P, which is along the tangent to its path. Also<br />
dt<br />
2<br />
dt<br />
gives the acceleration of the particle at P.<br />
5.24 FORMULAE OF DIFFERENTIATION<br />
<br />
<br />
(i) d <br />
( F G)<br />
d F dG (ii) d <br />
( F )<br />
d F d<br />
<br />
F (U.P. I semester, Dec. 2005)<br />
dt dt dt dt dt dt<br />
<br />
<br />
d<br />
<br />
dG d F d<br />
<br />
dG d F<br />
<br />
(iii) ( FG . ) F. . G(iv) ( F G)<br />
F G<br />
dt dt dt dt dt dt<br />
<br />
d d a d b d c<br />
(v) [ abc]<br />
bc a c ab<br />
<br />
dt dt <br />
dt <br />
dt<br />
<br />
<br />
d d a d b dc<br />
(vi) [ a ( b c)] ( b c)<br />
a c a b<br />
<br />
dt dt dt dt <br />
The order of the functions <br />
F,<br />
G <br />
is not to be changed.<br />
<br />
<br />
<br />
Q<br />
r<br />
r<br />
Tangent<br />
P (r)<br />
2
384 Vectors<br />
Example 12. A particle moves along the curve<br />
<br />
3 2 2 3<br />
r ( t 4 t) i ( t 4) t j (8t 3 t ) k ,<br />
where t is the time. Find the magnitude of the tangential components of its acceleration<br />
at t = 2.<br />
(Nagpur University, Summer 2005)<br />
Solution. We have, r =<br />
Velocity =<br />
<br />
3 2 2 3<br />
( t 4 t) i ( t 4) t j (8t 3 t ) k<br />
<br />
dr<br />
<br />
2 2<br />
(3t 4) i (2t 4) j (16t 9 t ) k<br />
dt<br />
<br />
At t = 2, Velocity = 8i 8 j 4<br />
k<br />
2 <br />
Acceleration = d r <br />
a = 6ti 2 j (1618)<br />
t k<br />
2<br />
dt<br />
<br />
At t = 2 a 12 i 2 j 20 k<br />
The direction of velocity is along tangent.<br />
So the tangent <strong>vector</strong> is velocity.<br />
Unit tangent <strong>vector</strong>,<br />
Tangential component of acceleration, a t<br />
= aT .<br />
<br />
<br />
<br />
<br />
<br />
v 8i 8 j 4k 8i 8 j 4k 2i 2 j k<br />
T = <br />
| v | 64 64 16<br />
12 3<br />
<br />
2i 2 j k<br />
24 4<br />
20 48<br />
= (12i 2 j 20 k).<br />
= = 16 Ans.<br />
3<br />
3 3<br />
<br />
<br />
Example 13. If da u a and db d <br />
u b then prove that [ a b ] u ( a b )<br />
dt<br />
dt<br />
dt<br />
(M.U. 2009)<br />
Solution. We have,<br />
d <br />
[ a b ] =<br />
dt<br />
<br />
<br />
db da <br />
<br />
a b = a ( u b) ( u a)<br />
b<br />
dt dt<br />
<br />
= a ( u b) b ( u a)<br />
<br />
= ( a. b) u ( a. u) b [( b. a) u ( b. u) a]<br />
(Vector triple product)<br />
<br />
= ( a. b) u ( u. a) b ( a. b) u ( u. b)<br />
a<br />
<br />
= ( u. b) a ( u. a)<br />
b<br />
<br />
= u ( a b )<br />
Proved.<br />
Example 14. Find the angle between the surface x 2 + y 2 + z 2 = 9 and z = x 2 + y 2 – 3 at<br />
(2, –1, 2). (M.D.U. Dec. 2009)<br />
Solution. Here, we have<br />
x 2 + y 2 + z 2 = 9 ...(1)<br />
z = x 2 + y 2 – 3 ...(2)<br />
Normal to (1) 1<br />
= (x 2 + y 2 + z 2 – 9)<br />
=<br />
<br />
<br />
<br />
ˆ <br />
<br />
x y z<br />
ˆ ˆ 2 2 2<br />
i j k ( x y z – 9)<br />
= 2 x î + 2 y ĵ + 2 z ˆk<br />
Normal to (1) at (2, – 1, 2), 1<br />
= 4î – 2 ĵ + 4 ˆk ...(3)
Vectors 385<br />
Normal to (2), 2<br />
= (z – x 2 – y 2 + 3)<br />
=<br />
<br />
<br />
<br />
ˆ <br />
<br />
x y z<br />
ˆ ˆ 2 2<br />
i j k ( z– x – y 3)<br />
= – 2 x î – 2 y ĵ + ˆk<br />
Normal to (2) at (2, – 1, 2), 2<br />
= – 4î + 2 ĵ + ˆk ...(4)<br />
1.<br />
2 = | 1|| 2|cos<br />
<br />
1.<br />
2<br />
(4 iˆ –2ˆj 4 kˆ).(– 4iˆ 2 ˆj kˆ) –16 –44<br />
cos = =<br />
=<br />
| 1|| 2|<br />
|4 iˆ –2ˆj 4 kˆ||– 4iˆ 2 ˆj kˆ|<br />
16 416 16 41<br />
= –16 –8<br />
=<br />
6 21 3 21<br />
–1 –8 <br />
= cos <br />
3 21 <br />
–1 –8 <br />
Hence the angle between (1) and (2) cos <br />
Ans<br />
3 21 <br />
EXERCISE 5.6<br />
2<br />
3<br />
t<br />
t<br />
1. The coordinates of a moving particle are given by x = 4t and y = 3 6 t . Find the<br />
2<br />
6<br />
velocity and acceleration of the particle when t = 2 secs. Ans. 4.47, 2.24<br />
2. A particle moves along the curve<br />
x = 2t 2 , y = t 2 – 4t and z = 3t – 5<br />
where t is the time. Find the components of its velocity and acceleration at time t = 1, in the<br />
<br />
8 14 14<br />
direction i 3 j 2 k.<br />
(Nagpur, Summer 2001) Ans. , <br />
7 7<br />
3. Find the unit tangent and unit normal <strong>vector</strong> at t = 2 on the curve x = t 2 – 1, y = 4t – 3,<br />
z = 2t 2 – 6t where t is any variable.<br />
Ans.<br />
1 1 <br />
(2 i 2 j k), (2i 2 k)<br />
3 3 5<br />
<br />
d<br />
<br />
dG d F <br />
4. Prove that ( F G)<br />
F G<br />
dt dt dt<br />
<br />
5. Find the angle between the tangents to the curve 2 3<br />
r t i 2 t j t k,<br />
at the points t = ± 1.<br />
Ans. 1 9<br />
cos<br />
<br />
<br />
<br />
17<br />
<br />
6. If the surface 5x 2 – 2byz = 9x be orthogonal to the surface 4x 2 y + z 3 = 4 at the point (1, –1, 2)<br />
then b is equal to<br />
(a) 0 (b) 1 (c) 2 (d) 3 (AMIETE, Dec. 2009) Ans. (b)<br />
5.25 SCALAR AND VECTOR POINT FUNCTIONS<br />
Point function. A variable quantity whose value at any point<br />
N<br />
^<br />
in a region of space depends upon the position of the point, is<br />
R<br />
Q<br />
called a point function. There are two types of point functions.<br />
n r<br />
(i) Scalar point function. If to each point P (x, y, z) of a<br />
region R in space there corresponds a unique scalar f (P), then f is<br />
P<br />
called a scalar point function. For example, the temperature<br />
distribution in a heated body, density of a body and potential due to gravity are the examples of<br />
a scalar point function.<br />
(ii) Vector point function. If to each point P (x, y, z) of a region R in space there corresponds<br />
a unique <strong>vector</strong> f (P), then f is called a <strong>vector</strong> point function. The velocity of a moving fluid,<br />
gravitational force are the examples of <strong>vector</strong> point function.<br />
(U.P., I Semester, Winter 2000)<br />
= c<br />
<br />
+ d = c
386 Vectors<br />
Vector Differential Operator Del i.e. <br />
The <strong>vector</strong> differential operator Del is denoted by . It is defined as<br />
<br />
= i j k<br />
x y z<br />
5.26 GRADIENT OF A SCALAR FUNCTION<br />
<br />
If (x, y, z) be a scalar function then i j k is called the gradient of the scalar<br />
x y z<br />
function .<br />
And is denoted by grad .<br />
<br />
Thus, grad = i j k<br />
x y z<br />
<br />
gard = i j k ( x, y, z)<br />
x y z<br />
gard = <br />
5.27 GEOMETRICAL MEANING OF GRADIENT, NORMAL<br />
( is read del or nebla)<br />
(U.P. Ist Semester, Dec 2006)<br />
If a surface (x, y, z) = c passes through a point P. The value of the function at each point<br />
on the surface is the same as at P. Then such a surface is called a level surface through P. For<br />
example, If (x, y, z) represents potential at the point P, then equipotential surface (x, y, z) = c<br />
is a level surface.<br />
Two level surfaces can not intersect.<br />
Let the level surface pass through the point P at which the value of the function is . Consider<br />
another level surface passing through Q, where the value of the function is + d.<br />
Let r and r r be the position <strong>vector</strong> of P and Q then PQ<br />
— <br />
r<br />
.dr =<br />
<br />
i j k <br />
.( idx jdy kdz <br />
<br />
)<br />
x y z<br />
=<br />
<br />
dx dy dz d<br />
x y z<br />
If Q lies on the level surface of P, then d = 0<br />
Equation (1) becomes . dr = 0. Then is to dr (tangent).<br />
Hence, is normal to the surface (x, y, z) = c<br />
Let = || N , where N is a unit normal <strong>vector</strong>. Let n be the perpendicular distance<br />
between two level surfaces through P and R. Then the rate of change of in the direction of the<br />
normal to the surface through P is<br />
<br />
.<br />
n<br />
...(1)<br />
d = lim lim<br />
dn n<br />
0 n<br />
n<br />
0<br />
=<br />
=<br />
| | Ndr .<br />
lim<br />
n<br />
n 0<br />
n<br />
0<br />
<br />
<br />
.<br />
dr<br />
n<br />
<br />
| | n<br />
lim | |<br />
n<br />
<br />
<br />
N . r | N || r<br />
|cos <br />
<br />
<br />
<br />
| r|cos<br />
n
Vectors 387<br />
<br />
|| =<br />
n<br />
Hence, gradient is a <strong>vector</strong> normal to the surface = c and has a magnitude equal to the<br />
rate of change of along this normal.<br />
5.28 NORMAL AND DIRECTIONAL DERIVATIVE<br />
(i) Normal. If (x, y, z) = c represents a family of surfaces for different values of the constant<br />
c. On differentiating , we get d = 0<br />
But d = .<br />
<br />
dr so . d r = 0<br />
The scalar product of two <strong>vector</strong>s and dr being zero, and<br />
each other. dr <br />
is in the direction of tangent to the given surface.<br />
Thus is a <strong>vector</strong> normal to the surface (x, y, z) = c.<br />
<br />
<br />
dr are perpendicular to<br />
(ii) Directional derivative. The component of in the direction of a <strong>vector</strong> d is equal to<br />
.d<br />
and is called the directional derivative of in the direction of d .<br />
<br />
r<br />
<br />
r<br />
lim<br />
=<br />
r<br />
0<br />
<br />
r<br />
where, r = PQ<br />
is called the directional derivative of at P in the direction of PQ.<br />
Let a unit <strong>vector</strong> along PQ be N .<br />
Now<br />
n<br />
r<br />
<br />
r<br />
n<br />
n<br />
= cos r = cos <br />
<br />
N.<br />
N<br />
<br />
lim N.<br />
N<br />
<br />
= r<br />
0<br />
<br />
<br />
n<br />
n<br />
<br />
NN<br />
.<br />
<br />
<br />
<br />
<br />
<br />
...(1)<br />
<br />
n<br />
<br />
From (1), r<br />
<br />
<br />
N.<br />
N<br />
<br />
<br />
<br />
<br />
<br />
= N . N | | = N . ( N | | )<br />
<br />
Hence,<br />
r<br />
, directional derivative is the component of in the direction N .<br />
<br />
= N . | |cos | |<br />
r<br />
Hence, is the maximum rate of change of .<br />
Example 15. For the <strong>vector</strong> field (i)<br />
<br />
A miˆ<br />
and (ii) A m r . Find . <br />
A and <br />
A .<br />
Draw the sketch in each case. (Gujarat, I Semester, Jan. 2009)<br />
<br />
<br />
Solution. (i) Vector A mi is represented in the figure (i).<br />
(ii) A = mr is represented in the figure (ii).<br />
(iii)<br />
. <br />
<br />
<br />
A = i j k .( xi y j zk) 1113<br />
x y z<br />
. <br />
A = 3 is represented on the number line at 3.<br />
(iv)<br />
<br />
A = i j k ( xi y j zk)<br />
x y z
388 Vectors<br />
<br />
i j k<br />
=<br />
<br />
x y z<br />
x y z<br />
are represented in the adjoining figure.<br />
^<br />
i<br />
^<br />
k<br />
^<br />
m . i<br />
k<br />
r<br />
<br />
j<br />
O<br />
O<br />
Number line<br />
j 0 1 2 3 O<br />
i<br />
i<br />
(i) (ii) (iii) (iv)<br />
m r<br />
Example 16. If = 3x 2 y – y 3 z 2 ; find grad at the point (1, –2, –1).<br />
(AMIETE, June 2009, U.P., I Semester, Dec. 2006)<br />
Solution. grad = <br />
2 3 2<br />
= i j k (3 x y y z )<br />
x y z<br />
<br />
2 3 2 <br />
2 3 2 2 3 2<br />
= i (3 x y y z ) j (3 x y yz) k (3 x y y z )<br />
x y z<br />
=<br />
= 0<br />
<br />
2 2 2 3<br />
i (6 xy) j (3x 3 y z ) k ( 2 yz)<br />
<br />
grad at (1, –2, –1) = i (6) (1) ( 2) j[(3) (1) 3(4) (1)] k ( 2)( 8) ( 1)<br />
<br />
= 12 i 9 j 16 k<br />
Ans.<br />
Example 17. If u = x + y + z, v = x 2 + y 2 + z 2 , w = yz + zx + xy prove that grad u,<br />
grad v and grad w are coplanar <strong>vector</strong>s. [U.P., I Semester, 2001]<br />
Solution. We have,<br />
grad u =<br />
<br />
<br />
i j k ( x y z)<br />
i j k<br />
x y z<br />
grad v =<br />
<br />
<br />
2 2 2<br />
i j k ( x y z ) 2xi 2yj<br />
2zk<br />
x y z<br />
grad w =<br />
<br />
<br />
i j k ( yz zx xy) i( z y) jz ( x) k( y x)<br />
x y z<br />
[For <strong>vector</strong>s to be coplanar, their scalar triple product is 0]<br />
Now, grad u.(grad v × grad w) =<br />
=<br />
=<br />
1 1 1 1 1 1<br />
2x 2y 2z 2 x y z<br />
z y z x y x z y z x y x<br />
1 1 1<br />
2 x y z x y z x y z<br />
z y z x y x<br />
1 1 1<br />
2( x y z) 1 1 1 0<br />
y z z x x y<br />
k<br />
[Applying R 2<br />
R 2<br />
+ R 3<br />
]<br />
j
Vectors 389<br />
Since the scalar product of grad u, grad v and grad w are zero, hence these <strong>vector</strong>s are<br />
coplanar <strong>vector</strong>s.<br />
Proved.<br />
Example 18. Find the directional derivative of x 2 y 2 z 2 at the point (1, 1, –1) in the direction<br />
of the tangent to the curve x = e t , y = sin 2t + 1, z = 1 – cos t at t = 0.<br />
(Nagpur University, Summer 2005)<br />
Solution. Let = x 2 y 2 z 2<br />
Directional Derivative of <br />
2 2 2<br />
= = i j k ( x y z )<br />
x y z<br />
<br />
2 2 2 2 2 2<br />
= 2xy z i 2yx z j 2zx y k<br />
Directional Derivative of at (1, 1, –1)<br />
Tangent <strong>vector</strong>,<br />
=<br />
Tangent(at t = 0) =<br />
<br />
2 2 2 2 2 2<br />
2(1)(1) ( 1) i 2(1)(1) ( 1) j 2( 1)(1) (1) k<br />
<br />
= 2 i 2 j 2k<br />
...(1)<br />
t<br />
<br />
<br />
r<br />
= xi yj zk e i (sin 2t 1) j (1<br />
cos t)<br />
k<br />
<br />
T dr <br />
t<br />
= e i 2cos 2tj<br />
sin tk<br />
dt<br />
<br />
0<br />
e i 2(cos 0) j (sin 0) k i 2 j<br />
...(2)<br />
( i 2 j)<br />
Required directional derivative along tangent = (2 i 2 j 2 k)<br />
1<br />
4<br />
[From (1), (2)]<br />
= 2 4 0 6 Ans.<br />
5 5<br />
Example 19. Find the unit normal to the surface xy 3 z 2 = 4 at (–1, –1, 2). (M.U. 2008)<br />
Solution. Let (x, y, z) = xy 3 z 2 = 4<br />
We know that is the <strong>vector</strong> normal to the surface (x, y, z) = c.<br />
<br />
Normal <strong>vector</strong> = = i <br />
x j <br />
y k<br />
<br />
<br />
z<br />
Now =<br />
Normal <strong>vector</strong> =<br />
<br />
i ( xyz) j ( xyz) k ( xyz)<br />
x y z<br />
<br />
3 2 3 2 3 2<br />
<br />
3 2 2 2 3<br />
yz i 3xy z j 2xyzk<br />
<br />
Normal <strong>vector</strong> at (–1, –1, 2) = 4i 12 j 4k<br />
Unit <strong>vector</strong> normal to the surface at (–1, –1, 2).<br />
<br />
4i 12 j 4k<br />
1 <br />
= ( i 3 j k ) Ans.<br />
| | 16 144 16 11<br />
Example 20. Find the rate of change of = xyz in the direction normal to the surface<br />
x 2 y + y 2 x + yz 2 = 3 at the point (1, 1, 1). (Nagpur University, Summer 2001)<br />
Solution. Rate of change of =<br />
<br />
<br />
= i j k ( x yz)<br />
iyz jxz kxy<br />
x y z
390 Vectors<br />
<br />
Rate of change of at (1, 1, 1) = ( i j k)<br />
Normal to the surface = x 2 y + y 2 x + yz 2 – 3 is given as -<br />
=<br />
=<br />
() (1, 1, 1)<br />
= 3i 4 j 2 k<br />
2 2 2<br />
i j k ( x y y x yz 3)<br />
x y z<br />
<br />
2 2 2<br />
i(2 xy y ) jx ( 2 xy z ) k2yz<br />
<br />
<br />
Unit normal = 3 i 4 j 2 k<br />
916<br />
4<br />
<br />
Required rate of change of =<br />
(3 i 4 j 2 k)<br />
( i j k).<br />
= 3 4 2 9 <br />
916<br />
4 29 29<br />
Ans.<br />
Example 21. Find the constants m and n such that the surface m x 2 – 2nyz = (m + 4)x will<br />
be orthogonal to the surface 4x 2 y + z 3 = 4 at the point (1, –1, 2).<br />
(M.D.U. Dec. 2009, Nagpur University, Summer 2002)<br />
Solution. The point P (1, –1, 2) lies on both surfaces. As this point lies in<br />
mx 2 – 2nyz = (m + 4)x, so we have<br />
m – 2n (–2) = (m + 4)<br />
m + 4n = m + 4 n = 1<br />
Let 1<br />
= mx 2 – 2yz – (m + 4)x and 2<br />
= 4x 2 y + z 3 – 4<br />
Normal to 1<br />
= 1<br />
=<br />
2<br />
i j k [ mx 2 yz ( m<br />
4) x]<br />
x y z<br />
<br />
= i(2mx m 4) 2zj2yk<br />
Normal to 1<br />
at (1, –1, 2) =<br />
<br />
<br />
i(2m m 4) 4 j 2k<br />
= ( m 4) i 4 j 2 k<br />
Normal to 2<br />
= 2<br />
=<br />
2 3<br />
<br />
i j k (4x y z 4)<br />
2 2<br />
= i8xy 4x j 3z k<br />
x y z<br />
<br />
Normal to 2<br />
at (1, –1, 2) = – 8i 4 j 12<br />
k<br />
Sinec 1<br />
and 2<br />
are orthogonal, then normals are perpendicular to each other.<br />
1<br />
. 2<br />
= 0<br />
<br />
[( m 4) i 4 j 2 k].[ 8i 4 j 12 k]<br />
= 0<br />
– 8 (m – 4) – 16 + 24 = 0<br />
m – 4 = –2 + 3 m = 5<br />
Hence m = 5, n = 1<br />
Ans.<br />
Example 22. Find the values of constants and so that the surfaces x 2 – yz = (+ 2) x,<br />
4x 2 y + z 3 = 4 intersect orthogonally at the point (1, – 1, 2).<br />
(AMIETE, II Sem., Dec. 2010, June 2009)<br />
Solution. Here, we have<br />
x 2 – yz = ( + 2) x ...(1)<br />
4x 2 y + z 3 = 4 ...(2)
Vectors 391<br />
<br />
Normal to the surface (1), λ x yz(λ2)<br />
x<br />
<br />
<br />
<br />
ˆ<br />
<br />
ˆ<br />
ˆ 2<br />
i j k x yz ( 2) x<br />
<br />
x y z<br />
<br />
iˆ(2x2) ˆj ( z) kˆ<br />
( y)<br />
Normal at (1, –1, 2) = î (2 – – 2) – ĵ (–2) + ˆk ...(3)<br />
Normal at the surface (2)<br />
= î ( – 2) + ĵ z (2) + ˆk <br />
<br />
ˆ<br />
<br />
ˆ<br />
ˆ 2 3<br />
i j k (4x yz<br />
4)<br />
x y z<br />
= î (8 × y) + ĵ (4x2 ) + ˆk (3z 2 )<br />
Normal at the point (1, –1, 2) = – 8 î + 4 ĵ + 12 ˆk ...(4)<br />
Since (3) and (4) are orthogonal so<br />
iˆ( 2) ˆj(2 ) kˆ. 8iˆ4ˆj12kˆ<br />
0<br />
<br />
8( 2) 4(2 ) 12 0 8 168120<br />
8 20160<br />
4( 2 54) 0<br />
2 540<br />
25 4<br />
...(5)<br />
Point (1, – 1, 2) will satisfy (1)<br />
<br />
2<br />
(1) ( 1)(2) ( 2)(1) + 2 = + 2 = 1<br />
Putting = 1 in (5), we get<br />
9<br />
25 4 <br />
2<br />
9<br />
Hence and =1<br />
Ans.<br />
2<br />
Example 23. Find the angle between the surfaces x 2 + y 2 + z 2 = 9 and z = x 2 + y 2 – 3 at<br />
the point (2, –1, 2). (Nagpur University, Summer 2002)<br />
Solution. Normal on the surface (x 2 + y 2 + z 2 – 9 = 0)<br />
=<br />
<br />
<br />
2 2 2<br />
i j k ( x y z 9) (2xi 2yj<br />
2 zk)<br />
x y z<br />
<br />
Normal at the point (2, –1, 2) = 4i 2 j 4 k<br />
...(1)<br />
Normal on the surface (z = x 2 + y 2 2 2<br />
– 3) = i j k ( x y z 3)<br />
x y z<br />
<br />
= 2xi 2yj<br />
k<br />
<br />
Normal at the point (2, –1, 2) = 4i 2<br />
j k<br />
...(2)<br />
Let be the angle between normals (1) and (2).<br />
<br />
(4 i 2 j 4 k).(4 i 2 j k)<br />
= 16 416 16 41cos<br />
<br />
16 + 4 – 4 = 6 21cos 16 = 6 21cos
392 Vectors<br />
cos =<br />
8<br />
3 21<br />
=<br />
cos<br />
1 8<br />
3 21<br />
Example 24. Find the directional derivative of 1 r in the direction r where r xi y j zk .<br />
Solution. Here, (x, y, z) =<br />
Now<br />
=<br />
1<br />
<br />
<br />
=<br />
r<br />
<br />
=<br />
Ans.<br />
<br />
(Nagpur University, Summer 2004, U.P., I Semester, Winter 2005, 2002)<br />
1<br />
<br />
2 2 2 2<br />
1 1<br />
( x y z )<br />
r 2 2 2<br />
x y z<br />
1<br />
<br />
<br />
2 2 2 2<br />
<br />
i j k ( x y z )<br />
x y z<br />
<br />
<br />
<br />
1 1 1<br />
<br />
2 2 2 2 2 2 2 2 2 2 2 2<br />
<br />
( x y z ) i ( x y z ) j ( x y z ) k<br />
x y z<br />
<br />
3 3<br />
1 <br />
<br />
2 2 2 1<br />
<br />
<br />
3<br />
2 2 2 2<br />
1<br />
( x y z ) 2 xi ( x y z )<br />
2<br />
2y<br />
j (<br />
2 2 2 )<br />
2 2<br />
2 2<br />
2 x y z <br />
z <br />
<br />
<br />
<br />
k<br />
<br />
( xi yj<br />
zk)<br />
= 2 2 2 3/2<br />
( x y z )<br />
<br />
and r = unit <strong>vector</strong> in the direction of xi yj<br />
zk<br />
<br />
xi yj<br />
zk<br />
=<br />
2 2 2<br />
x y z<br />
So, the required directional derivative<br />
=<br />
<br />
2 2 2<br />
xi yj zk xi yj zk x y z<br />
. r .<br />
<br />
( x y z ) ( x y z ) ( x y z )<br />
1 <br />
1<br />
x y z r<br />
=<br />
2 2 2 2<br />
2 2 2 3/2 2 2 2 1/2 2 2 2 2<br />
[From (1), (2)]<br />
Example 25. Find the direction in which the directional derivative of (x, y) =<br />
x<br />
2 2<br />
y<br />
xy<br />
...(1)<br />
...(2)<br />
(1, 1) is zero and hence find out component of velocity of the <strong>vector</strong> 3 2<br />
r ( t 1)<br />
i t j in<br />
the same direction at t = 1. (Nagpur University, Winter 2000)<br />
2 2<br />
x y <br />
Solution. Directional derivative = = i j k <br />
x y z xy <br />
<br />
2 2 2 2<br />
=<br />
xy .2 x ( x y ) y xy .2 y x ( y x<br />
i<br />
j<br />
) <br />
2 2 2 2 <br />
x y x y <br />
2 3 2 3<br />
=<br />
x y y i<br />
j<br />
xy x <br />
2 2 2 2 <br />
x y x y <br />
<br />
<br />
Directional Derivative at (1, 1) = i 0 j 0 0<br />
Since () (1, 1)<br />
= 0, the directional derivative of at (1, 1) is zero in any direction.<br />
Again r =<br />
<br />
3 2<br />
( t 1)<br />
i t<br />
j<br />
<br />
Ans.<br />
<br />
at
Vectors 393<br />
Velocity, v =<br />
dr <br />
2<br />
3t i 2tj<br />
dt<br />
<br />
<br />
Velocity at t = 1 is = 3i<br />
2 j<br />
The component of velocity in the same direction of velocity<br />
<br />
3i<br />
2 j<br />
9<br />
4<br />
=<br />
(3 i 2 j). 13<br />
9<br />
4 13<br />
Ans.<br />
<br />
Example 26. Find the directional derivative of (x, y, z) = x 2 y z + 4 x z 2 at (1, –2, 1) in<br />
<br />
the direction of 2i j 2 k . Find the greatest rate of increase of .<br />
(Uttarakhand, I Semester, Dec. 2006)<br />
Solution. Here, (x, y, z) = x 2 y z + 4xz 2<br />
Now, =<br />
=<br />
at (1, – 2, 1) =<br />
2 2<br />
i j k ( x yz 4 xz )<br />
x y z<br />
<br />
2 2 2<br />
(2xyz 4 z ) i ( x z) j ( x y 8 xz)<br />
k<br />
2<br />
<br />
{2(1) ( 2)(1) 4(1) } i (11) j {1( 2) 8(1)(1)} k<br />
<br />
= ( 4 4) i j ( 2 8) k = j 6 k<br />
<br />
Let a = unit <strong>vector</strong> = 2 i j 2 k 1 <br />
(2 i j 2 k)<br />
41<br />
4 3<br />
So, the required directional derivative at (1, –2, 1)<br />
1 <br />
= . a ( j 6 k). (2 i j 2 k)<br />
= 1 <br />
( 112)<br />
<br />
13<br />
3<br />
3 3<br />
Greatest rate of increase of = j 6k<br />
= 1<br />
36<br />
<br />
<br />
= 37 Ans.<br />
Example 27. Find the directional derivative of the function = x 2 – y 2 + 2z 2 at the point P<br />
(1, 2, 3) in the direction of the line PQ where Q is the point (5, 0, 4).<br />
(AMIETE, Dec. 20010, Nagpur University, Summer 2008, U.P., I Sem., Winter 2000)<br />
Solution. Directional derivative = <br />
=<br />
<br />
i j k ( x y 2 z ) 2xi 2yj<br />
4zk<br />
x y z<br />
<br />
2 2 2<br />
Directional Derivative at the point P (1, 2, 3) = 2i 4 j 12 k<br />
...(1)<br />
<br />
<br />
PQ = Q P = (5, 0, 4) – (1, 2, 3) = (4, –2, 1) ...(2)<br />
<br />
(4 i 2 j k)<br />
Directional Derivative along PQ = (2 i 4 j 12 k).<br />
[From (1) and (2)]<br />
16 41<br />
=<br />
8812 28<br />
<br />
21 21<br />
Ans.<br />
x<br />
Example 28. For the function (x, y) = 2 2 , find the magnitude of the directional<br />
x y<br />
derivative along a line making an angle 30° with the positive x-axis at (0, 2).<br />
(A.M.I.E.T.E., Winter 2002)
394 Vectors<br />
Solution. Directional derivative = <br />
<br />
x<br />
x y z<br />
x y<br />
<br />
= i j k 2 2<br />
<br />
2 2<br />
y x 2xy<br />
= i j<br />
2 2 2 2 2 2<br />
( x y ) ( x y )<br />
Directional derivative at the point (0, 2)<br />
4<br />
0 2(0) (2) i<br />
<br />
(0 4) (0 4) 4<br />
<br />
<br />
= i j<br />
2 2<br />
<br />
<br />
1 x(2 x) x(2 y)<br />
<br />
<br />
= i<br />
<br />
j<br />
2 2 2 2 2 <br />
2 2 2<br />
x y ( x y ) ( x y )<br />
Directional derivative at the point (0, 2) in the direction CA — 3 1 <br />
i.e. i j<br />
2 2 <br />
<br />
<br />
<br />
CA OB BA i cos 30 j sin 30<br />
i 3 1 <br />
<br />
<br />
= . i j<br />
<br />
3 1 <br />
<br />
4 2 2 <br />
<br />
<br />
i j<br />
<br />
<br />
2 2 <br />
3<br />
=<br />
Ans.<br />
8<br />
<br />
Example 29. Find the directional derivative of V where 2 2 2<br />
V xy i zy j xz k, at the<br />
point (2, 0, 3) in the direction of the outward normal to the sphere x 2 + y 2 + z 2 = 14 at the<br />
point (3, 2, 1). (A.M.I.E.T.E., Dec. 2007)<br />
<br />
Solution. V 2 = VV .<br />
<br />
2 2 2 2 2 2<br />
2<br />
,<br />
<br />
= ( xy i zy j xz k).( x y i zy j xz k)<br />
= x 2 y 4 + z 2 y 4 + x 2 z 4<br />
Directional derivative = 2<br />
V<br />
=<br />
2 4 2 4 2 4<br />
i j k ( x y z y x z )<br />
x y z<br />
=<br />
<br />
4 4 2 3 3 2 4 2 3<br />
(2xy 2 xz ) i (4x y 4 y z ) j (2y z 4 xz)<br />
k<br />
<br />
Directional derivative at (2, 0, 3) = (0 2281) i (0 0) j (0 4427)<br />
k<br />
<br />
= 324 i 432k 108 (3 i 4 k)<br />
...(1)<br />
Normal to x 2 + y 2 + z 2 – 14 = <br />
2 2 2<br />
= i j k ( x y z 14)<br />
x y z<br />
<br />
= (2xi 2yj<br />
2 zk)<br />
<br />
Normal <strong>vector</strong> at (3, 2, 1) = 6i 4 j 2 k<br />
...(2)<br />
Unit normal <strong>vector</strong> = 6 4 2 2(3 2 ) 3 <br />
i j k i j k i 2<br />
<br />
j k<br />
<br />
<br />
36 16 4 2 14 14<br />
<br />
3i 2 j k<br />
Directional derivative along the normal = 108(3 i 4 k). .<br />
14<br />
108 (9 4) 1404<br />
= <br />
14 14<br />
j<br />
—2<br />
1<br />
(0, 2) 30°<br />
i<br />
C —2<br />
3<br />
i<br />
1<br />
A<br />
[From (1), (2)]<br />
j<br />
Ans.
Vectors 395<br />
Example 30. Find the directional derivative of ( f) at the point (1, – 2, 1) in the direction<br />
of the normal to the surface xy 2 z = 3x + z 2 , where f = 2x 3 y 2 z 4 . (U.P., I Semester, Dec 2008)<br />
Solution. Here, we have<br />
f = 2x 3 y 2 z 4<br />
<br />
3 2 4<br />
f = i j k <br />
(2 x y z ) 2 2 4 3 4<br />
<br />
x y z<br />
= 3 2 3<br />
6 x y z i 4 xyz j 8x y z k<br />
<br />
<br />
<br />
2 2 4 3 4 3 2 3<br />
(f) = i j k (6x y z <br />
<br />
i 4x yz j 8 x y z k)<br />
x y z<br />
<br />
<br />
<br />
= 12xy 2 z 4 + 4x 3 z 4 + 24x 3 y 2 z 2<br />
Directional derivative of ( f )<br />
<br />
2 4 3 4 3 2 2<br />
= i j k (12xy z 4xz 24 x y z )<br />
x y z<br />
<br />
<br />
<br />
2 4 2 4 2 2 2<br />
= (12y z 12x z 72 x y z ) i (24xyz4<br />
48x3yz2)<br />
j<br />
+ (48xy 2 z 3 + 16x 3 z 3 + 48x 3 y 2 z) k<br />
Directional derivative at (1, – 2, 1) = (48 + 12 + 288) i + (– 48 – 96) j + (192 + 16 + 192) k <br />
= 348 i<br />
– 144 j 400k<br />
Normal to(xy 2 z – 3x – z 2 ) = (xy 2 z – 3x – z 2 )<br />
<br />
2 2<br />
= i j k <br />
( xy z –3 x – z )<br />
x y z<br />
<br />
<br />
<br />
2<br />
= 2<br />
( y z – 3) i (2 xyz) j ( xy – 2 z)<br />
k<br />
Normal at(1, – 2, 1) = i<br />
– 4 j 2k<br />
i –4j 2k<br />
Unit Normal Vector =<br />
= 1 ( – 4 2 )<br />
116<br />
4 21 i j k<br />
Directional derivative in the direction of normal<br />
1 <br />
= (348 i –144 j 400 k) ( i –4j 2 k)<br />
21<br />
1<br />
= (348 576 800) = 1724<br />
Ans.<br />
21<br />
21<br />
Example 31. If the directional derivative of = a x 2 y + b y 2 z + c z 2 x at the point<br />
x 1 y 3<br />
z<br />
(1, 1, 1) has maximum magnitude 15 in the direction parallel to the line ,<br />
2 2 1<br />
find the values of a, b and c. (U.P. I semester, Winter 2001)<br />
Solution. Given = a x 2 y + b y 2 z + c z 2 x<br />
<br />
= i j k <br />
x y z (a x2 y + b y 2 z + c z 2 x)<br />
=<br />
<br />
2 2 2<br />
i(2 axy cz ) jax ( 2 byz) kby ( 2 czx)<br />
<br />
at the point (1, 1, 1) = i(2 a c) j( a 2 b) kb ( 2 c)<br />
...(1)<br />
We know that the maximum value of the directional derivative is in the direction of . <br />
i.e. || = 15 (2a + c) 2 + (2b + a) 2 + (2c + b) 2 = (15) 2<br />
But, the directional derivative is given to be maximum parallel to the line
396 Vectors<br />
x 1 y 3 z<br />
= <br />
<br />
i.e., parallel to the <strong>vector</strong><br />
2 2 1<br />
2i 2 j k .<br />
On comparing the coefficients of (1) and (2)<br />
...(2)<br />
<br />
2a<br />
c<br />
= 2 b a 2 c b<br />
<br />
2 2 1<br />
2a + c = – 2b – a 3a + 2b + c = 0 ...(3)<br />
and 2b + a = – 2(2c + b)<br />
2b + a = – 4c – 2b a + 4b + 4c = 0 ...(4)<br />
Rewriting (3) and (4), we have<br />
3a 2b c<br />
0 <br />
<br />
<br />
b <br />
c<br />
a 4b 4c<br />
0<br />
4 11 10<br />
= k (say)<br />
a = 4k, b = –11k and c = 10k.<br />
Now, we have<br />
(2a + c) 2 + (2b + a) 2 + (2c + b) 2 = (15) 2<br />
(8k + 10k) 2 + (–22k + 4k) 2 + (20k – 11k) 2 = (15) 2<br />
k =<br />
a =<br />
<br />
5<br />
<br />
9<br />
20<br />
, b =<br />
9<br />
Example 32. If r xi y j zk,<br />
show that :<br />
<br />
r<br />
1<br />
(i) grad r =<br />
<br />
r<br />
r<br />
<br />
Solution. (i) r = xi yj zk r =<br />
r<br />
2r<br />
x<br />
r<br />
Similarly,<br />
y<br />
= 2x <br />
= y r<br />
grad r = r =<br />
and<br />
r<br />
(ii) grad .<br />
3<br />
r<br />
r<br />
x <br />
x<br />
r<br />
r<br />
z <br />
z<br />
r<br />
55<br />
and c =<br />
9<br />
<br />
50<br />
<br />
9<br />
Ans.<br />
(Nagpur University, Summer 2002)<br />
2 2 2<br />
x y z r 2 = x 2 + y 2 + z 2<br />
r r r<br />
i j k r i j k<br />
x y z<br />
x y z<br />
<br />
x y z xi yj<br />
zk r<br />
= i j k Proved.<br />
r r r r r<br />
1<br />
<br />
(ii) grad <br />
r<br />
= 1 1<br />
<br />
1 1 1<br />
i j k = i <br />
j k <br />
r x y zr<br />
xr yr zr<br />
<br />
= 1 r 1 r 1 r<br />
<br />
i j k<br />
2 2 <br />
2 <br />
r x r y<br />
r z<br />
<br />
=<br />
<br />
1 x 1 y 1 z<br />
i j k<br />
2 2 <br />
2 <br />
r r r r r r = xi yj<br />
zk r<br />
Proved.<br />
3 3<br />
r r<br />
2 2<br />
Example 33. Prove that f () r f ´´ () r f´( r)<br />
. (K. University, Dec. 2008)<br />
r<br />
Solution.
Vectors 397<br />
f () r<br />
<br />
i j k f () r<br />
x y z<br />
<br />
2 2 2 2 r r x r y r z <br />
r x y z 2r 2 x , and <br />
<br />
x x r y r z r <br />
r r r<br />
x y z <br />
if´( r) j f´() r k f´( r)<br />
f´( r)<br />
i j k<br />
x y z<br />
<br />
r r r<br />
<br />
<br />
<br />
xi<br />
yj<br />
zk<br />
f´( r) r<br />
2<br />
f () r [ f ( r)]<br />
<br />
xi yj zk <br />
i j k f´( r) x y z <br />
r<br />
<br />
<br />
<br />
x y z<br />
f´( r) <br />
f´() r <br />
f´( r)<br />
x <br />
r<br />
<br />
y <br />
r<br />
<br />
z <br />
r<br />
<br />
<br />
r<br />
r<br />
r1<br />
x<br />
r.1<br />
y<br />
r x r<br />
f´´( r) f´( r) x<br />
y <br />
y<br />
f´´( r)<br />
2 f´( r)<br />
<br />
xr r y<br />
r<br />
2<br />
<br />
r<br />
z<br />
r.1<br />
z<br />
r z f´´( r) f´( r)<br />
r<br />
2<br />
z r <br />
<br />
r<br />
2<br />
2<br />
2<br />
x<br />
y<br />
z<br />
r <br />
r <br />
r <br />
= x x f´´( r) f´( r)<br />
r y y<br />
<br />
f´´( r) f´( r)<br />
r z z<br />
<br />
2 <br />
´´ () ´( )<br />
r<br />
2 f r <br />
f r<br />
2<br />
r r r r r r r r<br />
r<br />
2 2 2 2<br />
2 2<br />
x x r x y<br />
r y z z<br />
r z<br />
= f´´( r) f´( r) f´´( r) f´()<br />
r<br />
3 <br />
<br />
3 f´´ () r <br />
f´( r)<br />
r r r r<br />
3<br />
r r r r<br />
2 2 2 2 2 2 2 2 2<br />
x y z y x z z x y<br />
f´´( r) f´( r) f´´( r) f´() r f´´( r) f´()<br />
r<br />
2 3 2 3 2 3<br />
r r r r r r<br />
<br />
2 2 2 2 2 2 2 2 2<br />
x y z y z z x x y <br />
= f´´( r) f´( r)<br />
2 2 2 <br />
3 3 3 <br />
r r r r r r <br />
2 2 2 2 2 2<br />
2 2<br />
x y x 2( x y z<br />
) r 2r<br />
f´´ () r f´( r)<br />
2 3<br />
= f´´( r) f´( r)<br />
2 3<br />
r<br />
r<br />
r r<br />
2<br />
= f´´() r f´( r)<br />
r<br />
Ans.<br />
EXERCISE 5.7<br />
1. Evaluate grad if = log (x 2 + y 2 + z 2 ) Ans.<br />
<br />
2( xi yj<br />
zk)<br />
2 2 2<br />
x y z<br />
2. Find a unit normal <strong>vector</strong> to the surface x 2 + y 2 + z 2 1<br />
= 5 at the point (0, 1, 2). Ans. ( ˆ 2 ˆ)<br />
5 j k<br />
(AMIETE, June 2010)<br />
3. Calculate the directional derivative of the function (x, y, z) = xy 2 + yz 3 at the point<br />
5<br />
(1, –1, 1) in the direction of (3, 1, –1) (A.M.I.E.T.E. Winter 2009, 2000) Ans.<br />
11<br />
4. Find the direction in which the directional derivative of f (x, y) = (x 2 – y 2 )/xy at (1, 1) is zero.<br />
(Nagpur Winter 2000)<br />
<br />
i j<br />
Ans.<br />
2
398 Vectors<br />
5. Find the directional derivative of the scalar function of (x, y, z) = xyz in the direction of the outer<br />
normal to the surface z = xy at the point (3, 1, 3).<br />
Ans.<br />
27<br />
11<br />
6. The temperature of the points in space is given by T(x, y, z) = x 2 + y 2 – z. A mosquito located at<br />
(1, 1, 2) desires to fly in such a direction that it will get warm as soon as possible. In what direction<br />
should it move?<br />
Ans.<br />
1 (2 i <br />
2 j <br />
<br />
k )<br />
3<br />
7. If (x, y, z) = 3xz 2 y – y 3 z 2 , find grad at the point (1, –2, –1) Ans.<br />
<br />
(16i 9j 4 k)<br />
8. Find a unit <strong>vector</strong> normal to the surface x 2 y + 2xz = 4 at the point (2, –2, 3).<br />
1<br />
Ans. ( i 2 j 2 k<br />
<br />
)<br />
3<br />
9. What is the greatest rate of increase of the function u = xyz 2 at the point (1, 0, 3)? Ans. 9<br />
10. If is the acute angle between the surfaces xyz 2 = 3x + z 2 and 3x 2 – y 2 + 2z = 1 at the point<br />
(1, –2, 1) show that cos = 3/7 6 .<br />
11. Find the values of constants a, b, c so that the maximum value of the directional directive of<br />
= axy 2 + byz + cz 2 x 3 at (1, 2, –1) has a maximum magnitude 64 in the direction parallel to the<br />
axis of z. Ans. a = b, b = 24, c = –8<br />
12. Find the values of and µ so that surfaces x 2 – µ y z = ( + 2)x and 4 x 2 y + z 3 = 4 intersect<br />
orthogonally at the point (1, –1, 2). Ans. = 9 , 1<br />
2 <br />
13. The position <strong>vector</strong> of a particle at time t is R = cos (t – 1) i + sinh (t – 1) j + at 2 k. If at t = 1,<br />
the acceleration of the particle be perpendicular to its position <strong>vector</strong>, then a is equal to<br />
1<br />
(a) 0 (b) 1 (c)<br />
2<br />
5.29 DIVERGENCE OF A VECTOR FUNCTION<br />
The divergence of a <strong>vector</strong> point function F <br />
<br />
(d)<br />
1<br />
2<br />
(AMIETE, Dec. 2009) Ans. (d)<br />
is denoted by div F and is defined as below.<br />
Let F = Fi 1 F 2 j Fk 3<br />
div F1 F2 F F = . F i j k ( iF1 jF2 kF3)<br />
= 3<br />
x y z<br />
x y z<br />
It is evident that div F is scalar function.<br />
5.30 PHYSICAL INTERPRETATION OF DIVERGENCE<br />
Let us consider the case of a fluid flow. Consider a small rectangular parallelopiped of<br />
dimensions dx, dy, dz parallel to x,y and z axes respectively.<br />
<br />
Let V Vx i Vy j Vz<br />
k<br />
be the velocity of the<br />
fluid at P(x, y, z).<br />
Mass of fluid flowing in through the face ABCD in unit time<br />
= Velocity × Area of the face = V x<br />
(dy dz )<br />
Mass of fluid flowing out across the face PQRS per unit time<br />
= V x<br />
(x + dx) (dy dz)<br />
Vx<br />
<br />
= Vx<br />
dx ( dydz)<br />
x<br />
<br />
<br />
Y<br />
Net decrease in mass of fluid in the parallelopiped<br />
corresponding to the flow along x-axis per unit time<br />
O<br />
Z<br />
A<br />
C<br />
D<br />
dz<br />
V x B<br />
S<br />
P<br />
R<br />
Q<br />
X
Vectors 399<br />
Vx<br />
<br />
= Vx<br />
dydz Vx<br />
dx dy dz<br />
x<br />
<br />
<br />
<br />
Vx<br />
= dx dy dz<br />
(Minus sign shows decrease)<br />
x<br />
Vy<br />
Similarly, the decrease in mass of fluid to the flow along y-axis = dx dy dz<br />
y<br />
Vz<br />
and the decrease in mass of fluid to the flow along z-axis = dx dy dz<br />
z<br />
V V<br />
x y Vz<br />
<br />
Total decrease of the amount of fluid per unit time = dx dy dz<br />
x y z<br />
<br />
V<br />
V<br />
x y Vz<br />
Thus the rate of loss of fluid per unit volume = <br />
x y z<br />
= i j k <br />
.( iV <br />
x jV y kV<br />
<br />
z)<br />
= . V div V<br />
x y z<br />
If the fluid is compressible, there can be no gain or loss in the volume element. Hence<br />
div V = 0 ...(1)<br />
and V is called a Solenoidal <strong>vector</strong> function.<br />
Equation (1) is also called the equation of continuity or conservation of mass.<br />
<br />
xi yj<br />
zk<br />
Example 34. If v <br />
,<br />
2 2 2<br />
x y z<br />
find the value of div v .<br />
(U.P., I Semester, Winter 2000)<br />
Solution. We have, v =<br />
<br />
xi yj<br />
zk<br />
2 2 2<br />
x y z<br />
div v =<br />
<br />
xi yj zk<br />
. v = i j k . <br />
<br />
x y z 2 2 2 1/2<br />
( x y z ) <br />
<br />
=<br />
x y z<br />
<br />
<br />
2 2 2 1/2 2 2 2 1/2 2 2 2 1/2<br />
x ( x y z ) y ( x y z ) z<br />
( x y z )<br />
<br />
1<br />
2 2 2 1/2 1<br />
<br />
2 2 2<br />
( x y z ) x. ( x y z )<br />
2<br />
.2x<br />
<br />
2<br />
<br />
=<br />
2 2 2<br />
( x y z )<br />
<br />
1 1<br />
2 2 2 2<br />
1<br />
2 2 2 1/2 1 2 2 2 1/2<br />
<br />
2 2 2<br />
( x y z ) y. ( x y z ) 2 2y<br />
<br />
( x y z ) z. ( x y z ) .2z<br />
<br />
<br />
2<br />
<br />
2<br />
<br />
<br />
<br />
<br />
2 2 2<br />
2 2 2<br />
( x y z )<br />
( x y z )<br />
=<br />
=<br />
2 2 2 2 2 2 2 2 2 2 2 2<br />
( x y z ) x ( x y z ) y ( x y z ) z<br />
<br />
<br />
2 2 2 3/2 2 2 2 3/2 2 2 2 3/2<br />
( x y z ) ( x y z ) ( x y z )<br />
2 2 2 2 2 2<br />
y z x z x y<br />
2 2 2 3/2<br />
( x y z )<br />
=<br />
2 2 2<br />
2( x y z ) 2<br />
<br />
2 2 2 3/2<br />
( x y z ) 2 2 2<br />
( x y z )<br />
<br />
Example 35. If u = x 2 + y 2 + z 2 , and r xi y j zk,<br />
then find div ( ur ) in terms of u.<br />
(A.M.I.E.T.E., Summer 2004)<br />
Ans.
400 Vectors<br />
Solution. div ( ur ) =<br />
<br />
<br />
2 2 2<br />
i j k .[( x y z )( xi yj<br />
zk)]<br />
x y z<br />
=<br />
<br />
<br />
2 2 2 2 2 2 2 2 2<br />
i j k .[( x y z ) xi( x y z ) yj( x y z ) zk]<br />
x y z<br />
=<br />
3 2 2 2 3 2 2 2 3<br />
( x xy xz ) ( x y y yz ) ( x z y z z )<br />
x y z<br />
= (3x 2 + y 2 + z 2 ) + (x 2 + 3y 2 + z 2 ) + (x 2 + y 2 + 3z 2 ) = 5 (x 2 + y 2 + z 2 ) = 5 u Ans.<br />
Example 36. Find the value of n for which the <strong>vector</strong><br />
<br />
r xi y j zk .<br />
Solution. Divergence F =<br />
=<br />
n<br />
r r is solenoidal, where<br />
n 2 2 2 n/2<br />
<br />
. F . r r .( x y z ) ( xi yj<br />
zk)<br />
<br />
<br />
2 2 2 n/2 2 2 2 n/2 2 2 2 n/2<br />
i j k .[( x y z ) xi( x y z ) yj( x y z ) zk]<br />
x y z<br />
= 2<br />
n (x 2 + y 2 + z 2 ) n/2 – 1 (2x 2 ) + (x 2 + y 2 + z 2 ) n/2 + 2<br />
n (x 2 + y 2 + z 2 ) n/2 – 1 (2y 2 )<br />
+ (x 2 + y 2 + z 2 ) n/2 + 2<br />
n (x 2 + y 2 + z 2 ) n/2 – 1 (2z 2 ) + (x 2 + y 2 + z 2 ) n/2<br />
= n(x 2 + y 2 + z 2 ) n/2 – 1 (x 2 + y 2 + z 2 ) + 3 (x 2 + y 2 + z 2 ) n/2<br />
= n(x 2 + y 2 + z 2 ) n/2 + 3(x 2 + y 2 + z 2 ) n/2 = (n + 3) (x 2 + y 2 + z 2 ) n/2<br />
n<br />
If r r is solenoidal, then (n + 3) (x 2 + y 2 + z 2 ) n/2 = 0 or n + 3 = 0 or n = –3. Ans.<br />
<br />
<br />
Example 37. Show that<br />
( a. r) a n( a. r)<br />
r<br />
<br />
. (M.U. 2005)<br />
n n n 2<br />
r <br />
r r<br />
Solution. We have,<br />
Let =<br />
<br />
<br />
x<br />
<br />
a.<br />
r<br />
r<br />
=<br />
n<br />
=<br />
<br />
But r 2 = x 2 + y 2 + z 2 r<br />
2r<br />
x<br />
<br />
( a i a j a k).( xi yj<br />
zk)<br />
1 2 3<br />
a.<br />
r ax 1 ay 2 az<br />
3<br />
<br />
n<br />
n<br />
r r<br />
n<br />
n1<br />
r . a ( ax a y az) nr ( r/ x)<br />
1 1 2 3<br />
2n<br />
n<br />
r<br />
r<br />
= 2x <br />
r x <br />
x<br />
r<br />
n<br />
=<br />
axa y az<br />
1 2 3<br />
n<br />
<br />
ar 1 ( ax 1 a2y az 3 ). nr . x a1<br />
nax ( 1 a2y a3zx<br />
)<br />
=<br />
2n<br />
= <br />
x<br />
r<br />
n<br />
n 2<br />
r r <br />
=<br />
<br />
i j k<br />
x y z<br />
=<br />
1 n<br />
<br />
( a1 i a<br />
n<br />
2 j a3k ) <br />
n 2 [( ax 1 a2y az 3 )( xi yj<br />
zk<br />
<br />
)]<br />
r<br />
r<br />
=<br />
a n<br />
( ar . ) r<br />
n n 2<br />
r r <br />
n<br />
2<br />
r
Vectors 401<br />
<br />
<br />
<br />
Example 38. Let r xi y j zk, r | r | and a is a constant <strong>vector</strong>. Find the value of<br />
<br />
<br />
a r<br />
div <br />
n<br />
r<br />
<br />
<br />
<br />
Solution. Let a = a 1 i a 2 j a 3 k<br />
=<br />
<br />
<br />
<br />
a r = ( a1i a2 j a3<br />
k) ( xi yj<br />
zk)<br />
<br />
<br />
a r<br />
<br />
| r | n<br />
<br />
a r<br />
div <br />
<br />
| r | n<br />
<br />
<br />
<br />
i j k<br />
a a a<br />
= 1 2 3<br />
=<br />
x y z<br />
<br />
= ( az 2 ayi 3 ) ( a1z a3x) j ( ay 1 axk 2 )<br />
<br />
( az ayi ) ( a z a x) j ( ay<br />
axk )<br />
2 3 1 3 1 2<br />
2 2 2 /2<br />
<br />
<br />
a r<br />
= .<br />
<br />
| r | n<br />
( x y z ) n<br />
<br />
( az 2 ayi 3 ) ( a1z a3x) j ( ay 1 axk 2 )<br />
= i j k .<br />
x y z<br />
2 2 2 /2<br />
<br />
( x y z ) n<br />
az 2 ay 3 az 1 ax 3 ( ay 1 ax 2 )<br />
<br />
<br />
2 2 2 n/2 2 2 2 n/2 2 2 2 n/2<br />
x ( x y z ) y ( x y z ) z<br />
( x y z )<br />
n ( azay)2 x n ( a z a x)2 y n ( ay<br />
ax)2z<br />
<br />
n2 n2 n2<br />
2 2 2<br />
2 2 2 2 2 2 2 2 2 2 2<br />
( x y z ) ( x y z ) ( x y z )<br />
2<br />
n<br />
[( azayx ) ( azaxy ) ( a y a xz ) ]<br />
=<br />
2 3 1 3 1 2<br />
=<br />
n 2 2 3 1 3 1 2<br />
2 2 2 2<br />
( x y z )<br />
n<br />
[ azxa xy ayz a xy ayz a zx]<br />
= 0 Ans.<br />
=<br />
n 2 2 3 1 3 1 2<br />
2 2 2 2<br />
( x y z )<br />
Example 39. Find the directional derivative of div ( u ) at the point (1, 2, 2) in the direction<br />
of the outer normal of the sphere x 2 + y 2 + z 2 <br />
4 4 4<br />
= 9 for u x i y j z k <br />
.<br />
<br />
Solution. div ( u ) = . u<br />
<br />
4 4 4 3 3 3<br />
= i j k .( x i y j z k) 4x 4y 4z<br />
x y z<br />
Outer normal of the sphere = (x 2 + y 2 + z 2 – 9)<br />
<br />
<br />
2 2 2<br />
= i j k ( x y z 9) 2xi 2yj<br />
2zk<br />
x y z<br />
Outer normal of the sphere at (1, 2, 2) = 2i 4 j 4 k<br />
...(1)<br />
Directional derivative =<br />
=<br />
<br />
3 3 3<br />
(4x 4y 4 z )<br />
<br />
<br />
<br />
i j k (4x 4y 4 z ) 12x i 12y j 12z k<br />
x y z<br />
<br />
3 3 3 2 2 2<br />
<br />
Directional derivative at (1, 2, 2) = 12 i 48 j 48 k<br />
...(2)
402 Vectors<br />
<br />
2i 4 j 4k<br />
Directional derivative along the outer normal = (12i 48 j 48 k).<br />
416 16<br />
[From (1), (2)]<br />
24 192 192<br />
= = 68 Ans.<br />
6<br />
Example 40. Show that div (grad r n ) = n (n + 1)r n – 2 , where<br />
2 2 2<br />
r = x y z<br />
2 1<br />
<br />
Hence, show that = 0. (U.P. I Semester, Dec. 2004, Winter 2002)<br />
r<br />
<br />
Solution. grad (r n <br />
n <br />
n n<br />
) = i r j r k r by definition<br />
x y z<br />
<br />
n 1 r <br />
n 1 r <br />
n<br />
1r<br />
<br />
n 1 r r <br />
r<br />
= inr j nr knr . = nr i j k<br />
x y z<br />
<br />
x y z<br />
<br />
n 1 x y z<br />
<br />
<br />
n2 n 2<br />
= nr i j k nr ( xi y j zk) nr r.<br />
r r r<br />
<br />
<br />
2 2 2 2 r r x <br />
<br />
r x y z 2r 2x etc.<br />
x x r<br />
<br />
<br />
<br />
<br />
Thus, grad (r n ) = n 2 n 2 n 2<br />
nr xi nr yj nr zk<br />
...(1)<br />
<br />
div grad r n = div [ n2 n2 n 2<br />
nr xi nr yj nr zk ]<br />
=<br />
=<br />
=<br />
<br />
<br />
n2 n 2 n 2<br />
i j k .( nr xi nr y j nr zk)<br />
x y z<br />
n 2 n 2 n 2<br />
( nr <br />
x) ( nr <br />
y) ( nr z)<br />
x y z<br />
n2 n3 r<br />
n2 n 3<br />
r<br />
<br />
nr nx ( n 2) r nr ny ( n 2) r <br />
x y<br />
<br />
nr nz ( n<br />
2) r<br />
<br />
n n r r r<br />
3 nr nn ( 2) r <br />
x y z<br />
<br />
<br />
x y z<br />
<br />
= 2 3<br />
[From (1)]<br />
(By definition)<br />
<br />
<br />
z<br />
<br />
n2 n3<br />
r<br />
n 2 n 3 x y z<br />
= 3 nr nn ( 2) r <br />
x y <br />
z <br />
<br />
r r r<br />
<br />
<br />
2 2 2 2 r r x <br />
<br />
r x y z 2r 2x etc.<br />
x x r<br />
<br />
<br />
<br />
= 3nr n – 2 + n (n – 2)r n – 4 [x 2 + y 2 + z 2 ]<br />
= 3nr n – 2 + n (n – 2) r n – 4 .r 2 ( r 2 = x 2 + y 2 + z 2 )<br />
= r n – 2 [3n + n 2 – 2n] = r n – 2 (n 2 + n) = n(n + 1) r n – 2<br />
If we put n = –1<br />
div grad (r – 1 ) = –1 (–1 + 1) r – 1 – 2<br />
2 1<br />
<br />
<br />
r<br />
= 0<br />
<br />
r <br />
1<br />
Ques. If r xi y j zk,<br />
and r = |r| find div .<br />
2 (U.P. I Sem., Dec. 2006) Ans. 2<br />
r<br />
<br />
r
Vectors 403<br />
EXERCISE 5.8<br />
<br />
<br />
<br />
1. If r = xi yj zk<br />
r <br />
and r = | r | , show that (i) div<br />
3<br />
| r |<br />
<br />
= 0,<br />
<br />
(ii) div (grad r n ) = n (n + 1) r n – 2 (AMIETE, June 2010) (iii) div (r ) = 3 + r grad .<br />
<br />
2. Show that the <strong>vector</strong> V = ( x3 y) i ( y 3 z) j ( x 2 z)<br />
k is solenoidal.<br />
(R.G.P.V., Bhopal, Dec. 2003)<br />
3. Show that .( A) = .A + (.A)<br />
4. If , , z are cylindrical coordinates, show that grad (log ) and grad are solenoidal <strong>vector</strong>s.<br />
5. Obtain the expression for 2 f in spherical coordinates from their corresponding expression in<br />
orthogonal curvilinear coordinates.<br />
Prove the following:<br />
<br />
6. .( F) ( ). F ( . F )<br />
<br />
7. (a) .() = 2 <br />
(b)<br />
( A<br />
R) (2 n) A n( A. R)<br />
R <br />
, | |<br />
n n n 2 r R<br />
r r r<br />
8. div ( f g) – div (g f) = f 2 g – g 2 f<br />
5.31 CURL (U.P., I semester, Dec. 2006)<br />
The curl of a <strong>vector</strong> point function F is defined as below<br />
Curl F <br />
<br />
<br />
curl F = F<br />
<br />
= i j k ( F1 i F2 j F3<br />
k)<br />
x y z<br />
<br />
i j k<br />
=<br />
1 2 3<br />
is a <strong>vector</strong> quantity.<br />
5.32 PHYSICAL MEANING OF CURL<br />
<br />
<br />
<br />
( F F i F j F k)<br />
1 2 3<br />
F<br />
3 F2 <br />
F<br />
3 F1<br />
F2 F1<br />
i j k <br />
x y z <br />
y z<br />
<br />
x z<br />
<br />
x y<br />
<br />
F F F<br />
(M.D.U., Dec. 2009, U.P. I Semester, Winter 2009, 2000)<br />
We know that V r,<br />
where is the angular velocity, V is the linear velocity and <br />
r<br />
is the position <strong>vector</strong> of a point on the rotating body.<br />
<br />
Curl V 1 i 2 j 3<br />
k <br />
<br />
= V<br />
<br />
r x i y j zk <br />
<br />
<br />
= ( r ) = [( 1i 2 j 3<br />
k) ( xi y j zk)]<br />
=<br />
<br />
i j k<br />
<br />
= <br />
[( 2z 3y) i ( 1z 3x) j ( 1y 2x) k]<br />
<br />
1 2 3<br />
x y z<br />
<br />
x y z<br />
<br />
= i j k 2z 3y i 1z 3x j 1y 2x k<br />
<br />
[( ) ( ) ( ) ]
404 Vectors<br />
=<br />
<br />
i j k<br />
<br />
x y z<br />
z y xz y x<br />
2 3 3 1 1 2<br />
<br />
= ( 1 2) i ( 2 2) j ( 3 3)<br />
k<br />
<br />
= 2( 1 i 2 j 3<br />
k) 2<br />
Curl V = 2 which shows that curl of a <strong>vector</strong> field is connected with rotational properties<br />
of the <strong>vector</strong> field and justifies the name rotation used for curl.<br />
If Curl F = 0, the field F is termed as irrotational.<br />
<br />
<br />
2 2 2<br />
Example 41. Find the divergence and curl of v ( xyz) i (3 x y) j ( xz y z)<br />
k at<br />
(2, –1, 1) (Nagpur University, Summer 2003)<br />
Solution. Here, we have<br />
<br />
2 2 2<br />
<br />
v = ( xyz) i (3 x y) j ( xz y z)<br />
k<br />
Div. <br />
v = <br />
Div 2 2 2<br />
v = ( xyz) (3 x y) ( xz y z)<br />
x y z<br />
= yz + 3x 2 + 2x z – y 2 = –1 + 12 + 4 – 1 = 14 at (2, –1, 1)<br />
Curl v =<br />
<br />
i j k<br />
<br />
x y z<br />
2 2 2<br />
xyz 3x y xz y z<br />
<br />
2<br />
= 2 yz i ( xy z ) j (6 xy xz)<br />
k<br />
Curl at (2, –1, 1)<br />
<br />
= 2( 1)(1) i {(2) ( 1) 1} j {6(2)( 1) 2(1)} k<br />
Example 42. If<br />
<br />
=<br />
<br />
2<br />
2 yz i ( z xy) j (6 xy xz)<br />
k<br />
= 2i 3 j 14k<br />
Ans.<br />
<br />
xi yj<br />
zk<br />
V <br />
x y z<br />
2 2 2 ,<br />
<br />
find the value of curl V .<br />
Solution. Curl V = V<br />
<br />
xi yj<br />
zk<br />
= i j k <br />
x y z 2 2 2 1/2<br />
( x y z ) <br />
<br />
=<br />
<br />
<br />
i j k<br />
<br />
x y z<br />
x y z<br />
(U.P., I Semester, Winter 2000)<br />
2 2 2 1/2 2 2 2 1/2 2 2 2 1/2<br />
( x y z ) ( x y z ) ( x y z )
Vectors 405<br />
=<br />
z y <br />
z <br />
i <br />
j<br />
2 2 2 1/2 2 2 2 1/2 2 2 2 1/2<br />
y <br />
( x y z ) z<br />
<br />
<br />
<br />
<br />
<br />
( x y z )<br />
<br />
x<br />
<br />
( x y z )<br />
<br />
<br />
x y x <br />
k<br />
<br />
2 2 2 1/2 2 2 2 1/2 2 2 2 1/2<br />
z <br />
( x y z ) x<br />
<br />
( x y z ) y<br />
<br />
<br />
( x y z ) <br />
yz yz . zx zx <br />
<br />
( x y z ) ( x y z ) ( x y z ) ( x y z<br />
) <br />
xy<br />
xy <br />
k <br />
0<br />
2 2 2 3/2 2 2 2 3/2 <br />
( x y z ) ( x y z<br />
) <br />
<br />
<br />
= i<br />
j<br />
<br />
2 2 2 3/2 2 2 2 3/2 2 2 2 3/2 2 2 2 3/2<br />
Example 43. Prove that<br />
2 2<br />
<br />
<br />
y z yz x i xz xy j xy xz z k <br />
Ans.<br />
( – 3 –2 ) (3 2 ) (3 –2 2 ) is both<br />
solenoidal and irrotational. (U.P., I Sem, Dec. 2008)<br />
Solution. Let F =<br />
2 2<br />
<br />
<br />
y z yz x i xz xy j xy xz z k <br />
( – 3 –2 ) (3 2 ) (3 –2 2)<br />
<br />
For solenoidal, we have to prove .F = 0.<br />
<br />
<br />
2 2<br />
<br />
Now, .F = i j k ( y – z 3 yz –2 x) i (3xz 2 xy) j (3 xy –2xz 2 z)<br />
k<br />
x y z<br />
<br />
<br />
<br />
= – 2 + 2x – 2x + 2 = 0<br />
Thus, F is solenoidal. For irrotational, we have to prove Curl F = 0.<br />
Now, Curl F =<br />
Thus, F is irrotational.<br />
<br />
i j k<br />
<br />
x y z<br />
2 2<br />
y – z 3 yz –2x 3xz 2xy 3 xy – 2xz 2z<br />
<br />
= (3z 2 y – 2y 3 z) i – (– 2z 3 y –3y 2 z)<br />
j <br />
<br />
= 0i 0 j 0 k = 0<br />
(3z 2 y – 2 y – 3) z k<br />
Hence, F is both solenoidal and irrotational.<br />
Proved.<br />
Example 44. Determine the constants a and b such that the curl of <strong>vector</strong><br />
2 2<br />
<br />
A = (2xy 3 yz) i ( x axz –4 z ) j –(3 xy byz)<br />
k is zero.<br />
(U.P. I Semester, Dec 2008)<br />
Solution. Curl A =<br />
<br />
<br />
<br />
<br />
2 2<br />
i j k [(2xy 3 yz) i ( x axz – 4 z ) j<br />
x y z<br />
<br />
<br />
<br />
=<br />
<br />
i j k<br />
<br />
x y z<br />
2 2<br />
2xy 3 yz x axz –4 z –3 xy – byz<br />
<br />
<br />
<br />
(3 xy byz) k]
406 Vectors<br />
<br />
= [– 3 x – bz – ax 8 z] i –[– 3 y –3 y] j [2 x az –2 x –3 zk ]<br />
<br />
= [– x(3 a) z(8– b)] i 6 y j z(–3 a)<br />
k<br />
= 0 (given)<br />
i.e., 3 + a = 0 and 8 – b = 0, – 3 + a = 0<br />
a = – 3, 3 b = 8 a = 3 Ans.<br />
Example 45. If a <strong>vector</strong> field is given by<br />
<br />
2 2<br />
F ( x – y x) i – (2 xy y)<br />
j . Is this field irrotational ? If so, find its scalar potential.<br />
Solution. Here, we have<br />
<br />
=<br />
2 2<br />
F<br />
Curl F =<br />
=<br />
=<br />
<br />
( x – y x) i – (2 xy y)<br />
j<br />
F <br />
<br />
<br />
<br />
<br />
<br />
<br />
2 2<br />
i j k ( x – y x) i – (2 xy y)<br />
j<br />
x y z<br />
<br />
i j k<br />
<br />
x y z<br />
2 2<br />
x – y x – 2 xy – y 0<br />
Hence, <strong>vector</strong> field F is irrotational.<br />
To find the scalar potential function <br />
F <br />
= <br />
<br />
<br />
(U.P. I Semester, Dec 2009)<br />
= i(0 –0)– j(0 –0) k(–2y 2 y)<br />
= 0<br />
d = dx dy <br />
<br />
<br />
dz = i j k ( i dx jdy kdz)<br />
x y z<br />
x y z<br />
<br />
<br />
<br />
<br />
<br />
= i j k ( d r )<br />
x y z<br />
<br />
<br />
= dr = Fdr <br />
<br />
2 2<br />
= [( x – y x) i<br />
–(2 xy y) j]( i dx jdy kdz)<br />
= (x 2 – y 2 + x)dx – (2xy + y)dy.<br />
2 2<br />
= [( x – y x) dx –(2 xy ydy ) ] c<br />
=<br />
Hence, the scalar potential is<br />
<br />
<br />
2 2<br />
x d x xdx ydy y dx 2 xy dy<br />
c =<br />
<br />
Example 46. Find the scalar potential function f for<br />
Solution. We have, A =<br />
3 2 2<br />
x 2<br />
<br />
<br />
x – y – xy c<br />
3 2 2<br />
3 2 2<br />
x x y 2<br />
– – xy c<br />
Ans.<br />
3 2 2<br />
Curl A = A <br />
=<br />
<br />
2 2<br />
y i 2xy j z k<br />
<br />
2 2<br />
A y i 2xyj z k .<br />
(Gujarat, I Semester, Jan. 2009)<br />
<br />
2 2<br />
i j k ( y i 2 xyj<br />
z k)<br />
x y z
Vectors 407<br />
=<br />
<br />
i j k<br />
<br />
x y z<br />
2 2<br />
y 2xy z<br />
<br />
= i(0) j(0) k(2y 2 y)<br />
= 0<br />
Hence, A is irrotational. To find the scalar potential function f.<br />
<br />
A<br />
= f<br />
df = f dx <br />
f dy <br />
f<br />
dz<br />
x y z<br />
<br />
= i j k f.<br />
dr<br />
x y z<br />
<br />
= Adr .<br />
f f f<br />
<br />
= i j k .( i dx jdy kdz)<br />
x y z<br />
<br />
= f.<br />
dr <br />
<br />
2 2<br />
= ( y i 2 xy j z k).( i dx jdy kdz)<br />
= y 2 dx + 2xy dy – z 2 dz = d (xy 2 ) – z 2 dz<br />
<br />
=<br />
f = d 2 2<br />
( xy ) z dz<br />
xy<br />
2<br />
3<br />
(A = f)<br />
z<br />
C<br />
Ans.<br />
3<br />
Example 47. A <strong>vector</strong> field is given by <br />
A = (x 2 + xy 2 ) i + (y 2 + x 2 y) j . Show that the field<br />
is irrotational and find the scalar potential.(Nagpur Univeristy, Summer 2003, Winter 2002)<br />
Solution. A is irrotational if curl A = 0<br />
<br />
i j k<br />
Curl A =<br />
Hence, A <br />
<br />
A<br />
<br />
<br />
A <br />
x y z<br />
2 2 2 2<br />
x xy y x y<br />
is irrotational. If is the scalar potential, then<br />
= grad <br />
d = dx <br />
dy <br />
dz<br />
x y z<br />
<br />
= i j k .( i dx jdy kdz)<br />
= grad . dr<br />
x y z<br />
<br />
<br />
0<br />
<br />
2 2 2 2<br />
<br />
= i(0 0) j(0 0) k(2xy 2 xy) 0<br />
[Total differential coefficient]<br />
= Adr . = [( x xy ) i ( y x y) j].( idx jdy kdz)<br />
= (x 2 + xy 2 ) dx + (y 2 + x 2 y) dy = x 2 dx + y 2 dy + (x dx)y 2 + (x 2 ) (y dy)<br />
=<br />
2 2 2 2<br />
xdx y dy [( xdx ) y ( x )( y dy )] =<br />
<br />
2 2<br />
3 3 2 2<br />
x y x y<br />
c Ans.<br />
3 3 2<br />
Example 48. Show that V( x, y, z) 2 x yzi ( x z 2 y)<br />
j x yk is irrotational and find a<br />
scalar function u(x, y, z) such that V <br />
= grad (u).<br />
Solution. V (x, y, z) =<br />
<br />
2 2<br />
2 xyzi ( x z 2 y)<br />
j x yk
408 Vectors<br />
Curl V =<br />
=<br />
=<br />
<br />
<br />
2 2<br />
i j k [2 x yzi ( x z 2 y) j x yk]<br />
x y z<br />
<br />
i j k<br />
<br />
x y z<br />
2 2<br />
2xyz x z 2y x y<br />
2 2<br />
<br />
( x x ) i (2xy 2 xy) j (2xz 2 xz) k 0<br />
Hence, V (x, y, z) is irrotational.<br />
To find corresponding scalar function u, consider the following relations given<br />
V <br />
= grad (u)<br />
<br />
or V = ( u)<br />
...(1)<br />
u u u<br />
du =<br />
dx <br />
dy <br />
dz (Total differential coefficient)<br />
x y z<br />
u u u<br />
<br />
= i j k .( i dx jdy k dz)<br />
x y z<br />
<br />
<br />
= udr<br />
. V.<br />
d r<br />
[From (1)]<br />
<br />
2 2<br />
= [2 xyzi ( xz 2 y) j x yk].( i dx jdy kdz)<br />
= 2 x y z dx + (x 2 z + 2y) dy + x 2 y dz<br />
= y(2x z dx + x 2 dz) + (x 2 z) dy + 2y dy<br />
= [yd (x 2 z) + (x 2 z) dy] + 2y dy = d(x 2 yz) + 2y dy<br />
Integrating, we get u = x 2 yz + y 2 Ans.<br />
<br />
<br />
Example 49. A fluid motion is given by v ( y z) i ( z x) j ( x y) k.<br />
Show that the<br />
motion is irrotational and hence find the velocity potential.<br />
(Uttarakhand, I Semester 2006; U.P., I Semester, Winter 2003)<br />
Solution. Curl v = v <br />
<br />
<br />
<br />
= i j k [( y z) i ( z x) j ( x y) k]<br />
x y z<br />
=<br />
<br />
i j k<br />
<br />
x y z<br />
y z z x x y<br />
<br />
= (1 1) i (1 1) j (1 1) k 0<br />
Hence, <br />
v is irrotational.<br />
To find the corresponding velocity potential , consider the following relation.<br />
v = <br />
d = dx <br />
dy <br />
dz<br />
[Total Differential coefficient]<br />
x y z
Vectors 409<br />
<br />
<br />
= i j k .( idx jdykdz)<br />
= i j k . d r .<br />
d r = vdr .<br />
x y z<br />
<br />
x y z<br />
<br />
= [( y z) i ( z x) j ( x y) k].( i dx jdy kdz)<br />
= (y + z) dx + (z + x) dy + (x + y) dz<br />
= y dx + z dx + z dy + x dy + x dz + y dz<br />
<br />
= xy + yz + zx + c<br />
= ( ydx x dy) ( zdy y dz) ( zdx x dz)<br />
Velocity potential = xy + yz + zx + c<br />
Example 50. A fluid motion is given by<br />
<br />
v = (y sin z – sin x) i + (x sin z + 2yz) j + (xy cos z + y 2 ) k <br />
is the motion irrotational? If so, find the velocity potential.<br />
Solution. Curl v = v<br />
=<br />
=<br />
<br />
<br />
<br />
i j k ( ysin zsin x) i +( xsin z+2 yz) j+( xy cos z + y ) k<br />
x y z<br />
<br />
i j k<br />
<br />
2<br />
<br />
x y z<br />
ysin z sin x xsin z 2yz xy cos z y<br />
= (x cos z + 2y – x cos z – 2y) i – [y cos z – y cos z] j + (sin z – sin z) k = 0<br />
Hence, the motion is irrotational.<br />
So,<br />
2<br />
Ans.<br />
v = where is called velocity potential.<br />
d = dx <br />
dy <br />
dz<br />
[Total differential coefficient]<br />
x y z<br />
<br />
= i j k .( i dx jdy kdz)<br />
= .d r = v.<br />
dr<br />
x y z<br />
<br />
= [(y sin z – sin x) i + (x sin z + 2yz) j + (xy cos z + y 2 ) k <br />
].[ idx jdy kdz]<br />
= (y sin z – sin x) dx + (x sin z + 2 y z) dy + (x y cos z + y 2 ) dz<br />
= (y sin z dx + x dy sin z + x y cos z dz) – sin x dx + (2 y z dy + y 2 dz)<br />
= d (x y sin z) + d (cos x) + d (y 2 z)<br />
<br />
2<br />
= d ( xy sin z) d (cos x) d( y z)<br />
= xy sin z + cos x + y 2 z + c<br />
Hence, Velocity potential = xy sin z + cos x + y 2 z + c. Ans.<br />
Example 51. Prove that<br />
<br />
Solution. Given F =<br />
Consider<br />
<br />
<br />
F =<br />
<br />
2<br />
F r r is conservative and find the scalar potential such that<br />
F = . (Nagpur University, Summer 2004)<br />
<br />
i j k<br />
<br />
x y z<br />
2 2 2<br />
r x r y r z<br />
2<br />
r r =<br />
<br />
2 2 2<br />
<br />
r 2 ( xi y j zk)<br />
= r xi r yj<br />
r zk
410 Vectors<br />
<br />
F <br />
= 0<br />
2 2 <br />
2 2 <br />
2 2 <br />
= i r z r y j r z r x k r y r x<br />
y z <br />
x z <br />
<br />
<br />
x y<br />
<br />
<br />
r r r r r r<br />
= i 2rz 2ry j 2rz 2rx k 2ry 2rx<br />
y z <br />
x z <br />
<br />
<br />
x y<br />
<br />
<br />
2 2 2 2 r x r y r z<br />
But r x y z<br />
, , , <br />
x r y r z r<br />
<br />
<br />
<br />
y z x z x y<br />
= i<br />
<br />
2rz 2ry j 2rz 2rx k 2ry 2rx<br />
r r r r r r<br />
<br />
<br />
= i(2yz 2 yz) j(2zx 2 zx) k(2xy 2 xy)<br />
= 0i 0 j 0k<br />
0<br />
F is irrotational F is conservative.<br />
Consider scalar potential such that F = .<br />
d = dx <br />
dy <br />
dz<br />
x y z<br />
<br />
= i j k .( i dx jdy kdz)<br />
x y z<br />
<br />
<br />
= i j k .( i dx jdy kdz)<br />
x y z<br />
<br />
= F.( idx jdy kdz)<br />
=<br />
= 2 2 2<br />
[Total differential coefficient]<br />
<br />
= .( idx jdy kdz)<br />
<br />
2<br />
<br />
r r.( idx jdy kdz)<br />
( = F )<br />
( x y z )( ix jy kz).( idx jdy kdz)<br />
= (x 2 + y 2 + z 2 ) (x dx + y dy + z dz)<br />
= x 3 dx + y 3 dy + z 3 dz + (x dx) y 2 + (x 2 ) (y dy)<br />
+ (x dx)z 2 + z 2 (y dy) + x 2 (z dz) + y 2 (z dz)<br />
3 3 3 2 2<br />
= x dx y dy z dz [( x dx) y ( y dy) x ]<br />
=<br />
<br />
4 4 4<br />
x y z 1 2 2 1 2 2 1 2 2<br />
x y xz y z c<br />
4 4 4 2 2 2<br />
<br />
2 2 2 2<br />
[( xdxz ) ( zdzx ) ] [( ydyz ) ( zdz) y ]<br />
= 1 4 (x4 + y 4 + z 4 + 2x 2 y 2 + 2x 2 z 2 + 2y 2 z 2 ) + c Ans.<br />
r<br />
Example 52. Show that the <strong>vector</strong> field F is irrotational as well as solenoidal. Find<br />
3<br />
| r |<br />
the scalar potential.<br />
(Nagpur University, Summer 2008, 2001, U.P. I Semester Dec. 2005, 2001)<br />
<br />
<br />
r xi y j zk<br />
Solution. F = <br />
2 2 2 3/2<br />
3<br />
| r |<br />
( x y z )<br />
<br />
Curl F xi yj<br />
zk<br />
= F = i j k <br />
x y z 2 2 2 3/2<br />
( x y z )
Vectors 411<br />
=<br />
<br />
i j k<br />
<br />
x y z<br />
x y z<br />
2 2 2 3/2 2 2 2 3/2 2 2 2 3/2<br />
( x y z ) ( x y z ) ( x y z )<br />
3 2yz<br />
3 2yz<br />
<br />
2 ( x y z ) 2 ( x y z )<br />
<br />
= i<br />
<br />
2 2 2 5/2 2 2 2 5/2<br />
= 0<br />
Hence, F is irrotational.<br />
F = , where is called scalar potential<br />
d = dx <br />
dy <br />
dz<br />
x y z<br />
<br />
= i j k .( i dx jdy kdz)<br />
x y z<br />
<br />
<br />
xi yj<br />
zk<br />
= .( <br />
idx jdy kdz )<br />
2 2 2 3/2<br />
( x y z )<br />
1 2xdx 2y dy 2zdz<br />
=<br />
2 2 2 3/2<br />
2<br />
<br />
( x y z )<br />
=<br />
Now, Div F <br />
= .F<br />
=<br />
3 2xz<br />
3<br />
2xz<br />
j <br />
<br />
2 <br />
( x y z ) 2 <br />
( x y z )<br />
3 2xy<br />
3<br />
2xy<br />
k <br />
2 <br />
( x y z ) 2 <br />
( x y z )<br />
1<br />
<br />
2 2 2 2<br />
<br />
<br />
<br />
2 2 2 5/2 2 2 2 5/2<br />
2 2 2 5/2 2 2 2 5/2<br />
<br />
<br />
<br />
<br />
<br />
<br />
[Total differential coefficient]<br />
<br />
= . dr Fdr .<br />
xdx y dy zdz<br />
= 2 2 2 3/2<br />
( x y z )<br />
1<br />
2<br />
1 1<br />
( x y z ) <br />
1<br />
2<br />
1 <br />
| r |<br />
2 2 2<br />
( x y z ) 2<br />
<br />
xi yj<br />
zk<br />
= i j k . (<br />
2 2 2 )<br />
3/2<br />
x y z<br />
x y z<br />
x y z<br />
<br />
<br />
x ( x y z ) y ( x y z ) z<br />
( x y z )<br />
2 2 2 3/2 3<br />
2 2 2 1/2<br />
( x y z ) (1) x<br />
( x y z ) (2 x)<br />
2<br />
<br />
2 2 2 3<br />
( x y z )<br />
=<br />
2 2 2 3/2 2 2 2 3/2 2 2 2 3/2<br />
Ans.<br />
2 2 2 3/2 3<br />
2 2 2 1/2<br />
( x y z ) (1) y<br />
( x y z ) (2 y)<br />
<br />
2<br />
<br />
2 2 2 3<br />
( x y z )<br />
2 2 2 3/2 3<br />
2 2 2 1/2<br />
( x y z ) (1) z<br />
( x y z ) (2 z)<br />
<br />
2<br />
<br />
2 2 2 3<br />
( x y z )
412 Vectors<br />
=<br />
= 0<br />
2 2 2 1/2<br />
( x y z )<br />
2 2 2 3<br />
( x y z )<br />
[x 2 + y 2 + z 2 – 3x 2 + x 2 + y 2 + z 2 – 3y 2 + x 2 + y 2 + z 2 – 3z 2 ]<br />
Hence, F is solenoidal.<br />
Proved.<br />
<br />
2 2 2 2<br />
Example 53. Given the <strong>vector</strong> field V ( x y 2 xz) i ( xz xy yz) j ( z x ) k find<br />
curl V. Show that the <strong>vector</strong>s given by curl V at P 0<br />
(1, 2, –3) and P 1<br />
(2, 3, 12) are orthogonal.<br />
<br />
<br />
Solution. Curl V = V<br />
<br />
<br />
2 2 2 2<br />
= i j k [( x y 2 xz) i( xz xy yz) j( z x ) k]<br />
x y z<br />
curl V =<br />
<br />
i j k<br />
<br />
x y z<br />
2 2 2 2<br />
x y 2xz xz xy yz z x<br />
<br />
= ( x y) i (2x2 x) j ( z y 2 yk ) = ( x y) i ( y z)<br />
k<br />
<br />
curl V at P 0<br />
(1, 2, –3) = (1 2) i (23) k 3<br />
i k<br />
<br />
curl V at P 1<br />
(2, 3, 12) = (2 3) i (312) k 5i 15k<br />
The curl V <br />
at (1, 2, –3) and (2, 3, 12) are perpendicular since<br />
<br />
( 3 i k).( 5i 15 k)<br />
= +15 – 15 = 0 Proved.<br />
Example 54. Find the constants a, b, c, so that<br />
<br />
F = ( x2 y az) i ( bx 3 y z) j (4xcy 2 z)<br />
k<br />
...(1)<br />
is irrotational and hence find function such that F <br />
= .<br />
(Nagpur University, Summer 2005, Winter 2000; R.G.P.V., Bhopal 2009)<br />
Solution. We have,<br />
<br />
F <br />
=<br />
As F is irrotational, 0<br />
F <br />
<br />
i j k<br />
<br />
x y z<br />
( x 2 y az) ( bx 3 y z) (4x cy 2) z<br />
<br />
= ( c 1) i (4 a) j ( b 2) k<br />
<br />
<br />
i.e., ( c 1) i (4 a) j ( b 2) k 0i 0 j 0k<br />
c + 1 = 0, 4 – a = 0 and b – 2 = 0<br />
i.e., a = 4, b = 2, c = –1<br />
Putting the values of a, b, c in (1), we get<br />
<br />
F = ( x 2y 4 z) i (2x 3 y z) j(4x y 2 z)<br />
k
Vectors 413<br />
Now we have to find such that F <br />
= <br />
We know that<br />
d = dx <br />
dy <br />
dz<br />
x y z<br />
[Total differential coefficient]<br />
=<br />
<br />
i j k .( i dx jdy kdz)<br />
x y z<br />
<br />
=<br />
<br />
<br />
i j k .( i dx jdy kdz)<br />
= .( idx jdy kdz)<br />
x y z<br />
<br />
= F.( i dx jdy kdz)<br />
<br />
= [( x 2y 4 z) i (2x3 y z) j (4x y 2 z) k)].( i dx jdy kdz)<br />
= (x + 2y + 4z) dx + (2x – 3y – z) dy + (4x – y + 2z) dz<br />
= x dx – 3y dy + 2z dz + (2y dx + 2x dy) + (4z dx + 4x dz) + (–z dy – y dz)<br />
( zdy<br />
ydz)<br />
= xdx3 ydy 2 zdz (2ydx 2 xdy) (4zdx 4 xdz)<br />
<br />
=<br />
2 2<br />
x 3y<br />
+ z 2 + 2xy + 4zx – yz + c Ans.<br />
2 2<br />
Example 55. Let V (x, y, z) be a differentiable <strong>vector</strong> function and (x, y, z) be a scalar<br />
function. Derive an expression for div ( V <br />
) in terms of .V , div V and .<br />
(U.P. I Semester, Winter 2003)<br />
<br />
Solution. Let V = V 1 i V 2 j V 3 k<br />
div ( V <br />
) = .( F)<br />
<br />
<br />
= i j k .[ V1 i V2 j V3<br />
k]<br />
= ( V1) ( V2) ( V3)<br />
x y z<br />
x y z<br />
V1 V2<br />
V3<br />
<br />
= V1 V2 V3<br />
x x y y <br />
z z<br />
<br />
<br />
<br />
V1 V2<br />
V3<br />
<br />
= V1 V2 V3<br />
x y z <br />
x y z<br />
<br />
<br />
<br />
= i j k .( V1 i V2 j V3<br />
k)<br />
i j k .( V1 i V2 j V3<br />
k)<br />
x y z<br />
x y z<br />
<br />
<br />
= ( . V) ( ). V (div V) (grad ).<br />
V<br />
Ans.<br />
Example 56. If A is a constant <strong>vector</strong> and R = x î + y ĵ + z ˆk , then prove that<br />
<br />
<br />
<br />
<br />
Curl A.<br />
R<br />
A A R<br />
(K. University, Dec. 2009)<br />
<br />
<br />
<br />
Solution. Let A = A1 î + A 2<br />
ĵ + A 3<br />
ˆk , R = xî + y ĵ + z ˆk<br />
<br />
A.<br />
R (A 1 î + A 2<br />
ĵ + A 3<br />
ˆk ) . (x î + y ĵ + z ˆk ) = A 1<br />
x + A 2<br />
y + A 3<br />
z<br />
<br />
[ A. R]<br />
R = (A1 x + A 2<br />
y + A 3<br />
z) (x î + y ĵ + z ˆk )<br />
= (A 1<br />
x 2 + A 2<br />
xy + A 3<br />
zx) î + (A 1<br />
xy + A 2<br />
y 2 + A 3<br />
yz) ĵ + (A 1 xz + A 2 yz + A 3 z2 ) ˆk
414 Vectors<br />
iˆ<br />
ˆj kˆ<br />
<br />
Curl ( A. R)<br />
R<br />
= <br />
x y z<br />
2 2 2<br />
Ax Axy A zx Axy A y A yz Axz A yz Az<br />
L.H.S. = A<br />
R<br />
1 2 3 2 2 3 1 2 3<br />
= (A 2<br />
z – A 3<br />
y) î – [A 1<br />
z – A 3<br />
x) ĵ [A 1 y – A 2 x] ˆk ... (1)<br />
<br />
<br />
= (A 1 î + A 2<br />
ĵ + A 3<br />
k) ×(x î + y ĵ + z ˆk )<br />
iˆ<br />
ˆj k<br />
A A A<br />
= 1 2 3<br />
x y z<br />
Example 57. Suppose that UV ,<br />
= (A 2<br />
z – A 3<br />
y) î – (A 1<br />
z – A 3<br />
x) ĵ + (A y – A x) 1 2<br />
ˆk<br />
= R.H.S. [From (1)]<br />
<br />
<br />
and f are continuously differentiable fields then<br />
Prove that, div ( U V) Vcurl . U U.<br />
curl V . (M.U. 2003, 2005)<br />
<br />
Solution. Let U = ui 1 u2 ju3k,<br />
V v1i v2 j<br />
v3k<br />
<br />
<br />
U V =<br />
<br />
i j k<br />
u u u<br />
1 2 3<br />
v v v<br />
1 2 3<br />
<br />
= ( uv 2 3 uv 3 2) i ( u1v3 u3v1) j + ( uv 1 2 uv 2 1)<br />
k<br />
<br />
<br />
div ( U V)<br />
= i j k .[( u2v3u3v2) i( uv 1 3 uv 3 1) j+ ( u1v2u2v1) k]<br />
x y z<br />
<br />
= ( uv 2 3uv 3 2) ( uv 1 3uv 3 1) ( uv 1 2uv<br />
2 1)<br />
x y z<br />
v3 u2 v2 u3<br />
v3<br />
u1<br />
v1<br />
u3<br />
<br />
= u2 v3 u3 v2 u1 v3<br />
x x x x <br />
u3 v1<br />
y y<br />
y<br />
y<br />
<br />
<br />
v2 u1 v1 u2<br />
u1 v2 u2 v1<br />
z z z z<br />
<br />
<br />
<br />
u3 u2 u3<br />
u1 u2 u1<br />
<br />
= v1 v2 v3<br />
<br />
y z <br />
x z <br />
<br />
x y<br />
<br />
<br />
v3 v2 v3<br />
v1 v1 v2<br />
<br />
u1 u2<br />
u3<br />
<br />
y z <br />
x z <br />
<br />
y x<br />
<br />
<br />
u3 u2 u 1 u3 <br />
<br />
u2 u1<br />
= ( v1i v2 j v3<br />
k).<br />
i j<br />
k<br />
<br />
y z z x <br />
<br />
x y<br />
<br />
<br />
<br />
v2 v3 v3<br />
v1 v1 v2<br />
<br />
( u1 i u2 j u3<br />
k).<br />
i j k <br />
z y x z <br />
<br />
y x<br />
<br />
<br />
<br />
<br />
= V.( U) U.( V) V.curl U U.curl<br />
V<br />
Proved.
Vectors 415<br />
Example 58. Prove that<br />
<br />
<br />
( F G)<br />
= F( . G) G( . F) ( G. ) F ( F. ) G (M.U. 2004, 2005)<br />
Solution.<br />
<br />
( F G)<br />
= i ( F G)<br />
x<br />
=<br />
F G F G<br />
<br />
i G F i G i F<br />
<br />
x x x x<br />
<br />
=<br />
F F G<br />
G<br />
( iG . ) i G i F ( i. F)<br />
<br />
x x x x<br />
<br />
=<br />
F F G<br />
G<br />
( Gi . ) Gi. Fi. ( F. i)<br />
x x x x<br />
=<br />
G <br />
F <br />
F G<br />
F i Gi. ( G. i) ( F. i)<br />
x x x x<br />
<br />
= F ( G) G ( . F) ( G. ) F ( F. )<br />
G<br />
Proved.<br />
Questions for practice:<br />
Prove that<br />
<br />
( FG . ) = ( G. ) F ( F. ) G G( F) F ( G)<br />
Example 59. Prove that, for every field V ; div curl V = 0.<br />
(Nagpur University, Summer 2004; AMIETE, Sem II, June 2010)<br />
<br />
Solution. Let V = V1 i V2 j V3<br />
k<br />
=<br />
<br />
div (curl V ) = .( V<br />
)<br />
<br />
i j k<br />
<br />
.<br />
x y z<br />
V V V<br />
1 2 3<br />
V3 V2 V<br />
3 V1 V 2 V1<br />
<br />
= i j k .<br />
i j<br />
k <br />
x y z y z x z x y<br />
<br />
V3 V2 V3<br />
V1 V2 V1<br />
<br />
= <br />
x y z y<br />
<br />
x z<br />
<br />
z x y<br />
<br />
2 2 2 2 2 2<br />
V3 V2 V3<br />
V1 V2 V1<br />
= <br />
xy xz yx yz zx zy<br />
<br />
2 2 2 2<br />
2 2<br />
V1 V <br />
1 V2 V <br />
2 V3 V <br />
3<br />
=<br />
<br />
yz zy zx xz xy yx<br />
<br />
= 0 Ans.<br />
Example 60. If a <br />
is a constant <strong>vector</strong>, show that<br />
<br />
<br />
<br />
a ( r)<br />
= ( a. r) ( a. ) r.<br />
(U.P., Ist Semester, Dec. 2007)<br />
<br />
Solution. a = a1 i a2 j a3k,<br />
r r1 i r2 j r3k
416 Vectors<br />
<br />
<br />
r =<br />
<br />
<br />
i j k<br />
<br />
x y z<br />
r r r<br />
=<br />
1 2 3<br />
<br />
i j k<br />
a ( r)<br />
=<br />
a1 a2 a3<br />
r3 r2 r3<br />
r1 r2 r1<br />
<br />
y z x z x y<br />
r<br />
r r<br />
r r r<br />
<br />
i j k<br />
y z<br />
<br />
x z<br />
<br />
x y<br />
<br />
<br />
3 2 3 1 2 1<br />
r2 r1 r3 r1 r2 r1 r3<br />
r2<br />
<br />
a2 a2 a3 a3 i a1 a1 a3 a3<br />
j<br />
= <br />
x y <br />
x z <br />
x y y z<br />
<br />
<br />
<br />
<br />
r3 r1 r3<br />
r2<br />
<br />
a1 a1 a2 a2<br />
k<br />
x z y z<br />
<br />
<br />
r <br />
1 r <br />
2 r3 r <br />
1 r<br />
<br />
2 r3<br />
<br />
= a1i a2 i a3 i a1 j a2 j a3<br />
j <br />
x x x y y y<br />
<br />
r <br />
1 r <br />
2 r3 <br />
r <br />
1 r<br />
<br />
2 r3<br />
<br />
ak 1 a2k a3k <br />
a1 i a1 j ak 1 <br />
z z z <br />
x x x<br />
<br />
r <br />
1 r <br />
2 r3 r <br />
1 r<br />
<br />
2 r3<br />
<br />
a2 i a2 j a2k a3 i a3 j a3k<br />
<br />
y y y z z z<br />
<br />
<br />
= i j k ( a11 r a2r2 a33 r ) a1 a2 a3 ( r1 i r2 j<br />
r3<br />
k)<br />
x y z x y z<br />
<br />
<br />
= ( a. r) ( a. )<br />
r<br />
Proved.<br />
Example 61. If r is the distance of a point (x, y, z) from the origin, prove that<br />
1 1<br />
Curl k grad <br />
grad kgrad<br />
.<br />
= 0, where k is the unit <strong>vector</strong> in the direction OZ.<br />
r r (U.P., I Semester, Winter 2000)<br />
Solution. r 2 = (x – 0) 2 + (y – 0) 2 + (z – 0) 2 = x 2 + y 2 + z 2<br />
<br />
1<br />
r<br />
= (x 2 + y 2 + z 2 ) – 1/2<br />
grad 1 1 2 2 2 1/2<br />
= = i j k ( x y z )<br />
r<br />
<br />
r x y z<br />
=<br />
1 ( 2 2 2 3/2<br />
x y z (2 xi 2 yj 2 zk<br />
)<br />
2<br />
k × grad 1 r<br />
2 2 2 3/2<br />
<br />
= – ( x y z ) ( xi yj<br />
zk)<br />
=<br />
2 2 2 3/2<br />
<br />
k [ ( x y z ) ( xi yj<br />
zk)]<br />
curl<br />
1 <br />
k<br />
grad <br />
r <br />
=<br />
=<br />
2 2 2 3/2<br />
( x y z ) ( xj<br />
yi)<br />
1 <br />
k<br />
grad <br />
r
Vectors 417<br />
<br />
= i j k <br />
x y z × [–(x2 + y 2 + z 2 ) –3/2 <br />
( xj yi)<br />
]<br />
<br />
i j k<br />
<br />
= x y z<br />
y<br />
x<br />
0<br />
2 2 2 3/2 2 2 2 3/2<br />
( x y z ) ( x y z )<br />
3 ( x)(2) z 3 y (2 z) 3 ( x)(2 x)<br />
= <br />
i j <br />
2 2 2 5/2 2 2 2 5/2 <br />
2 2 2 5/2<br />
2 ( x y z ) 2( x y z ) 2 ( x y z )<br />
1 ( 3/2) ( y)(2 y) 1 <br />
k<br />
2 2 2 3/2 2 2 2 5/2 2 2 2 3/2 <br />
( x y z ) ( x y z ) ( x y z ) <br />
2 2 2 2 2 2 2 2<br />
3xz 3 yz (3x x y z 3 y x y z<br />
) <br />
=<br />
i <br />
j <br />
k<br />
2 2 2 5/2 2 2 2 5/2 2 2 2 5/2<br />
( x y z ) ( x y z ) ( x y z )<br />
<br />
2 2 2<br />
<br />
3xz i 3 yzj ( x y 2 z ) k<br />
=<br />
2 2 2 5/2<br />
( x y z )<br />
...(1)<br />
k . grad 1 r = <br />
2 2 2 3/2<br />
z<br />
k.[ ( x y z ) ( xi yj zk)]<br />
<br />
2 2 2 3/2<br />
( x y z )<br />
1 z<br />
grad k.grad<br />
= i j k 2 2 2 3/2<br />
r x y z<br />
( x y z )<br />
=<br />
<br />
<br />
3 i( z)(2 x) 3 j( z)(2 y)<br />
<br />
<br />
2<br />
2 2 2 5/2<br />
2<br />
2 2 2 5/2<br />
( x y z ) ( x y z<br />
)<br />
3 ( z)(2 z) 1 <br />
<br />
<br />
k<br />
2 2 2 5/2 2 2 2 3/2<br />
2<br />
<br />
( x y z ) ( x y z<br />
) <br />
<br />
2 2 2 2 2 2 2<br />
= 3xz i 3 yzj(3 z x y z ) k 3xz i 3 yzj( x y 2 z ) k<br />
<br />
2 2 2 5/2 2 2 2 5/2<br />
( x y z ) ( x y z )<br />
...(2)<br />
Adding (1) and (2), we get<br />
1 1<br />
Curl kgrad <br />
grad k.grad<br />
<br />
<br />
r r = 0 Proved.<br />
<br />
<br />
a r (2 n) a n( a. r)<br />
r<br />
Example 62. Prove that <br />
.<br />
n n n 2<br />
r r r<br />
<br />
(M.U. 2009, 2005, 2003, 2002; AMIETE, II Sem. June 2010)<br />
Solution. We have,<br />
<br />
a r<br />
r<br />
n<br />
<br />
1<br />
i j k<br />
a a a<br />
=<br />
n 1 2 3<br />
r<br />
<br />
x y z<br />
1 1 1<br />
<br />
= ( az 2 ayi 3 ) ( a3x a1z) j ( ay 1 axk 2 )<br />
n n n<br />
r r r
418 Vectors<br />
<br />
i j k<br />
<br />
<br />
( a r ) <br />
n<br />
= x y z<br />
r<br />
azay ax az ay<br />
ax<br />
2 3 3 1 1 2<br />
n n n<br />
r r r<br />
<br />
ay 1 ax 2 ax 3 az 1 <br />
ay 1 ax<br />
2 az 2 ay<br />
3 <br />
= i j<br />
n <br />
n n n<br />
y r z<br />
<br />
<br />
r x r z<br />
<br />
<br />
r <br />
ax 3 az 1 az 2 ay 3 <br />
k <br />
n<br />
n<br />
x<br />
<br />
<br />
r y<br />
<br />
<br />
r <br />
Now, r 2 = x 2 + y 2 + z 2 <br />
r<br />
r<br />
x<br />
2r<br />
= 2x <br />
x<br />
x<br />
r<br />
Similarly,<br />
r<br />
y<br />
= ,<br />
r<br />
z <br />
y<br />
r<br />
z<br />
r<br />
<br />
a r <br />
<br />
n<br />
1<br />
y <br />
1 <br />
= i . ( n<br />
r<br />
nr a1y a2x)<br />
a<br />
n 1<br />
r <br />
r <br />
<br />
n<br />
1<br />
z <br />
1 <br />
nr ( ax 3 az 1 ) ( a1<br />
)<br />
r<br />
n + two similar terms<br />
<br />
r <br />
=<br />
<br />
n 2 a1 n<br />
2 a1<br />
i ( a<br />
2 1y a2xy) ( axz<br />
2 3 a1z<br />
) <br />
n<br />
n n<br />
n<br />
r r r r <br />
+ two similar terms<br />
=<br />
<br />
2a1<br />
n 2 2 n<br />
<br />
i a<br />
2 1( y z ) ( axy<br />
2 2 a3xz)<br />
n n<br />
n<br />
<br />
+ two similar terms<br />
r r r<br />
<br />
Adding and subtracting<br />
n 2<br />
ax to third and from second term, we get<br />
n 2<br />
r <br />
1<br />
<br />
a r<br />
<br />
<br />
2a1 na1<br />
2 2 2 n 2<br />
<br />
n<br />
r<br />
= i ( x y z ) ( ax<br />
2 2 1 axy 2 a3xz)<br />
n n<br />
n <br />
<br />
r r r<br />
<br />
<br />
+ two similar terms<br />
2a1 na1<br />
2 n<br />
<br />
= i r xax (<br />
2 2 1 a2y a3<br />
z)<br />
n n<br />
n <br />
+ two similar terms<br />
r r r<br />
<br />
2a1 na1<br />
n<br />
<br />
<br />
2a2 na2<br />
n<br />
<br />
= i x( ax<br />
2 1 ay 2 az 3 )<br />
n n n <br />
<br />
j<br />
y( a<br />
2 2 y az 3 ax 1 )<br />
n n n <br />
<br />
r r r<br />
r r r<br />
<br />
2a3 na3<br />
n<br />
<br />
k<br />
z( az<br />
2 3 ax 1 a2<br />
y)<br />
n n n <br />
<br />
r r r<br />
<br />
2 n n<br />
= ( a<br />
n 1 i a2 j a3k ) <br />
n ( a1 i a2 j a3k<br />
) ( 2 1 2 3 )( <br />
ax a y az xi yj<br />
zk )<br />
n <br />
r<br />
r<br />
r<br />
2 n n<br />
<br />
= ( a1 i a2 j a3k) ( ax<br />
2 1 a2y az 3 )( xi yj<br />
zk)<br />
n<br />
n <br />
r<br />
r<br />
2 n n <br />
= a ( a. r)<br />
r<br />
n n<br />
2<br />
Proved.<br />
r r<br />
Example 63. If f and g are two scalar point functions, prove that<br />
div (f g) = f 2 g + f g. (U.P., I Semester, compartment, Winter 2001)
Vectors 419<br />
g g g<br />
<br />
Solution. We have, g = i j k<br />
x y z<br />
g g g<br />
<br />
f g = f i f j f k<br />
x y z<br />
g g g<br />
<br />
div (f g) = f <br />
f <br />
f <br />
x x y y z z<br />
<br />
<br />
2 2 2<br />
g g g f g f g f g<br />
= f<br />
<br />
<br />
2 2 2<br />
x y z<br />
<br />
<br />
x x y y z z<br />
<br />
<br />
2 2 2<br />
f f f g g g<br />
<br />
= f <br />
g i j k . i j k<br />
2 2 2<br />
x y z<br />
<br />
<br />
x y z x y z<br />
<br />
= f 2 g + f.g Proved.<br />
Example 64. For a solenoidol <strong>vector</strong> F , show that curl curl curl curl F = 4<br />
F .<br />
(M.D.U., Dec. 2009)<br />
Solution. Since <strong>vector</strong> F is solenoidal, so div F = 0 ... (1)<br />
We know that curl curl F = grad div ( F – 2<br />
F ) ... (2)<br />
Using (1) in (2), grad div F = grad (0) = 0 ... (3)<br />
On putting the value of grad div F in (2), we get<br />
curl curl F = – 2<br />
F ... (4)<br />
Now, curl curl curl curl F = curl curl (– 2<br />
F ) [Using (4)]<br />
= – curl curl ( 2<br />
F ) = – [grad div ( 2<br />
F ) – 2 ( 2<br />
F ) ] [Using (2)]<br />
= – grad ( . 2<br />
F ) + 2<br />
( 2<br />
F ) = – grad ( 2<br />
. F<br />
<br />
) + 4<br />
F [ . F = 0]<br />
= 0 + 4 F = 4 F [Using (1)] Proved.<br />
EXERCISE 5.9<br />
1. Find the divergence and curl of the <strong>vector</strong> field V = (x 2 – y 2 ) i + 2xy j + (y 2 – xy) k .<br />
Ans. Divergence = 4x, Curl = (2y – x) i + y j + 4y k <br />
2. If a is constant <strong>vector</strong> and r is the radius <strong>vector</strong>, prove that<br />
<br />
<br />
(i) ( a. r)<br />
a (ii) div ( r a) 0 (iii) curl ( r <br />
a) 2<br />
a<br />
where r<br />
<br />
= xi yj zk and a a1i a2 j a3k<br />
.<br />
3. Prove that:<br />
(i) .(A) = .A + (.A)<br />
(ii) (A.B) = (A.)B + (B.)A + A × ( × B) + B × ( × A) (R.G.P.V. Bhopal, June 2004)<br />
(iii) × (A × B) = (B.)A – B(.A) – (A.)B + A(.B)<br />
4. If F = (x + y + 1) i + j – (x + y) k , show that F.curl F = 0.<br />
(R.G.P.V. Bhopal, Feb. 2006, June 2004)<br />
Prove that<br />
<br />
<br />
5. ( F) ( ) F ( F )<br />
6. .( F G) G.( F) F.( G)<br />
<br />
7. Evaluate div ( A r)<br />
if curl A <br />
= 0. 8. Prove that curl ( a r)<br />
= 2a
420 Vectors<br />
9. Find div F and curl F where F = grad (x 3 + y 3 + z 3 – 3xyz). (R.G.P.V. Bhopal Dec. 2003)<br />
Ans. div F = 6(x + y + z), curl F = 0<br />
10. Find out values of a, b, c for which v = (x + y + az) i + (bx + 3y – z) j + (3x + cy + z) k <br />
is irrotational.<br />
Ans. a = 3, b = 1, c = –1<br />
11. Determine the constants a, b, c, so that F = (x + 2y + az) i + (bx – 3y – z) j + (4x + cy + 2z) k is<br />
irrotational. Hence find the scalar potential such that F = grad .<br />
(R.G.P.V. Bhopal, Feb. 2005) Ans. a = 4, b = 2, c = 1<br />
Choose the correct alternative:<br />
Potential =<br />
12. The magnitude of the <strong>vector</strong> drawn in a direction perpendicular to the surface<br />
x 2 + 2y 2 + z 2 = 7 at the point (1, –1, 2) is<br />
2 2<br />
x 3y<br />
2<br />
<br />
z 2xy yz 4zx<br />
<br />
2 2<br />
<br />
2<br />
3<br />
(i)<br />
(ii) (iii) 3 (iv) 6 (A.M.I.E.T.E., Summer 2000) Ans. (iv)<br />
3<br />
2<br />
13.If u = x 2 – y 2 + z 2 and V xi yj zk<br />
then ( uV ) is equal to<br />
(i) 5u (ii) 5| V | (iii) 5( u | |) (iv) 5( u | |) (A.M.I.E.T.E., June 2007)<br />
14.A unit normal to x 2 + y 2 + z 2 = 5 at (0, 1, 2) is equal to<br />
1 <br />
(i) ( )<br />
5 i j k<br />
1 <br />
(ii) ( )<br />
5 i j k<br />
1 <br />
(iii) ( 2 )<br />
5 j <br />
k 1 <br />
(iv) ( )<br />
5 i j k<br />
<br />
<br />
V <br />
V <br />
(A.M.I.E.T.E., Dec. 2008)<br />
15.The directional derivative of = x y z at the point (1, 1, 1) in the direction i is:<br />
1<br />
(i) –1 (ii) <br />
1<br />
(iii) 1 (iv)<br />
Ans. (iii)<br />
3<br />
3<br />
(R.G.P.V. Bhopal, II Sem., June 2007)<br />
<br />
<br />
16.If r xi y j zk and r = | r | then (r) is:<br />
(i) (r) r <br />
(ii)<br />
<br />
()<br />
r r<br />
r<br />
(iii)<br />
<br />
()<br />
r r<br />
r<br />
(iv) None of these<br />
Ans. (iii)<br />
(R.G.P.V. Bhopal, II Semester, Feb. 2006)<br />
17. If <br />
r = xi yj zk is position <strong>vector</strong>, then value of (log r) is (U.P., I Sem, Dec 2008)<br />
<br />
<br />
<br />
r<br />
r<br />
(i) <br />
(ii) <br />
(iii) – r (iv) none of the above. Ans. (ii)<br />
2<br />
3<br />
r<br />
r<br />
r<br />
<br />
18. If r xi y j zk and | r |<br />
= r, then div r is:<br />
(i) 2 (ii) 3 (iii) –3 (iv) –2 Ans. (ii)<br />
(R.G.P.V. Bhopal, II Semester, Feb. 2006)<br />
<br />
2<br />
<br />
2<br />
<br />
2<br />
<br />
19. If V xy i 2yx zj 3yz k then curl V at point (1, –1, 1) is<br />
<br />
<br />
<br />
<br />
(i) ( j 2 k)<br />
(ii) ( i 3 k)<br />
(iii) ( i 2 k)<br />
(iv) ( i 2 j k)<br />
(R.G.P.V. Bhopal, II Semester, Feb 2006)<br />
Ans. (iii)<br />
20. If A is such that A = 0 then A is called<br />
(i) Irrotational (ii) Solenoidal (iii) Rotational (iv) None of these<br />
(A.M.I.E.T.E., Dec. 2008)<br />
21. If F is a conservative force field, then the value of curl F is<br />
(i) 0 (ii) 1 (iii) F (iv) –1 (A.M.I.E.T.E., June 2007)
Vectors 421<br />
5.33 LINE INTEGRAL<br />
<br />
Let y,<br />
F ( x, z)<br />
be a <strong>vector</strong> function and a curve AB.<br />
Line integral of a <strong>vector</strong> function F along the curve AB is defined as integral of the component<br />
of F along the tangent to the curve AB.<br />
Component of F along a tangent PT at P<br />
= Dot product of F<br />
<br />
and unit <strong>vector</strong> along PT<br />
<br />
<br />
<br />
dr dr<br />
= F is aunit <strong>vector</strong> along tangent PT<br />
ds ds<br />
<br />
<br />
<br />
<br />
dr<br />
Line integral = F from A to B along the curve<br />
ds<br />
<br />
<br />
<br />
<br />
dr<br />
Line integral = F <br />
ds<br />
c<br />
ds<br />
=<br />
<br />
F<br />
<br />
dr<br />
<br />
<br />
c<br />
<br />
Note (1) Work. If F represents the variable force acting on a particle along arc AB, then the<br />
total work done =<br />
<br />
B<br />
F<br />
<br />
A<br />
<br />
dr<br />
(2) Circulation. If V represents the velocity of a liquid then<br />
V<br />
<br />
c<br />
<br />
dr is called the circulation<br />
of V round the closed curve c.<br />
If the circulation of V round every closed curve is zero then V is said to be irrotational there.<br />
(3) When the path of integration is a closed curve then notation of integration is in place<br />
<br />
of .<br />
22.If 2<br />
[(1 – x) (1 – 2x)] is equal to<br />
(i) 2 (ii) 3 (iii) 4 (iv) 6 (A.M.I.E.T.E., Dec. 2009) Ans. (iii)<br />
23.If R = xi + yj + zk and A <br />
is a constant <strong>vector</strong>, curl ( A R ) is equal to<br />
(i) R (ii) 2 R (iii) A (iv) 2 A (A.M.I.E.T.E., Dec. 2009) Ans. (iv)<br />
1<br />
24. If r is the distance of a point (x, y, z) from the origin, the value of the expression ˆj grad 2<br />
equals<br />
(i) 2 2 2<br />
3<br />
<br />
2 ˆ<br />
( x y z ) ( ˆjz kx)<br />
(ii)<br />
(iii) zero (iv)<br />
<br />
2<br />
Example 65. If a force F 2x yiˆ 3xyj<br />
ˆ displaces a particle in the xy-plane from (0, 0) to<br />
(1, 4) along a curve y = 4 x 2 . Find the work done.<br />
Solution. Work done = F <br />
. dr<br />
<br />
<br />
c r xiˆ yj ˆ<br />
<br />
<br />
2 <br />
<br />
=<br />
ˆ ˆ ˆ ˆ<br />
(2 x yi 3 xyj).( dx i dyj)<br />
dr dxiˆdyj<br />
ˆ<br />
<br />
=<br />
<br />
c<br />
c<br />
2<br />
(2 x y dx 3 xy dy)<br />
3<br />
<br />
2 2 2<br />
( x y z ) 2 ( ˆjz<br />
iz ˆ )<br />
3<br />
<br />
2 2 2 2 ˆ<br />
( x y z ) ( ˆjy<br />
kx)<br />
(AMIETE, Dec. 2010) Ans. (ii)
422 Vectors<br />
Putting the values of y and dy, we get<br />
<br />
1 2 2 2<br />
= [2 x (4 x ) dx 3 x (4 x )8 xdx]<br />
0<br />
5<br />
1<br />
1 4<br />
x 104<br />
= 104 x dx 104<br />
<br />
0 <br />
5 <br />
5<br />
0<br />
<br />
2<br />
y 4 x <br />
<br />
dy 8 xdx <br />
<br />
Ans.<br />
Example 66. Evaluate<br />
<br />
C<br />
<br />
F.<br />
drwhere F x<br />
2ˆ i xyj ˆand C is the boundary of the square in the<br />
plane z = 0 and bounded by the lines x = 0, y = 0, x = a and y = a.<br />
(Nagpur University, Summer 2001)<br />
<br />
<br />
Solution. F. dr F. dr F. dr F. dr F.<br />
dr<br />
C OA AB BC CO<br />
<br />
Here r xiˆ yj ˆ, dr dxiˆ dyj ˆ, F x<br />
2ˆ i xyj ˆ<br />
On OA, y = 0 2<br />
F.<br />
dr x dx<br />
<br />
F.<br />
dr = x 2 dx + xydy ...(1)<br />
<br />
<br />
F . dr =<br />
OA<br />
On AB, x = a dx = 0<br />
(1) becomes<br />
<br />
F.<br />
dr = aydy<br />
2<br />
a<br />
3<br />
F <br />
. dr<br />
<br />
= a y a<br />
aydy a <br />
<br />
<br />
Ab<br />
0<br />
2 2<br />
...(3)<br />
0<br />
On BC, y = a dy = 0<br />
(1) becomes 2<br />
F.<br />
dr x dx<br />
<br />
<br />
<br />
F . dr =<br />
BC<br />
On CO, x = 0, F. dr 0<br />
(1) becomes<br />
<br />
<br />
<br />
<br />
<br />
3<br />
0<br />
3<br />
0 2<br />
x – a<br />
xdx <br />
a<br />
3 3<br />
...(4)<br />
<br />
<br />
a<br />
F . dr = 0 ...(5)<br />
CO<br />
3 3 3 3<br />
On adding (2), (3), (4) and (5), we get F <br />
. dr<br />
<br />
=<br />
a a – a 0 a<br />
Ans.<br />
C 3 2 3 2<br />
Example 67. A <strong>vector</strong> field is given by<br />
<br />
F = (2y 3) iˆ<br />
xzj ˆ ( yz – x) kˆ<br />
. Evaluate F <br />
. dr<br />
<br />
along the path c is x = 2t,<br />
C<br />
y = t, z = t 3 from t = 0 to t = 1. (Nagpur University, Winter 2003)<br />
<br />
<br />
<br />
C<br />
<br />
<br />
a<br />
3 3<br />
a 2<br />
x a<br />
xdx <br />
0<br />
3 3<br />
...(2)<br />
0<br />
Solution. F . dr = (2y 3) dx ( xz) dy ( yz – x)<br />
dz<br />
C C<br />
<br />
3<br />
Since x 2t y t z t <br />
<br />
<br />
dx dy dz 2<br />
2 1 3t<br />
<br />
<br />
dt dt dt <br />
Y<br />
O<br />
B<br />
A<br />
X
Vectors 423<br />
1 3 4 2<br />
= (2 t 3) (2 dt ) (2)( t t ) dt ( t – 2)(3 t t dt ) =<br />
0<br />
<br />
2<br />
t 2 5 3 7 6 4<br />
= 4 6 t t t – t <br />
2 5 7 4 <br />
1<br />
0<br />
<br />
1 4 6 3<br />
(4t 6 2t 3 t – 6 t ) dt<br />
0<br />
2 2 5 3 7 3 4<br />
= 2t 6 t t t – t<br />
5 7 2<br />
<br />
<br />
<br />
2 3 3<br />
= 2 6 – = 7.32857. Ans.<br />
5 7 2<br />
Example 68. The acceleration of a particle at time t is given by<br />
a<br />
= 18 cos 3tiˆ<br />
8sin 2tj ˆ 6 tk ˆ .<br />
If the velocity v and displacement r be zero at t = 0, find v and r at any point t.<br />
2 <br />
Solution. Here, a d r<br />
=<br />
2<br />
dt<br />
On integrating, we have<br />
<br />
= 18 cos 3tiˆ<br />
8sin 2tj ˆ 6 tkˆ<br />
.<br />
<br />
v = dr<br />
iˆ<br />
18 cos 3t dt ˆj 8sin 2t dt k ˆ 6t dt<br />
dt<br />
<br />
2<br />
v = 6sin 3tiˆ 4cos 2tj ˆ 3t kˆ<br />
<br />
c ...(1)<br />
At t = 0, v = 0 <br />
Putting t = 0 and <br />
v = 0 in (1), we get<br />
<br />
0 = 4ˆj <br />
c c 4ˆj<br />
<br />
dr<br />
2<br />
v = 6sin 3tiˆ 4(cos 2t 1) ˆj 3tkˆ<br />
dt<br />
Again integrating, we have<br />
2<br />
r<br />
= iˆ<br />
6sin 3 ˆ 4(cos 2 1) ˆ<br />
t dt j t dt k3t dt<br />
3<br />
r = 2cos 3 tiˆ(2sin 2t 4 t)<br />
ˆj t kˆ<br />
c <br />
...(2)<br />
At, t = 0, r = 0<br />
Putting t = 0 and r = 0 in (2), we get<br />
<br />
<br />
0 = 2iˆC ˆ<br />
1 C1<br />
2i<br />
Hence, r<br />
3<br />
= 2(1cos 3) t iˆ<br />
2(sin 2t 2 t)<br />
ˆj t kˆ<br />
Ans.<br />
Example 69. If A 2 2<br />
(3x 6 y) iˆ<br />
–14yzj ˆ 20 xz kˆ<br />
, evaluate the line integral Adr . from<br />
(0, 0, 0) to (1, 1, 1) along the curve C.<br />
x = t, y = t 2 , z = t 3 . (Uttarakhand, I Semester, Dec. 2006)<br />
Solution. We have,<br />
<br />
<br />
A . dr =<br />
C<br />
=<br />
<br />
<br />
C<br />
C<br />
2 2<br />
[(3x 6 y) iˆ –14yzj ˆ 20 xz kˆ].[ iˆ dx ˆjdy kˆ<br />
dz]<br />
2 2<br />
[(3x 6 y) dx –14yzdy 20 xz dz]<br />
If x = t, y = t 2 , z = t 3 , then points (0, 0, 0) and (1, 1, 1) correspond to t = 0 and t = 1 respectively.<br />
<br />
<br />
Now, A . dr =<br />
C<br />
=<br />
<br />
t 1 2 2 2 3 2 3 2 3<br />
t 0<br />
[(3t 6 t ) d ( t)–14 t t d ( t ) 20 t ( t ) d ( t )]<br />
t 1 2 5 7 2<br />
[9 t dt –14 t .2tdt 20 t .3 t dt]<br />
=<br />
t 0<br />
<br />
1<br />
1 2 6 9<br />
0<br />
1<br />
0<br />
<br />
<br />
(9 t –28 t 60 t ) dt
424 Vectors<br />
=<br />
<br />
3 7 10<br />
t t t<br />
<br />
9 – 28 60 <br />
3 7 10<br />
<br />
<br />
<br />
1<br />
0<br />
= 3 – 4 + 6 = 5 Ans.<br />
Example 70. Evaluate<br />
<br />
S<br />
<br />
2<br />
A. nˆ<br />
ds where A ( x y ) iˆ<br />
– 2xj ˆ 2yzkˆ<br />
and S is the surface of<br />
the plane 2x + y + 2z = 6 in the first octant. (Nagpur University, Summer 2000)<br />
Solution. A <strong>vector</strong> normal to the surface “S” is given by<br />
<br />
(2x y 2) z = ˆ<br />
<br />
ˆ<br />
ˆ <br />
i j k (2x y 2) z 2iˆ ˆj 2kˆ<br />
<br />
x y z<br />
And ˆn = a unit <strong>vector</strong> normal to surface S<br />
Z<br />
2iˆ<br />
ˆj 2kˆ<br />
2 1 2 N<br />
=<br />
i ˆ ˆ j k ˆ<br />
41<br />
4 3 3 3<br />
K<br />
kˆ<br />
2 1 2 2<br />
n<br />
. n ˆ = ˆ <br />
k.<br />
iˆ<br />
ˆj kˆ<br />
–<br />
<br />
3 3 3 3<br />
O<br />
M Y<br />
3<br />
A . nds<br />
dx dy<br />
R<br />
ˆ = A.<br />
nˆ<br />
S<br />
R<br />
k ˆ L<br />
. n<br />
Where R is the projection of S.<br />
X<br />
Now, A . nˆ<br />
= [( x y 2 ) iˆ – 2xj ˆ 2 yzkˆ].<br />
2<br />
iˆ 1 ˆj 2 kˆ<br />
<br />
3 3 3 <br />
2 2 2 4 2 2 4<br />
= ( x y ) – x yz y yz<br />
...(1)<br />
3 3 3 3 3<br />
Putting the value of z in (1), we get<br />
on the plane 2x y 2z<br />
6, <br />
<br />
A.<br />
nˆ<br />
2 2 4 6<br />
2 x y<br />
<br />
<br />
= y y <br />
(6 2 x y)<br />
3 3 2 z <br />
<br />
2<br />
<br />
<br />
A.<br />
nˆ<br />
= 2 y ( y 6 –2 x – y) 4 y (3 – x)<br />
...(2)<br />
3 3<br />
M<br />
Hence,<br />
A <br />
. nds<br />
dx dy<br />
ˆ = A.<br />
n<br />
S<br />
R<br />
| k ˆ<br />
...(3)<br />
. n |<br />
<br />
Putting the value of A.<br />
nˆ<br />
from (2) in (3), we get<br />
A<br />
<br />
. nds 4 3<br />
3 62x<br />
ˆ = (3 – ). 2 (3 )<br />
S y x dx dy y x dydx<br />
R 3 2<br />
<br />
0 0<br />
=<br />
<br />
3<br />
0<br />
<br />
2<br />
y <br />
2(3– x)<br />
<br />
2 <br />
6–2x<br />
0<br />
dx<br />
= <br />
3 (3 – )(6 –2 ) 2 4 3<br />
(3 – )<br />
3<br />
0 0<br />
4<br />
3<br />
x x dx x dx<br />
(3 – x)<br />
<br />
= 4. –(0–81) 81<br />
4(–1) <br />
0<br />
<br />
iy ˆ ˆjx<br />
Example 71. Compute F. dr,<br />
where F <br />
c<br />
2 2<br />
and c is the circle x 2 + y 2 = 1 traversed<br />
x y<br />
counter clockwise.<br />
O<br />
2x + 3y = 6<br />
L<br />
Ans.<br />
X
Vectors 425<br />
Solution. r = ix ˆ ˆjy kz ˆ , dr idx ˆ ˆjdy kdz ˆ<br />
<br />
F . dr<br />
iy ˆ ˆjx<br />
= ( ˆ ˆ ˆ )<br />
c idx jdy kdz<br />
c 2 2<br />
x y<br />
ydx xdy<br />
= ( ydx xdy)<br />
c 2 2<br />
x y<br />
c<br />
Parametric equation of the circle are x = cos , y = sin .<br />
Putting x = cos , y = sin , dx = – sin d , dy = cos d in (1), we get<br />
Fd<br />
<br />
r<br />
2<br />
= sin ( sin d ) cos (cos d )<br />
C <br />
=<br />
...(1) [ x 2 + y 2 = 1]<br />
0<br />
2<br />
2 2<br />
2<br />
= 2 0<br />
(sin cos ) d d <br />
0 0<br />
<br />
2 Ans.<br />
<br />
2 3 2 2 2<br />
Example 72. Show that the <strong>vector</strong> field F 2 x( y z ) iˆ<br />
2x yj ˆ 3xzkˆ<br />
is conservative.<br />
Find its scalar potential and the work done in moving a particle from (–1, 2, 1) to (2, 3, 4).<br />
(A.M.I.E.T.E. June 2010, 2009)<br />
Solution. Here, we have<br />
<br />
2 3 ˆ 2 ˆ 2 2<br />
F 2 x( y z ) i 2x yj 3x z kˆ<br />
<br />
<br />
Curl F F<br />
iˆ<br />
ˆj kˆ<br />
<br />
<br />
x y z<br />
2 3 2 2 2<br />
2 xy ( z ) 2x y 3x z<br />
(00) i(6xz 6 xz ) ˆj(4xy 4 xy)<br />
k = 0<br />
Hence, <strong>vector</strong> field F is irrotational.<br />
To find the scalar potential function <br />
<br />
F <br />
<br />
d dx dy<br />
dz ˆ<br />
<br />
ˆ<br />
ˆ<br />
<br />
i j k . idx ˆ ˆjdy kdz ˆ<br />
<br />
x y z x y z<br />
<br />
<br />
ˆ<br />
<br />
ˆ<br />
<br />
i j kˆ<br />
<br />
<br />
<br />
. d r . d r F.<br />
dr<br />
x y z <br />
2 3 2 2 2<br />
2 xy ( z ) iˆ2x yj ˆ3 x z kˆ<br />
( idx ˆ ˆjdy kdz<br />
ˆ )<br />
<br />
<br />
=<br />
2 2 ˆ<br />
<br />
<br />
2 3 2 2 2<br />
2 x( y z ) dx2x ydy3x z dz<br />
2 3 2 2 2<br />
2 ( ) 2 3 <br />
<br />
xy z dx x ydy x z dz C<br />
2 2 3 2 2<br />
(2 xy dx 2 x ydy ) (2 xz dx 3 x z dz ) + C = x 2 y 2 + x 2 z 3 + C<br />
Hence, the scalar potential is x 2 y 2 + x 2 z 3 + C<br />
Now, for conservative field<br />
Work done =<br />
<br />
(2,3, 4) (2,3, 4)<br />
<br />
F.<br />
d r d<br />
( 1,2,1) ( 1, 2,1)<br />
<br />
<br />
(2,3,4) (2,3,4)<br />
2 2 2 3<br />
x y x z c<br />
( 1,2,1) <br />
( 1,2,1)<br />
<br />
= (36 + 256) – (2 – 1) = 291 Ans.
426 Vectors<br />
<br />
Example 73. A <strong>vector</strong> field is given by F (sin y) iˆ x(1 cos y) ˆj.<br />
Evaluate the line integral<br />
over a circular path x 2 + y 2 = a 2 , z = 0. `(Nagpur University, Winter 2001)<br />
Solution. We have,<br />
F<br />
<br />
C<br />
Work done = . dr<br />
F<br />
<br />
C<br />
<br />
= [(sin ) ˆ (1 cos ) ˆ].[ ˆ ˆ<br />
y i x y j dxi dyj]<br />
( z = 0 hence dz = 0)<br />
<br />
C<br />
. d r = sin ydx x (1 cos y ) dy (sin y dx x cos y dy xdy )<br />
C<br />
C<br />
<br />
= d ( xsin y)<br />
x dy<br />
C<br />
(where d is differential operator).<br />
The parametric equations of given path<br />
x 2 + y 2 = a 2 are x = a cos , y = a sin ,<br />
Where varies form 0 to 2<br />
F<br />
<br />
C<br />
. d r =<br />
=<br />
<br />
<br />
<br />
C<br />
2<br />
2<br />
d [ a cos sin ( a sin )] a cos . a cos d <br />
0 0<br />
2<br />
2<br />
2 2<br />
d [ a cos sin ( a sin )] a cos .<br />
d <br />
0 0<br />
= [ a cos sin ( asin )] a cos d<br />
<br />
<br />
<br />
<br />
2<br />
2<br />
2 2<br />
0 0<br />
2<br />
2 21cos 2 a sin 2 <br />
= 0 a <br />
d<br />
0<br />
<br />
2 2<br />
<br />
2<br />
<br />
0<br />
2<br />
a<br />
2<br />
= .2a<br />
2<br />
Example 74. Determine whether the line integral<br />
Ans.<br />
2 2 2 2<br />
(2 xyz ) dx ( x z z cos yz) dy (2x yz y cos yz)<br />
dz is independent of the path of<br />
<br />
integration ? If so, then evaluate it from (1, 0, 1) to<br />
<br />
0, ,1 .<br />
2 <br />
2 2 2 2<br />
Solution. (2 xy z ) dx ( x z z cos yz) dy (2x yz y cos yz)<br />
dz<br />
=<br />
<br />
c<br />
<br />
c<br />
2 2 2 2<br />
[(2 xy ziˆ) ( x z z cos yz) ˆj (2x yz y cos yz) kˆ].( idx ˆ ˆjdy kdz ˆ )<br />
= Fdr<br />
<br />
<br />
c<br />
This integral is independent of path of integration if<br />
F<br />
= 0<br />
iˆ<br />
ˆj kˆ<br />
F =<br />
F<br />
<br />
<br />
x y z<br />
2 2 2 2<br />
2xyz x z z cos yz 2x yz y cos yz<br />
= (2x 2 z + cos yz – yz sin yz – 2x 2 z – cos yz + yz sin yz) = iˆ<br />
–(4 xyz – 4 xyz) ˆj (2 xz –2 xz ) k<br />
= 0<br />
Hence, the line integral is independent of path.<br />
d = dx <br />
dy <br />
dz<br />
x y z<br />
(Total differentiation)<br />
2<br />
2 2 ˆ
Vectors 427<br />
<br />
B A<br />
<br />
ˆ<br />
ˆ<br />
=<br />
ˆ<br />
<br />
ˆ<br />
<br />
<br />
i j k ( idx ˆ ˆjdy kdz)<br />
= dr F<br />
d r<br />
x y z<br />
2 2 2 2<br />
= [(2 xyz ) iˆ( x z z cos yz) ˆj (2x yz ycos yz) kˆ]. ( idx ˆ ˆjdy kdz ˆ )<br />
= 2xyz 2 dx + (x 2 z 2 + z cos y z) dy + (2x 2 yz + y cos yz) dz<br />
= [(2x dx) yz 2 + x 2 (dy) z 2 + x 2 y (2z dz)] + [(cos yz dy) z + (cos yz dz) y]<br />
= d (x 2 yz 2 ) + d (sin yz)<br />
=<br />
<br />
<br />
2 2 2 2<br />
d ( x yz ) d (sin yz) x yz sin yz<br />
= (B) – (A)<br />
2 2 2 2<br />
= [ x yz sin yz] [ x yz sin yz]<br />
(1, 0,1) =<br />
<br />
0 sin ( 1) [0 0]<br />
(0, ,1)<br />
<br />
2<br />
2<br />
<br />
<br />
<br />
= 1 Ans.<br />
<br />
Example 75. Evaluate A. nˆ<br />
dS, where A 18 ziˆ<br />
–12ˆj 3ykˆ<br />
<br />
and S is the part of the<br />
S<br />
plane 2x + 3y + 6z = 12 included in the first octant. (Uttarakhand, I semester, Dec. 2006)<br />
Solution. Here, A = 18 ziˆ<br />
–12ˆj 3ykˆ<br />
Given surface f (x, y, z) = 2x + 3y + 6z – 12<br />
<br />
ˆ<br />
Normal <strong>vector</strong> = f =<br />
ˆ<br />
<br />
ˆ<br />
<br />
i j k (2x 3y 6 z –12) = 2iˆ3ˆj 6kˆ<br />
x y z<br />
ˆn = unit normal <strong>vector</strong> at any point (x, y, z) of 2x + 3y + 6z = 12<br />
2iˆ3ˆj 6kˆ<br />
1<br />
=<br />
(2iˆ3ˆj 6 kˆ<br />
)<br />
4936<br />
7<br />
dx dy dx dy dxdy 7<br />
dS = dx dy<br />
nˆ . kˆ 1<br />
ˆ ˆ ˆ ˆ 6 6<br />
(2i 3 j 6 k).<br />
k<br />
7 7<br />
Now, A <br />
. ndS<br />
1 7<br />
<br />
ˆ = (18 ˆ –12 ˆ 3 ˆ). (2ˆ 3ˆ<br />
6 ˆ<br />
zi j yk i j k)<br />
dx dy<br />
7 6<br />
dx dy<br />
= (36 z –36<br />
18 y)<br />
= (6 z –6<br />
3 y)<br />
dx dy<br />
6 <br />
Putting the value of 6z = 12 – 2x – 3y, we get<br />
Y<br />
=<br />
=<br />
=<br />
=<br />
1<br />
6 (12 –2 x)<br />
3<br />
0 0<br />
<br />
<br />
(12–2 x –3 y –6<br />
3 y)<br />
dxdy<br />
1<br />
6 (12 –2 x)<br />
3<br />
0 0<br />
1<br />
6 (12 –2 x)<br />
3<br />
(6 –2 x)<br />
dx<br />
0 0<br />
1<br />
6<br />
(12–2 x )<br />
(6 –2 x) dx( y)<br />
3<br />
0 0<br />
<br />
(6 –2 x)<br />
dxdy<br />
dy<br />
6 1<br />
1 6<br />
2<br />
= (6 –2 x) (12 –2 x)<br />
dx = (4 x –36x 72) dx<br />
0<br />
0<br />
3<br />
3<br />
<br />
3<br />
6<br />
14x<br />
2<br />
<br />
–18x<br />
72x<br />
3<br />
3<br />
= 1 [4 36 2–18 36 72 6] = 72 [4 –9 6] 24<br />
Ans.<br />
3<br />
3<br />
0<br />
O<br />
B<br />
2x + 3y = 12<br />
A<br />
X
428 Vectors<br />
EXERCISE 5.10<br />
1. Find the work done by a force yiˆ xj ˆ which displaces a particle from origin to a point ( iˆ ˆj).<br />
Ans. 1<br />
2. Find the work done when a force 2 2<br />
F ( x – y x) iˆ(2 xy y)<br />
ˆj<br />
moves a particle from origin to<br />
(1, 1) along a parabola y 2 = x. Ans. 2 3<br />
3. Show that<br />
<br />
<br />
3 ˆ <br />
2 ˆ <br />
2 ˆ is a conservative field. Find its scalar potential such that<br />
V (2 xy z ) i x j 3xz k<br />
V = grad . Find the work done by the force V in moving a particle from (1, – 2, 1) to (3, 1, 4).<br />
Ans. x 2 y + xz 3 , 202<br />
4. Show that the line integral 2<br />
(2xy 3) dx ( x 4 z) dy 4ydz<br />
<br />
c<br />
where c is any path joining (0, 0, 0) to (1, – 1, 3) does not depend on the path c and evaluate the line<br />
integral. Ans. 14<br />
2 2<br />
x y<br />
5. Find the work done in moving a particle once round the ellipse 1 , z = 0, under the field of<br />
25 16<br />
force given by F = (2x – y + z) î + (x + y – z2 ) ĵ + (3x – 2y + 4z) k ˆ. Is the field of force conservative?<br />
(A.M.I.E.T.E., Winter 2000) Ans. 40 <br />
6.<br />
If 4<br />
= (y 2 – 2xyz 3 ) î + (3 + 2xy – x 2 z 3 ) ĵ + (z 3 – 3x 2 yz 2 z 2 2 3<br />
) k ˆ, find . Ans. 3y xy x yz<br />
<br />
4<br />
R . dR<br />
7. is independent of the path joining any two point if it is. (A.M.I.E.T.E., June 2010)<br />
C<br />
Ans. (i)<br />
(i) irrotational field (ii) solenoidal field (iii) rotational field (iv) <strong>vector</strong> field.<br />
5.34 SURFACE INTEGRAL<br />
A surface r = f(u, v) is called smooth if f (u, v) posses continous<br />
first order partial derivative.<br />
Let F be a <strong>vector</strong> function and S be the given surface.<br />
Surface integral of a <strong>vector</strong> function F over the surface S is defined<br />
as the integral of the components of F <br />
along the normal to the<br />
surface.<br />
Component of F along the normal<br />
= F . ˆn , where n is the unit normal <strong>vector</strong> to an element ds and<br />
grad f<br />
ˆn =<br />
|grad f |<br />
Surface integral of F over S<br />
Note. (1) Flux =<br />
<br />
ds =<br />
dx dy<br />
( nˆ<br />
kˆ<br />
)<br />
= Fn ˆ<br />
= ( Fnˆ<br />
) ds<br />
<br />
S<br />
<br />
<br />
( F nˆ<br />
) ds where, F represents the velocity of a liquid.<br />
S<br />
<br />
<br />
If<br />
( Fn ˆ)<br />
ds = 0, then F is said to be a solenoidal <strong>vector</strong> point function.<br />
S<br />
<br />
<br />
Example 76. Evaluate<br />
( ˆ ˆ ˆ<br />
yzi zxj xyk)<br />
ds where S is the surface of the sphere<br />
S<br />
x 2 + y 2 + z 2 = a 2 in the first octant. (U.P., I Semester, Dec. 2004)<br />
Solution. Here, = x 2 + y 2 + z 2 – a 2
Vectors 429<br />
Vector normal to the surface = =<br />
iˆ<br />
ˆj kˆ<br />
<br />
x y z<br />
<br />
ˆ 2 2 2 2<br />
ˆ<br />
=<br />
ˆ<br />
<br />
ˆ<br />
<br />
i j k ( x y z a ) 2xiˆ 2yj ˆ 2zk<br />
x y z<br />
ˆn<br />
2xiˆ 2yj ˆ 2zkˆ<br />
xiˆ yj ˆ zkˆ<br />
= <br />
<br />
| | 2 2 2 2 2 2<br />
4x 4y 4z x y z<br />
xiˆ<br />
yj ˆ zkˆ<br />
=<br />
[ x 2 + y 2 + z 2 = a 2 ]<br />
a<br />
Here, F = yz iˆ<br />
zxj ˆ xy kˆ<br />
<br />
ˆ ˆ ˆ 3<br />
Fn ˆ = ( ˆ ˆ ˆ<br />
xi yj zk xyz<br />
yz i zxj xy k)<br />
<br />
<br />
a <br />
<br />
a<br />
Now, F<br />
n ˆ ds<br />
<br />
2 2<br />
dx dy a a x 3xyz dx dy<br />
S<br />
= ( Fn ˆ)<br />
<br />
S<br />
| kˆ<br />
. nˆ<br />
<br />
|<br />
00<br />
z <br />
a <br />
a <br />
a<br />
2<br />
x<br />
2<br />
2<br />
a a<br />
2<br />
x<br />
2<br />
a y <br />
= 3 xy dy dx 3 x dx<br />
0<br />
0 0 2 <br />
<br />
=<br />
0<br />
2 2 4<br />
a<br />
4 4 4<br />
3 a 2 2 3 a x x 3 a a 3a<br />
2<br />
x a x dx 0<br />
2 <br />
2 4 <br />
<br />
2 2 4 <br />
8<br />
0 <br />
( ) .<br />
3<br />
Example 77. Show that Fnds ˆ ,<br />
S<br />
where<br />
2<br />
F = 4 xz î – y 2 ĵ + yz ˆk<br />
and S is the surface of the cube bounded by the planes,<br />
x= 0, x = 1, y = 0, y = 1, z = 0, z = 1.<br />
Solution.<br />
<br />
<br />
<br />
= <br />
ˆ<br />
Fnds ˆ<br />
S<br />
Fnds<br />
<br />
<br />
OABC<br />
Fnds ˆ<br />
Fnds<br />
ˆ<br />
<br />
DEFG<br />
<br />
<br />
OAGF<br />
<br />
Fnds ˆ<br />
Fnds<br />
ˆ<br />
BCED<br />
<br />
Fnds<br />
<br />
OCEF<br />
<br />
ABDG<br />
ˆ<br />
...(1)<br />
Now,<br />
<br />
DEFG<br />
Fnds<br />
<br />
<br />
OABC<br />
<br />
2<br />
= (4 ˆ ˆ ˆ<br />
xzi y j yz k)( k)<br />
dx dy = 1 1<br />
OABC<br />
2<br />
(4 xziˆ<br />
y ˆj yz kˆ)<br />
kˆ<br />
dx dy<br />
=<br />
<br />
DEFG<br />
2<br />
1<br />
1 y<br />
<br />
1<br />
= dx [ x]<br />
0<br />
0<br />
0<br />
2 ˆ<br />
1 1<br />
<br />
yz dx dy y (1) dxdy<br />
0 0<br />
1 1<br />
<br />
2 2 2<br />
(4 ˆ ˆ<br />
xz i y j yzk) ( j)<br />
dxdz =<br />
OAGF<br />
0 0<br />
Ans.<br />
S.No. Surface Outward ds<br />
normal<br />
1 OABC – k dx dy z = 0<br />
2 DEFG k dx dy z = 1<br />
3 OAGF – j dx dz y = 0<br />
4 BCED j dx dz y = 1<br />
5 ABDG i dy dz x = 1<br />
6 OCEF – i dy dz x = 0<br />
yz dx dy 0<br />
(as z = 0)<br />
2<br />
y dxdz 0<br />
(as y = 0)<br />
OAGF
430 Vectors<br />
(4 xz iˆ y ˆj yzk)<br />
ˆ<br />
jdxdz =<br />
BCED<br />
2 ˆ<br />
=<br />
2 ˆ<br />
<br />
BCED<br />
1 dx 1 dz x 1 1<br />
0 0<br />
0 z 0<br />
(4 xziˆ y ˆj yzk)<br />
idydz<br />
ˆ<br />
<br />
=<br />
ABDG<br />
=<br />
2<br />
( y ) dxdz<br />
(as y = 1)<br />
( ) ( ) 1<br />
<br />
2<br />
1<br />
1<br />
z 1<br />
<br />
0 <br />
4( y) <br />
4(1) <br />
2<br />
2 <br />
2<br />
(4 ˆ ˆ )( ˆ<br />
xz i y j yzk i)<br />
dydz =<br />
OCEF<br />
On putting these values in (1), we get<br />
=<br />
F<br />
nds ˆ<br />
S<br />
1. Evaluate A <br />
. n ˆ ds ,<br />
S<br />
2 ˆ<br />
0<br />
1 1<br />
<br />
4xz dy dz 4 (1) zdydz<br />
1 1<br />
0 0<br />
1<br />
0 01 2 0 = 3 2<br />
2<br />
EXERCISE 5.11<br />
0 0<br />
where A = 2<br />
( x y ) iˆ<br />
2xj ˆ 2yzkˆ<br />
4xz dy dz 0<br />
(as x = 0)<br />
Proved.<br />
and S is the surface of the plane<br />
2x + y + 2z = 6 in the first octant. Ans. 81<br />
2. Evaluate A <br />
. n ˆ ds ,<br />
S<br />
where A = 2<br />
ziˆ xj ˆ 3y zkˆ<br />
and S is the surface of the cylinder x 2 + y 2 = 16<br />
included in the first octant between z = 0 and z = 5. Ans. 90<br />
3. If r =<br />
tiˆ t ˆ j ( t 1) k<br />
2<br />
ˆ<br />
and S =<br />
2ˆ<br />
tkˆ<br />
evaluate<br />
2ti<br />
6 ,<br />
2 <br />
r S dt.<br />
Ans. 12<br />
0<br />
4. Evaluate FndS<br />
<br />
ˆ , where, F = 18 ziˆ12 ˆj 3ykˆ<br />
and S is the surface of the plane 2x + 3y + 6z = 12<br />
S<br />
in the first octant. Ans. 24<br />
5. Evaluate Fnds<br />
<br />
ˆ ,<br />
2<br />
where, F = 2yxiˆ yzj ˆ xkˆ<br />
over the surface S of the cube bounded by the<br />
S<br />
coordinate planes and planes x = a, y = a and z = a.<br />
<br />
2<br />
6. If F 2yiˆ<br />
3ˆj xkˆ<br />
and S is the surface of the parabolic cylinder y<br />
2<br />
= 8x in the first octant bounded<br />
<br />
by the planes y = 4, and z = 6, then evaluate FndS ˆ .<br />
Ans. 132<br />
S<br />
5.35 VOLUME INTEGRAL<br />
Let F be a <strong>vector</strong> point function and volume V enclosed by a closed surface.<br />
The volume integral =<br />
F <br />
dv<br />
V<br />
Example 78. If F = 2 z î – x ĵ + y ˆk , evaluate<br />
the surfaces<br />
x = 0, y = 0, x = 2, y = 4, z = x 2 , z = 2.<br />
Solution.<br />
(2 ziˆ<br />
xj ˆ yk)<br />
dx dy dz<br />
<br />
= ˆ<br />
<br />
=<br />
=<br />
F <br />
dv<br />
V<br />
2 4 2<br />
0 0 x<br />
2<br />
dx dy (2 ziˆ<br />
xj ˆ ykˆ<br />
) dz<br />
=<br />
<br />
2 4 4 3 2<br />
dx dy [4iˆ 2xj ˆ 2 ykˆ<br />
x iˆ x ˆj x ykˆ]<br />
<br />
0 0<br />
Ans.<br />
1 4<br />
2 a<br />
Fdv<br />
<br />
where, v is the region bounded by<br />
V<br />
<br />
2 4 2<br />
2<br />
dx ˆ ˆ ˆ<br />
0 dy [ z i xzj yzk] 0<br />
x<br />
2
Vectors 431<br />
=<br />
<br />
2<br />
0<br />
<br />
2 2<br />
ˆ ˆ 2 ˆ 4 ˆ 3<br />
4 2<br />
ˆ<br />
x y<br />
dx yi xyj yk x yi x yj kˆ<br />
<br />
<br />
<br />
2 <br />
2 4 3 2<br />
= ˆ ˆ ˆ ˆ ˆ ˆ<br />
(16i 8xj 16k 4xi 4x j 8 xk)<br />
dx<br />
0<br />
2<br />
<br />
5 3<br />
2 4 4 8<br />
= ˆ ˆ ˆ x<br />
16 4 16 ˆ ˆ<br />
x <br />
xi x j xk i x j kˆ<br />
<br />
<br />
5 3 0<br />
128 64<br />
= 32 ˆ 16 ˆ 32 ˆ ˆ 16 ˆ ˆ 32 iˆ<br />
32kˆ<br />
i j k i j k = = 32 (3iˆ<br />
5 kˆ<br />
)<br />
5 3 5 3 15<br />
EXERCISE 5.12<br />
1. If F <br />
= 2<br />
(2x 3) z iˆ2xyj ˆ 4 xkˆ<br />
, then evaluate FdV , where V is bounded by the plane<br />
V<br />
4<br />
0<br />
Ans.<br />
x = 0, y = 0, z = 0 and 2x + 2y + z = 4. Ans. 8 3<br />
2. Evaluate dV ,<br />
V where = 45 x 2 y and V is the closed region bounded by the planes<br />
4x + 2y + z = 8, x = 0, y = 0, z = 0 Ans. 128<br />
3. If F = (2x 2 <br />
– 3z) iˆ<br />
2xyj ˆ 4 xkˆ<br />
, then evaluate FdV<br />
, where V is the closed region bounded<br />
V<br />
8<br />
by the planes x = 0, y = 0, z = 0 and 2x + 2y + z = 4.<br />
Ans. ( ˆ ˆ)<br />
3 j k<br />
4. Evaluate (2 x<br />
y) dV , where V is closed region bounded by the cylinder z = 4 – x 2 and the planes<br />
V<br />
x = 0, y = 0, y = 2 and z = 0. Ans. 80 3<br />
5. If F 2<br />
= 2 ˆ ˆ ˆ<br />
<br />
xz i xj y k,<br />
evaluate<br />
F dV over the region bounded by the surfaces x = 0, y = 0,<br />
y = 6 and z = x 2 , z = 4. Ans. (16iˆ3ˆj 48 kˆ<br />
)<br />
5.36 GREEN’S THEOREM (For a plane)<br />
<br />
Statement. If (x, y), (x, y), and be continuous functions over a region R bounded<br />
y<br />
x<br />
by simple closed curve C in x – y plane, then<br />
( dx<br />
dy)<br />
= <br />
<br />
dx dy<br />
C<br />
R<br />
<br />
x<br />
y<br />
(AMIETE, June 2010, U.P., I Semester, Dec. 2007)<br />
Proof. Let the curve C be divided into two curves C 1<br />
(ABC) and C 1<br />
(CDA).<br />
Let the equation of the curve C 1<br />
(ABC) be y = y 1<br />
(x) and equation of the curve C 2<br />
(CDA) be<br />
y = y 2<br />
(x).<br />
Let us see the value of<br />
<br />
dx dy = xc<br />
y<br />
y<br />
2<br />
( x)<br />
<br />
dy dx<br />
R y<br />
xa y y ( x)<br />
<br />
1 y<br />
<br />
= c 2 ( )<br />
( , )<br />
y <br />
x y<br />
y x<br />
dx<br />
a<br />
y y ( x)<br />
1<br />
c<br />
a<br />
c<br />
= ( x, y2) ( x, y1)<br />
dx = ( x, y2) dx ( x, y1)<br />
dx<br />
a<br />
a<br />
c<br />
= <br />
<br />
( x, y2) dx ( x, y1)<br />
dx<br />
<br />
c<br />
a<br />
<br />
= <br />
<br />
( x , y ) dx ( x , y ) dx<br />
<br />
<br />
c2 c1<br />
<br />
= – ( , )<br />
<br />
c<br />
c<br />
<br />
a<br />
x y dx
432 Vectors<br />
Thus,<br />
dx<br />
c<br />
=<br />
Similarly, it can be shown that<br />
<br />
dx dy<br />
...(1)<br />
R y<br />
<br />
dy<br />
c<br />
= dx dy ...(2)<br />
x<br />
On adding (1) and (2), we get<br />
<br />
<br />
( dx<br />
dy)<br />
= <br />
dx dy<br />
R<br />
Proved.<br />
x<br />
y<br />
<br />
Note. Green’s Theorem in <strong>vector</strong> form<br />
where,<br />
<br />
<br />
Fdr<br />
<br />
= ( F ) ˆ<br />
c <br />
R<br />
k dR<br />
<br />
F iˆ ˆj, r xiˆ yj ˆ,<br />
kˆ<br />
is a unit <strong>vector</strong> along z-axis and dR = dx dy.<br />
<br />
Example 79. A <strong>vector</strong> field F is given by F sin yiˆ x (1 cos y) ˆj.<br />
Evaluate the line integral Fdr<br />
<br />
where C is the circular path given by x 2 + y 2 = a 2 .<br />
C<br />
<br />
Solution. F sin yiˆ x(1 cos y)<br />
ˆj<br />
Fdr<br />
<br />
<br />
C<br />
= [sin ˆ (1 cos ) ˆ] ( ˆ ˆ<br />
yi x y j idx jdy)<br />
C<br />
= sin ydx x (1 cos y)<br />
dy<br />
C<br />
On applying Green’s Theorem, we have<br />
<br />
<br />
( dx dy)<br />
c<br />
= <br />
dx dy<br />
s<br />
<br />
x<br />
y<br />
<br />
= [(1<br />
cos y) cos y]<br />
dx dy<br />
s<br />
where s is the circular plane surface of radius a.<br />
= dx dy<br />
s<br />
= Area of circle = a 2 . Ans.<br />
2 2<br />
Example 80. Using Green’s Theorem, evaluate ( x ydx x dy),<br />
where c is the boundary<br />
described counter clockwise of the triangle with vertices (0, 0), (1, 0), (1, 1).<br />
(U.P., I Semester, Winter 2003)<br />
Solution. By Green’s Theorem, we have<br />
Y<br />
A<br />
<br />
<br />
(1, 1)<br />
=<br />
( dx dy)<br />
c<br />
= dx dy<br />
R<br />
<br />
x<br />
y<br />
<br />
2 2<br />
( x y dx <br />
2<br />
x dy)<br />
= (2 x x ) dxdy<br />
c<br />
1 2<br />
(2 x x ) dx dy =<br />
0 0<br />
x<br />
=<br />
<br />
<br />
R<br />
1 2<br />
0<br />
(2 x x ) dx[ y]x<br />
1 2<br />
(2 x x )( x ) dx =<br />
0<br />
2 1<br />
= <br />
3 4 = 5<br />
12<br />
Example 81. State and verify Green’s Theorem in the plane for<br />
c<br />
0<br />
1 2 3<br />
(2 x x ) dx =<br />
0<br />
<br />
2 2<br />
(0, 0)<br />
(1, 0)<br />
<br />
3 4<br />
2 x x <br />
<br />
<br />
3 4 <br />
<br />
1<br />
0<br />
Ans.<br />
(3 x –8 y ) dx (4 y – 6 xy)<br />
dy where C is the boundary of the region bounded by x 0, y 0 and 2x – 3y = 6.<br />
(Uttarakhand, I Semester, Dec. 2006)<br />
O<br />
y = x<br />
X
Vectors 433<br />
Solution. Statement: See Article 24.4 on page 576.<br />
Here the closed curve C consists of straight lines OB, BA and AO, where coordinates of A and<br />
B are (3, 0) and (0, – 2) respectively. Let R be the region bounded by C.<br />
Then by Green’s Theorem in plane, we have<br />
<br />
2 2<br />
[(3 x –8 y ) dx (4 y –6 xy) dy ]<br />
=<br />
<br />
<br />
x<br />
<br />
y<br />
2 2<br />
(4 y –6 xy)– (3 x –8 y ) dxdy<br />
R<br />
<br />
<br />
...(1)<br />
<br />
= (– 6y 16 y) dx dy 10 ydxdy<br />
R<br />
2<br />
3 0 3 y<br />
= 10 dx 1 ydy 10 dx <br />
<br />
x<br />
2 <br />
0 (2 –6) 0<br />
3 1<br />
(2 x –6)<br />
3<br />
<br />
=<br />
3<br />
5 (2 x – 6) 5 3<br />
= – – (0 6)<br />
9 3<br />
2 54<br />
0<br />
Now we evaluate L.H.S. of (1) along OB, BA and AO.<br />
Along OB, x = 0, dx = 0 and y varies form 0 to –2.<br />
Along BA, x = 1 (6 3 y),<br />
dx 3 dy and y varies from –2 to 0.<br />
2 2<br />
and along AO, y = 0, dy = 0 and x varies from 3 to 0.<br />
L.H.S. of (1) =<br />
=<br />
<br />
OB<br />
<br />
2 2<br />
[(3 x – 8 y ) dx (4 y –6 xy) dy]<br />
3<br />
2 2 2 2<br />
BA<br />
R<br />
<br />
0<br />
<br />
<br />
=<br />
5 3<br />
– (2 – 6)<br />
9<br />
dx x<br />
0<br />
5<br />
– (216) –20 ...(2)<br />
54<br />
[(3 x – 8 y ) dx (4 y –6 xy) dy] [(3 x – 8 y ) dx (4 x –6 xy) dy ]<br />
<br />
<br />
AO<br />
2 2<br />
[(3 x – 8 y ) dx (4 y –6 xy) dy ]<br />
–2 0 3 2 23<br />
<br />
0<br />
2<br />
= 4 ydy (6 3 y) –8 y dy [4 y –3(6 3 y) y] dy 3x dx<br />
0 –2<br />
<br />
4<br />
<br />
2<br />
<br />
<br />
3<br />
=<br />
2 –2 0 9<br />
2 2 2 3 0<br />
[2 y ] <br />
<br />
0<br />
<br />
(63 y) –12y 4 y –18 y –9 y dy ( x )<br />
3<br />
–2<br />
8<br />
<br />
<br />
<br />
0 9<br />
2 2 <br />
= 2[4] <br />
(6 3 y) –21 y –14 y dy (0 –27)<br />
–2<br />
8<br />
<br />
<br />
<br />
0<br />
<br />
3<br />
9(6 3 y) 3 2<br />
216<br />
3 2<br />
=<br />
8 – 7 y – 7 y –27 –19 7(– 2) 7(–2)<br />
8 3 3 8<br />
<br />
<br />
<br />
–2<br />
= – 19 + 27 – 56 + 28 = – 20 ...(3)<br />
With the help of (2) and (3), we find that (1) is true and so Green’s Theorem is verified.<br />
2 2 2 2<br />
Example 82. Apply Green’s Theorem to evaluate [(2 x y ) dx ( x y ) dy],<br />
where C<br />
is the boundary of the area enclosed by the x-axis and the upper half of circle x 2 + y 2 = a 2 .<br />
(M.D.U. Dec. 2009, U.P., I Sem., Dec. 2004)<br />
2 2 2 2<br />
Solution. [(2 x y ) dx ( x y ) dy]<br />
<br />
C<br />
By Green’s Theorem, we’ve ( dx<br />
dy)<br />
=<br />
C<br />
<br />
S<br />
C<br />
<br />
<br />
dx dy<br />
x<br />
y<br />
2
434 Vectors<br />
=<br />
a a<br />
2<br />
x<br />
2<br />
<br />
a<br />
0<br />
a a<br />
2<br />
x<br />
2<br />
a<br />
0<br />
2 2 2 2 <br />
( x y ) (2 x y ) dx dy<br />
x<br />
y<br />
<br />
a a<br />
2<br />
x<br />
2<br />
= (2x<br />
2 y)<br />
dx dy = 2 dx ( x y ) dy<br />
a<br />
2<br />
x<br />
2<br />
2<br />
= 2 a y <br />
dx<br />
xy <br />
a<br />
2 = 2<br />
<br />
=<br />
<br />
a<br />
0<br />
<br />
a<br />
a<br />
2 2 a 2 2<br />
a 2 x a x dx ( a x ) dx<br />
a<br />
<br />
a<br />
0<br />
<br />
2 2<br />
2 2 a x <br />
<br />
x a x <br />
dx<br />
2 <br />
<br />
<br />
<br />
<br />
<br />
<br />
3<br />
a<br />
3<br />
2 x<br />
3 a<br />
2 2<br />
= 0 2 a ( a x ) dx = 2 a x<br />
2<br />
a<br />
3<br />
<br />
0<br />
3 3 <br />
= 4a<br />
3<br />
0<br />
<br />
a<br />
a<br />
( ) 2 ( ) , is even<br />
<br />
<br />
<br />
<br />
0, is odd <br />
f x dx f x dx f<br />
a 0<br />
<br />
f<br />
Example 83. Evaluate y x<br />
dx dy,<br />
where C C<br />
C 2 2 2 2<br />
1UC2<br />
x y x y<br />
with C : x2 + y2 = 1<br />
1<br />
and C 2<br />
: x = 2, y = 2. (Gujarat, I Semester, Jan 2009)<br />
y<br />
x<br />
Solution. dx <br />
C 2 2 2 2<br />
x y x y<br />
dy<br />
Y<br />
x y <br />
= <br />
<br />
dx dy<br />
x<br />
2 2<br />
y<br />
2 2<br />
x y x y <br />
<br />
<br />
<br />
2 2 2 2<br />
( x y )1–2 x( x) ( x y )1–2 y ( y)<br />
<br />
X<br />
<br />
<br />
= <br />
<br />
dx dy<br />
2 2 2 2 2 2 x = – 2<br />
( x y ) ( x y ) <br />
<br />
2 2 2 2 2 2<br />
x y –2 x x y –2y<br />
<br />
= <br />
<br />
dx dy<br />
2 2 2 2 2 2 <br />
( x y ) ( x y ) <br />
2 2 2 2<br />
= y – x x – y <br />
<br />
dx dy<br />
2 2 2 2 2 2<br />
( x y ) ( x y ) <br />
= 0<br />
dx dy 0<br />
( x<br />
2 y<br />
2 )<br />
2<br />
5.37 AREA OF THE PLANE REGION BY GREEN’S THEOREM<br />
Proof. We know that<br />
Mdx Ndy = <br />
N M <br />
– dx dy<br />
A<br />
<br />
x y<br />
C<br />
<br />
...(1)<br />
<br />
N <br />
<br />
M <br />
On putting N = x 1<br />
and M = – y 1<br />
in (1), we get<br />
x <br />
y <br />
– ydx xdy = [1 –(–1)] dx dy<br />
A<br />
= 2 dx dy = 2 A<br />
C<br />
Area = 1 ( – )<br />
2<br />
xdy ydx<br />
C<br />
Example 84. Using Green’s theorem, find the area of the region in the first quadrant bounded<br />
by the curves<br />
y = x, y = 1 x<br />
, y = (U.P. I, Semester, Dec. 2008)<br />
x 4<br />
Solution. By Green’s Theorem Area A of the region bounded by a closed curve C is given by<br />
O<br />
Y<br />
y = 2<br />
x 2 + y 2 = 1<br />
y = – 2<br />
Ans.<br />
X<br />
x = 2<br />
Ans.
Vectors 435<br />
A = 1 Y<br />
( xdy – ydx)<br />
2<br />
C<br />
x<br />
Here, C consists of the curves C 1<br />
: y = , C<br />
4 2<br />
: y = 1 x<br />
and C 3<br />
: y = x So<br />
1 1 1 <br />
A <br />
<br />
<br />
<br />
( I1 I2 I3<br />
)<br />
C<br />
<br />
2C 2 C <br />
1 C <br />
2 C3<br />
<br />
2<br />
<br />
2<br />
<br />
<br />
C<br />
x 1<br />
3 C 1<br />
Along C 1<br />
: y = , dy dx , x :0to 2<br />
x<br />
y = —<br />
4 4<br />
4<br />
1 x <br />
I 1<br />
= ( xdy – ydx) x dx – dx 0<br />
C <br />
1 C<br />
<br />
<br />
1<br />
4 4 <br />
Along C 2<br />
: y = 1 , dy – 1<br />
O (0, 0)<br />
dx, x :2to1<br />
2<br />
x x<br />
1<br />
1 1<br />
I 2<br />
= ( xdy – ydx)<br />
x – <br />
dx – <br />
dx<br />
C<br />
2<br />
2<br />
2 <br />
x 2 <br />
<br />
=<br />
1<br />
[– 2log x] 2log 2<br />
2<br />
Along C 3<br />
: y = x, dy = dx ; x : 1 to 0 ;<br />
I 3<br />
=<br />
<br />
C3<br />
<br />
( xdy – ydx) ( xdx – xdx) 0<br />
A = 1 ( I 1<br />
1 I2 I3) (0 2log 2+0) log 2<br />
Ans.<br />
2 2<br />
EXERCISE 5.13<br />
2 2<br />
1. Evaluate<br />
[(3x 6 yz) dx(2y 3 xz) dy (1 4 xyz ) dz]<br />
from (0, 0, 0) to (1, 1, 1) along the path c<br />
c<br />
given by the straight line from (0, 0, 0) to (0, 0, 1) then to (0, 1, 1) and then to (1, 1, 1).<br />
2 2 3<br />
2. Verify Green’s Theorem in plane for ( x 2 xy) dx ( y x y) dy,<br />
where c is a square with the<br />
C<br />
1<br />
vertices P (0, 0), Q (1, 0), R (1, 1) and S (0, 1).<br />
Ans. <br />
2<br />
2 2<br />
3. Verify Green’s Theorem for ( x 2 xy) dx ( x y 3) dy around the boundary c of the region<br />
c<br />
y 2 = 8 x and x = 2.<br />
2 2 2 2<br />
4. Use Green’s Theorem in a plane to evaluate the integral<br />
[(2 x y ) dx ( x y ) dy]<br />
,<br />
c<br />
where c is the boundary in the xy-plane of the area enclosed by the x-axis and the semi-circle x 2 + y 2 =1<br />
in the upper half xy-plane. Ans. 4 3<br />
5. Apply Green’s Theoem to evaluate [( y sin x) dy cos xdy],<br />
c<br />
where c is the plane triangle enclosed<br />
<br />
2 x<br />
2<br />
8<br />
by the lines y = 0, x = and y .<br />
Ans. <br />
2<br />
<br />
4<br />
6. Either directly or by Green’s Therorem, evaluate the line integral <br />
e x (cos y dx sin ydy),<br />
c<br />
<br />
where c is the rectangle with vertices (0, 0), (, 0,), , and 0, .<br />
Ans. 2 (1 – e – )<br />
2 2<br />
(AMIETE II Sem June 2010)<br />
2<br />
7. Verify the Green’s Theorem to evaluate the line integral (2 y dx 3 x dy),<br />
where c is the boundary<br />
c<br />
of the closed region bounded by y = x and y = x 2 .<br />
(U.P., I Semester, Dec. 20005, AMIETE Summer 2004, Winter 2001) Ans. 27 4<br />
y = x<br />
B (1, 1)<br />
y = — 1 x<br />
A (2, —)<br />
1<br />
2<br />
X
436 Vectors<br />
8. Evaluate F . nds ˆ. ,<br />
s<br />
and s is the region of the plane 2x + 2y + z = 6<br />
<br />
where 2<br />
F xyiˆ x ˆ j ( x z)<br />
kˆ<br />
in the first octant. (A.M.I.E.T.E., Summer 2004, Winter 2001) Ans. 27 4<br />
9. Verify Green’s Theorem for<br />
2 2<br />
( xy y ) dx x dy<br />
<br />
C <br />
where C is the boundary by y = x and y = x 2 .<br />
(AMIETE, June 2010)<br />
5.38 STOKE’S THEOREM (Relation between Line Integral and Surface Integral)<br />
(Uttarakhand, I Sem. 2008, U.P., Ist Semester, Dec. 2006)<br />
Statement. Surface integral of the component of curl F along the normal to the surface S,<br />
taken over the surface S bounded by curve C is equal to the line integral of the <strong>vector</strong> point function<br />
F taken along the closed curve C.<br />
Mathematically<br />
<br />
<br />
F . d r = curl Fnds ˆ<br />
S<br />
where ˆn = cos î + cos ĵ + cos ˆk is a unit<br />
external normal to any surface ds,<br />
Proof. Let r = xiˆ<br />
yj ˆ<br />
zkˆ<br />
dr = idx ˆ ˆj dy kdz ˆ<br />
F = Fi ˆ<br />
ˆ<br />
1 F ˆ 2 j F3 k<br />
<br />
On putting the values of F,<br />
d r in the statement of the theorem<br />
<br />
c<br />
( Fi ˆ ˆ ˆ ˆ ˆ ˆ<br />
1 F2 j Fk 3 ) ( idx j dy kdz)<br />
= <br />
i j k <br />
S<br />
<br />
x y z<br />
( Fiˆ ˆ ˆ ˆ ˆ ˆ<br />
<br />
1 F 2 j F3 k). (cos i cos j cos k)<br />
ds<br />
( F1 dx <br />
F3 F2 1 3<br />
2 1<br />
F2 dy F3<br />
dz)<br />
=<br />
ˆ F F<br />
ˆ<br />
F F<br />
<br />
i j kˆ<br />
<br />
<br />
.<br />
S<br />
<br />
<br />
y z z x x y<br />
<br />
( iˆcos ˆj cos kˆ<br />
cos )<br />
ds<br />
3 2 1 3<br />
2 1<br />
= F<br />
F F F<br />
F F<br />
<br />
cos cos cos ds<br />
S y z<br />
<br />
<br />
z x<br />
<br />
x y<br />
<br />
...(1)<br />
Let us first prove<br />
F c<br />
1 dx<br />
1 1<br />
= F<br />
F<br />
<br />
cos cos ds<br />
S<br />
<br />
<br />
<br />
z<br />
y<br />
<br />
...(2)<br />
Let the equation of the surface S be z = g (x, y). The projection of the surface on x – y plane<br />
is region R.<br />
F 1 ( x , y , z ) dx<br />
c<br />
= F 1 [ x , y , g ( x , y )] dx<br />
c<br />
<br />
= F1 ( x, y, g)<br />
dx dy [By Green’s Theorem]<br />
R y<br />
F1 F1<br />
g <br />
= dx dy<br />
R<br />
<br />
y z y<br />
<br />
...(3)<br />
The direction consines of the normal to the surface z = g(x, y) are given by<br />
cos <br />
= cos cos <br />
<br />
g<br />
g<br />
1<br />
x<br />
y
Vectors 437<br />
And dx dy = projection of ds on the xy-plane = ds cos <br />
Putting the values of ds in R.H.S. of (2)<br />
F<br />
z<br />
F<br />
y<br />
<br />
1 1<br />
cos cos ds<br />
S<br />
<br />
=<br />
=<br />
R<br />
<br />
<br />
<br />
<br />
z<br />
cos y<br />
<br />
<br />
<br />
<br />
F1 cos F <br />
1 dx dy<br />
=<br />
R<br />
F1 F1<br />
dx dy<br />
cos cos <br />
z<br />
y<br />
cos <br />
F1 g<br />
F1<br />
<br />
<br />
dx dy<br />
R<br />
z y y<br />
<br />
F1 F1<br />
g <br />
= dx dy<br />
R<br />
<br />
y z y<br />
<br />
...(4)<br />
From (3) and (4), we get<br />
F c<br />
1 = F1 F1<br />
<br />
cos cos ds<br />
S<br />
<br />
z<br />
y<br />
<br />
...(5)<br />
Similarly, F c<br />
2 2 2<br />
= F<br />
F<br />
<br />
cos cos ds<br />
S<br />
<br />
x<br />
z<br />
<br />
<br />
<br />
...(6)<br />
and F c<br />
3 3 3<br />
= F<br />
F<br />
<br />
cos cos ds<br />
S<br />
<br />
y<br />
x<br />
<br />
...(7)<br />
On adding (5), (6) and (7), we get<br />
( F1 dx F2 dy F3<br />
dz)<br />
c<br />
=<br />
<br />
S<br />
F1 F1 F2 F2<br />
cos cos cos cos <br />
z y x z<br />
F3 F3<br />
<br />
cos cos ds Proved.<br />
y<br />
x<br />
<br />
5.39 ANOTHER METHOD OF PROVING STOKE’S THEOREM<br />
The circulation of <strong>vector</strong> F around a closed curve C is equal to the flux of<br />
the curve of the <strong>vector</strong> through the surface S bounded by the curve C.<br />
<br />
F<br />
d r<br />
<br />
c = curl Fnds curl FdS <br />
S S<br />
Proof : The projection of any curved surface over xy-plane can be treated as kernal of the<br />
surface integral over actual surface<br />
Now,<br />
S<br />
<br />
<br />
( F)<br />
kˆ<br />
dS<br />
= ( ) ( )<br />
<br />
= [( iˆ)( F ˆj) ( ˆj)( Fiˆ)]<br />
dx dy<br />
<br />
F i j dx dy<br />
[ kˆ iˆ<br />
ˆj]<br />
<br />
S<br />
<br />
<br />
F F dx dy<br />
<br />
= ( y) ( x)<br />
S<br />
S<br />
<br />
x<br />
y<br />
Fx<br />
dx Fy<br />
dy [By Green’s theorem]<br />
= [ ]<br />
=<br />
S<br />
[ ˆ ˆ ] ( ˆ ˆ<br />
iFx<br />
jFy<br />
idx j dy)<br />
S<br />
= F <br />
. dr<br />
<br />
c<br />
<br />
curl F ndS ˆ = . d r .<br />
S<br />
c F<br />
where, F = F iˆ F ˆj F kˆand<br />
dr dx iˆ dyj ˆ dz kˆ<br />
x y z<br />
<br />
<br />
Example 85. Evaluate by Strokes theorem ( yz dx zx dy xy dz)<br />
where C is the curve<br />
C<br />
x 2 + y 2 = 1, z = y 2 . (M.D.U., Dec 2009)<br />
Solution. Here we have<br />
yz dx zx dy xy dz<br />
= ( yziˆ zxj ˆ xykˆ<br />
).( idx ˆ ˆjdy kdz)
438 Vectors<br />
= Fdx .<br />
Curl F =<br />
iˆ<br />
ˆj kˆ<br />
<br />
x y z<br />
yz zx xy<br />
= curl F.<br />
ds<br />
= (x – x) î + (y – y) ĵ + (z – z) ˆk<br />
= 0 = 0 Ans.<br />
Example 86. Using Stoke’s theorem or otherwise, evaluate<br />
2 2<br />
[(2 x y) dx yz dy y z dz]<br />
<br />
c<br />
where c is the circle x 2 + y 2 = 1, corresponding to the surface of sphere of unit radius.<br />
(U.P., I Semester, Winter 2001)<br />
2 2<br />
Solution. [(2 x y) dx yz dy y z dz]<br />
c<br />
2 2 ˆ ˆ<br />
= [(2 x y) iˆ yz ˆj y z k] ( iˆdx ˆjdy k dz)<br />
By Stoke’s theorem<br />
Curl F =<br />
<br />
c<br />
<br />
<br />
<br />
F<br />
d r = Curl F n ds<br />
S<br />
...(1)<br />
iˆ<br />
ˆj kˆ<br />
F <br />
=<br />
Putting the value of curl F in (1), we get<br />
= ˆ<br />
ˆ<br />
= kˆ nˆ<br />
ds<br />
<br />
x y z<br />
2 2<br />
2x y yz y z<br />
= (– 2 yz 2 yz) iˆ<br />
–(0–0) ˆj (0 1)<br />
kˆ kˆ<br />
dx dy<br />
dx dy <br />
k n = dx dy<br />
nk ˆ ˆ = Area of the circle = <br />
ds <br />
( nk ˆ ˆ<br />
<br />
<br />
) <br />
<br />
<br />
Example 87. Evaluate F . d r, where F ( x,<br />
y i xj z kand C is the curve of<br />
C<br />
intersection of the plane y + z = 2 and the cylinder x 2 + y 2 = 1. (Gujarat, I sem. Jan. 2009)<br />
<br />
2 2<br />
Solution. F. dr curl F . nds ˆ curl (– y iˆ<br />
x ˆj z kˆ<br />
<br />
) nds ˆ<br />
...(1)<br />
Normal <strong>vector</strong> = <br />
C S S<br />
2 2 ˆ<br />
F (x, y, z) =<br />
Curl F =<br />
F <br />
2 2<br />
, z) – y ˆ ˆ ˆ<br />
y iˆ xj ˆ z k<br />
(By Stoke’s Theorem)<br />
iˆ<br />
ˆj kˆ<br />
<br />
x y z<br />
2 2<br />
– y x z<br />
= iˆ(0 –0)– ˆj (0 –0) kˆ(12 y) (1<br />
2 y)<br />
kˆ<br />
<br />
= ˆ<br />
<br />
ˆ<br />
ˆ <br />
i j k ( y z – 2) ˆj kˆ<br />
<br />
x y z<br />
ˆj<br />
kˆ<br />
Unit normal <strong>vector</strong> ˆn =<br />
2<br />
dx dy<br />
ds =<br />
ˆ . kˆ<br />
O<br />
3y + z = 2<br />
1<br />
x 2 + y 2 = 1
Vectors 439<br />
On putting the values of curl F,<br />
nˆ<br />
and ds in (1), we get<br />
=<br />
=<br />
=<br />
<br />
F<br />
<br />
. dr<br />
<br />
ˆ ˆ<br />
=<br />
ˆ j k dx dy<br />
(1 2 ) .<br />
C y k<br />
S<br />
2 ˆj<br />
kˆ<br />
<br />
. kˆ<br />
2 <br />
<br />
1<br />
2ydxdy<br />
(1 2 y)<br />
dxdy 2<br />
1<br />
<br />
2 1 = (1 2 r sin ) rd d r<br />
0 <br />
0<br />
Y<br />
2<br />
<br />
<br />
2<br />
1 2<br />
<br />
( r 2r sin ) d d r<br />
0 0<br />
1<br />
2 3<br />
2<br />
2<br />
r<br />
2r<br />
1 2 <br />
d sin sin d <br />
2 3<br />
<br />
2 3<br />
<br />
<br />
0 0<br />
<br />
0<br />
2<br />
<br />
2 2 2 <br />
= – cos – – 0<br />
2 3<br />
<br />
0 3 3<br />
<br />
= Ans.<br />
Example 88. Apply Stoke’s Theorem to find the value of<br />
<br />
c<br />
( ydx z dy xdz)<br />
where c is the curve of intersection of x 2 + y 2 + z 2 = a 2 and x + z = a. (Nagpur, Summer 2001)<br />
<br />
Solution. ( ydx z dy xdz)<br />
c<br />
ˆ<br />
ˆ<br />
= ( ˆ ˆ ) ( ˆ ˆ<br />
yi zj xk idx j dy kdz)<br />
c<br />
=<br />
= curl ( yiˆ zj ˆ xkˆ<br />
) nds ˆ<br />
<br />
S<br />
<br />
C<br />
( yiˆ<br />
zj ˆ xkˆ<br />
) dr<br />
(By Stoke’s Theorem)<br />
ˆ<br />
ˆ<br />
= <br />
ˆ<br />
<br />
ˆ<br />
<br />
i j k ( yiˆ zj ˆ xk)<br />
nˆ<br />
ds<br />
S<br />
<br />
x y z<br />
= ˆ ˆ ˆ<br />
( i j k)<br />
nˆ<br />
ds<br />
S<br />
...(1)<br />
where S is the circle formed by the intersection of x 2 + y 2 + z 2 = a 2 and x + z = a.<br />
<br />
ˆ<br />
<br />
ˆ<br />
ˆ <br />
<br />
i j k ( x z a)<br />
ˆn = | = x y z<br />
|<br />
| |<br />
iˆ<br />
kˆ<br />
ˆn = <br />
2 2<br />
Putting the vlaue of ˆn in (1), we have<br />
ˆ ˆ<br />
= ( ˆ ˆ ˆ<br />
i k <br />
i j k)<br />
ds<br />
S<br />
<br />
2 2<br />
<br />
1 1 <br />
= <br />
ds<br />
S<br />
<br />
2 2 <br />
2 2<br />
2 2<br />
a a<br />
= ds <br />
2<br />
S<br />
<br />
2 2 2<br />
Example 89. Directly or by Stoke’s Theorem, evaluate<br />
=<br />
iˆ<br />
kˆ<br />
1<br />
1<br />
<br />
2 2<br />
2 2 2 2 a a <br />
Use<br />
r R p a <br />
<br />
2 2 <br />
Ans.<br />
ˆ , ˆ ˆ ˆ<br />
curl vnds v iy jz kx,<br />
s is<br />
the surface of the paraboloid z = 1 – x 2 – y 2 , z 3 > 0 and ˆn is the unit <strong>vector</strong> normal to s.<br />
s<br />
<br />
<br />
O<br />
r ddr<br />
X
440 Vectors<br />
Solution.<br />
<br />
v =<br />
iˆ<br />
ˆj kˆ<br />
iˆ<br />
ˆj kˆ<br />
x y z<br />
y z x<br />
Obviously ˆn = k ˆ.<br />
Therefore<br />
<br />
( v)<br />
nˆ<br />
= (– iˆ<br />
– ˆj – kˆ). kˆ<br />
–1<br />
Hence<br />
( v)<br />
nds ˆ = ( 1) dx dy<br />
S<br />
=<br />
S<br />
<br />
S<br />
dx dy<br />
= – (1) 2 = – . (Area of circle = r 2 ) Ans.<br />
Example 90. Use Stoke’s Theorem to evaluate<br />
v ,<br />
c dr<br />
2<br />
where v y iˆ xyj ˆ xzkˆ<br />
, and c<br />
is the bounding curve of the hemisphere x 2 + y 2 + z 2 = 9, z > 0, oriented in the positive<br />
direction.<br />
Solution. By Stoke’s theorem<br />
<br />
( v)<br />
nds ˆ =<br />
S<br />
<br />
v <br />
c<br />
<br />
dr = (curl v) nˆ ds ( v)<br />
nds<br />
ˆ<br />
S S<br />
iˆ<br />
ˆj kˆ<br />
ˆ ˆ<br />
ˆ<br />
(0 0) i ( z 0) j ( y 2 y)<br />
k<br />
v =<br />
x y z zj ˆ ykˆ<br />
2<br />
y xy xz<br />
2 2 2<br />
<br />
i j k ( x y z 9)<br />
ˆn = | = x y z<br />
|<br />
| |<br />
=<br />
2 xiˆ 2 yj ˆ 2 zkˆ xiˆ yj ˆ zkˆ xiˆ yj ˆ zkˆ<br />
<br />
<br />
2 2 2 2 2 2<br />
4x 4y 4 z x y z 3<br />
<br />
ˆ ˆ ˆ<br />
( v)<br />
nˆ<br />
2<br />
= ( ˆ ˆ xi yj zk yz yz yz<br />
zj yk)<br />
<br />
3 3 3<br />
ˆ ˆ ˆ<br />
ˆn<br />
kds ˆ = dx dy <br />
xi yj zk . k ˆ<br />
z<br />
dx = dx dy ds = dx dy<br />
3<br />
3<br />
ds = 3 dx dy<br />
z<br />
<br />
2yz<br />
3<br />
<br />
dx dy = 2 ydxdy<br />
3 z <br />
<br />
=<br />
= 2r sin r d dr<br />
3<br />
3<br />
2 r <br />
0<br />
0<br />
<br />
2<br />
3<br />
<br />
2 sin d r dr<br />
<br />
0 0<br />
= 2( cos )<br />
= – 2 (– 1 + 1) 9 = 0 Ans.<br />
3 <br />
Example 91. Evaluate the surface integral curl F . nˆ<br />
dS by transforming it into a line<br />
S<br />
integral, S being that part of the surface of the paraboloid z = 1 – x 2 – y 2 for which<br />
<br />
z0and<br />
F yiˆ<br />
zj ˆ xkˆ<br />
. (K. University, Dec. 2008)<br />
<br />
2
Vectors 441<br />
Solution.<br />
<br />
<br />
F =<br />
iˆ<br />
ˆj kˆ<br />
iˆ<br />
ˆj kˆ<br />
x y z<br />
y z x<br />
Obviously ˆn = k ˆ.<br />
Therefore<br />
<br />
( F)<br />
nˆ<br />
= (– iˆ<br />
– ˆj – kˆ). kˆ<br />
–1<br />
Hence<br />
( F)<br />
nˆ<br />
ds<br />
= ( 1) dx dy<br />
S<br />
<br />
=<br />
S<br />
<br />
S<br />
dx dy<br />
= – (1) 2 = – . (Area of circle = r 2 ) Ans.<br />
Example 92. Evaluate Fdr<br />
<br />
2 2<br />
by Stoke’s Theorem, where F y iˆ x ˆ j ( x z)<br />
kˆ<br />
and<br />
C<br />
C is the boundary of triangle with vertices at (0, 0, 0), (1, 0, 0) and (1, 1, 0).<br />
(U.P., I Semester, Winter 2000)<br />
Solution. We have, curl F =<br />
=<br />
iˆ<br />
ˆj kˆ<br />
<br />
x y z<br />
2 2<br />
F <br />
0. iˆ ˆj 2( x y) kˆ<br />
.<br />
y x ( x z)<br />
We observe that z co-ordinate of each vertex of the triangle is zero.<br />
Therefore, the triangle lies in the xy-plane.<br />
ˆn = ˆk<br />
<br />
curl Fn ˆ [ ˆj 2( x yk ) ˆ]. kˆ<br />
2( x y).<br />
In the figure, only xy-plane is considered.<br />
The equation of the line OB is y = x<br />
By Stoke’s theorem, we have<br />
C<br />
<br />
Fdr <br />
=<br />
<br />
S<br />
<br />
<br />
(curl Fnˆ<br />
) ds<br />
1 x<br />
= 2( )<br />
2<br />
1<br />
x y dx dy<br />
y <br />
x0 = 2<br />
y 0 xy<br />
dx<br />
0<br />
2 0<br />
2<br />
1 2<br />
= 2 x <br />
2<br />
x<br />
dx<br />
0<br />
2 <br />
=<br />
1 x<br />
1<br />
2 dx = 2<br />
0 2<br />
x dx <br />
3<br />
x <br />
=<br />
0 <br />
3 <br />
x<br />
1<br />
0<br />
= 1 .<br />
3<br />
Example 93. Evaluate Fdr<br />
<br />
<br />
by Stoke’s Theorem, where<br />
2 2<br />
F ( x y ) iˆ 2 xy ˆj<br />
C and C<br />
is the boundary of the rectangle x = a, y = 0 and y = b. (U.P., I Semester, Winter 2002)<br />
Solution. Since the z co-ordinate of each vertex of the given rectangle is zero, hence the given<br />
rectangle must lie in the xy-plane.<br />
Here, the co-ordinates of A, B, C and D are (a, 0), (a, b), (– a, b) and (– a, 0) respectively.<br />
Curl F =<br />
iˆ<br />
ˆj kˆ<br />
<br />
= – 4 y k<br />
x y z<br />
2 2<br />
x y 2xy<br />
0<br />
Ans.
442 Vectors<br />
Here,<br />
nˆ kˆ<br />
, so by Stoke’s theorem, we’ve<br />
=<br />
S<br />
Fdr<br />
<br />
<br />
C<br />
= curl ˆ<br />
S<br />
( 4 ykˆ)()<br />
kˆ<br />
d xdy<br />
=<br />
a<br />
<br />
2<br />
y <br />
= 4 dx<br />
2<br />
=<br />
<br />
a<br />
b<br />
0<br />
F n ds<br />
a<br />
b<br />
4 <br />
xa y 0<br />
a<br />
2 2<br />
<br />
2b dx 4ab<br />
a<br />
Example 94. Apply Stoke’s Theorem to calculate<br />
c<br />
<br />
ydxdy<br />
4ydx 2zdy 6ydz<br />
where c is the curve of intersection of x 2 + y 2 + z 2 = 6 z and z = x + 3.<br />
Solution.<br />
<br />
<br />
<br />
F dr = 4 ydx 2z dy 6 ydz<br />
c c<br />
=<br />
F =<br />
F <br />
=<br />
<br />
c<br />
(4 yiˆ 2zj ˆ 6 ykˆ) ( idx ˆ ˆjdy kdz ˆ )<br />
4 yiˆ<br />
2 zj ˆ 6<br />
ykˆ<br />
iˆ<br />
ˆj kˆ<br />
(6 2) iˆ(0 0) ˆj (0 4) kˆ<br />
x y z 4iˆ<br />
4kˆ<br />
4y 2z 6 y<br />
Ans.<br />
S is the surface of the circle x 2 + y 2 + z 2 = 6z, z = x + 3, ˆn is normal to the plane x – z + 3 = 0<br />
<br />
ˆ ˆ ˆ <br />
i j k ( x z 3)<br />
<br />
ˆn = | = x y z<br />
|<br />
| |<br />
( F)<br />
nˆ<br />
=<br />
ˆ ˆ<br />
(4 ˆ 4 ˆ i k<br />
i k)<br />
= 4 4 = 4 2<br />
2 2<br />
=<br />
iˆkˆ<br />
iˆkˆ<br />
<br />
1<br />
1 2<br />
Fdr<br />
<br />
= (curl ) ˆ<br />
c F nds = 4 2 ( dx dz)<br />
S<br />
S<br />
= 4 2 (area of circle)<br />
Centre of the sphere x 2 + y 2 + (z – 3) 2 = 9, (0, 0, 3) lies on the plane z = x + 3. It means that<br />
the given circle is a great circle of sphere, where radius of the circle is equal to the radius of the<br />
sphere.<br />
Radius of circle = 3, Area = (3) 2 = 9 <br />
( F)<br />
nˆ<br />
ds = 4 2(9 ) 36 2 Ans.<br />
S<br />
Example 95. Verify Stoke’s Theorem for the function<br />
F ziˆ x ˆ j ykˆ<br />
, where C is the unit<br />
2 2<br />
circle in xy-plane bounding the hemisphere z = ( 1x y ). (U.P., I Semester Comp. 2002)<br />
Solution. Here F = ziˆ xj ˆ ykˆ<br />
.<br />
...(1)<br />
Also, r = xiˆ<br />
yj ˆ zkˆ<br />
dr = dxiˆ dyj ˆ dz kˆ<br />
.<br />
<br />
F<br />
dr = z dx + x dy + y dz.<br />
<br />
F<br />
dr = ( zdx x dy ydz).<br />
C C<br />
...(2)
Vectors 443<br />
On the circle C, x 2 + y 2 = 1, z = 0 on the xy-plane. Hence on C, we<br />
have z = 0 so that dz = 0. Hence (2) reduces to<br />
F<br />
dr = xdy .<br />
C C<br />
...(3)<br />
Now the parametric equations of C, i.e., x 2 + y 2 = 1 are<br />
x = cos , y = sin . ...(4)<br />
Using (4), (3) reduces to F <br />
. dr<br />
<br />
C<br />
=<br />
2<br />
cos cos d<br />
=<br />
0 2<br />
<br />
2<br />
0<br />
1 cos 2 <br />
2<br />
1 sin 2 <br />
=<br />
2<br />
<br />
2<br />
= ...(5)<br />
0<br />
Let P(x, y, z) be any point on the surface of the hemisphere x 2 + y 2 + z 2 = 1, O origin is the<br />
centre of the sphere.<br />
Radius = OP =<br />
xiˆ<br />
yj ˆ zkˆ<br />
Normal = xiˆ<br />
yj ˆ zkˆ<br />
ˆn<br />
xiˆ<br />
yj ˆ zkˆ<br />
=<br />
xiˆ yj ˆ zkˆ<br />
2 2 2<br />
x y z<br />
(Radius is to tangent i.e. Radius is normal)<br />
x = sin cos , y = sin sin , z = cos ...(6)<br />
ˆn = sin cos î + sin sin ĵ + cos ˆk<br />
iˆ<br />
ˆj kˆ<br />
Also, Curl F = / x / y /<br />
z iˆ<br />
ˆj kˆ<br />
...(7)<br />
z x y<br />
<br />
=<br />
=<br />
d <br />
<br />
Curl Fn ˆ = ( iˆ ˆj kˆ).(sin cos iˆsin sin ˆj sin kˆ)<br />
= sin cos + sin sin + cos <br />
/2 2<br />
Curl Fn ˆ dS = ( ˆ ˆ ˆ<br />
i j k)<br />
0<br />
<br />
0<br />
s<br />
=<br />
<br />
. (sin cos î + sin sin ĵ + cos ˆk ) sin d d<br />
/2 2<br />
<br />
sin d (sin cos sin sin cos )<br />
d<br />
0 0<br />
[ dS = Elementary area on hemisphere = sin d d]<br />
/2<br />
2<br />
sin d<br />
[sin sin sin ( cos ) cos ]<br />
0<br />
0 =<br />
/2<br />
(0 0 2 sin cos ) d =<br />
0<br />
= – (/2) [– 1 – 1] = .<br />
From (5) and (8),<br />
C<br />
<br />
<br />
Fdr =<br />
/2<br />
sin 2 d<br />
=<br />
0<br />
<br />
/2<br />
0<br />
<br />
<br />
<br />
sin d<br />
<br />
cos 2<br />
2<br />
<br />
<br />
<br />
/2<br />
curl F<br />
ndS ˆ , which verifies Stokes’s theorem.<br />
S<br />
Example 96. Verify Stoke’s theorem for the <strong>vector</strong> field F (2 x – y) iˆ<br />
– yz ˆj – y zk over<br />
0<br />
2 2 ˆ<br />
the upper half of the surface x 2 + y 2 + z 2 = 1 bounded by its projection on xy- plane.<br />
(Nagpur University, Summer 2001)<br />
Solution. Let S be the upper half surface of the sphere x 2 + y 2 + z 2 = 1. The boundary C or S<br />
is a circle in the xy plane of radius unity and centre O. The equation of C are x 2 + y 2 = 1,<br />
z = 0 whose parametric form is<br />
x = cos t, y = sin t, z = 0, 0 < t < 2<br />
<br />
<br />
F . dr =<br />
C<br />
<br />
C<br />
2 2<br />
[(2 x – y) iˆ – yz ˆj – y zkˆ].[ iˆ dx ˆjdy kdz ˆ ]
444 Vectors<br />
2 2<br />
= [(2 x – y) dx – yz dy – y z dz]<br />
C<br />
= (2 x – y) dx,<br />
C<br />
since on C, z = 0 and 2z = 0<br />
2<br />
dx 2<br />
= (2cos t –sin t) dt (2cos t –sin t)(–sin t)<br />
dt<br />
0 dt<br />
0<br />
2 2<br />
2 1–cos 2t<br />
<br />
= (– sin 2t sin t) dt – sin 2t dt<br />
0 <br />
0<br />
<br />
<br />
<br />
2 <br />
2<br />
cos 2t t sin 2t<br />
1 1<br />
= – –<br />
2 2 4<br />
<br />
<br />
0 2 2<br />
iˆ<br />
ˆj kˆ<br />
...(1)<br />
<br />
Curl F =<br />
<br />
x y z = (– 2yz 2 yz) iˆ<br />
(0–0) ˆj (01)<br />
kˆ kˆ<br />
2 x – y<br />
2<br />
– yz<br />
2<br />
– y z<br />
<br />
Z<br />
Curl F . nˆ<br />
= kˆ. nˆ<br />
nk ˆ.<br />
ˆ<br />
<br />
Curl F.<br />
nds ˆ = ˆ. ˆ ˆ. ˆ dx dy<br />
. .<br />
S nkds nk<br />
S R ˆ ˆ<br />
Where R is the projection of S on xy-plane.<br />
O<br />
Y<br />
=<br />
2<br />
1 1– x<br />
dx dy =<br />
2<br />
–1 – 1– x<br />
<br />
1 2 1 2<br />
2 1– x dx 4 1– x dx<br />
–1 0<br />
1<br />
2 –1<br />
x<br />
1 1<br />
<br />
= 4 1– x sin x 4 .<br />
2 2<br />
<br />
0 2 2<br />
<br />
<br />
From (1) and (2), we have<br />
<br />
<br />
F . dr = Curl F . nds ˆ which is the Stoke's theorem.<br />
C <br />
<br />
Example 97. Verify Stoke’s Theorem for 2 ˆ ˆ<br />
2<br />
F ( x y 4) i 3 xyj (2 xz z ) kˆ<br />
over the surface of hemisphere x 2 + y 2 + z 2 = 16 above the xy-plane.<br />
<br />
Solution. Fdr , where c is the boundary of the circle x 2 + y 2 + z 2 = 16<br />
c<br />
(bounding the hemispherical surface)<br />
2 2 ˆ<br />
= [( x y 4) iˆ 3 xyj ˆ (2 xz z ) k]( idx ˆ ˆjdy)<br />
Putting<br />
<br />
<br />
c<br />
2<br />
= [( x y 4) dx 3 xy dy)]<br />
c<br />
x = 4 cos , y = 4 sin , dx = – 4 sin d , dy = 4 cos d <br />
=<br />
<br />
2<br />
2 2<br />
0<br />
<br />
[(16 cos 4sin 4) ( 4sin d ) (192 sin cos d )]<br />
2 2 2 2<br />
= 16 [ 4cos sin sin sin 12 sin cos ] d <br />
<br />
<br />
0<br />
2<br />
2 2<br />
= 16 (8 sin cos sin sin ) d <br />
=<br />
=<br />
0<br />
<br />
2<br />
16 sin d<br />
<br />
0<br />
2<br />
<br />
2 2<br />
=<br />
0<br />
16 4 sin d<br />
<br />
To evaluate surface integral<br />
F <br />
=<br />
1<br />
<br />
64 = – 16 .<br />
22 <br />
iˆ<br />
ˆj kˆ<br />
<br />
x y z<br />
X<br />
<br />
<br />
<br />
<br />
<br />
2 2<br />
x y 4 3xy 2 xz z<br />
<br />
<br />
2<br />
0<br />
2<br />
0<br />
C<br />
...(2)<br />
Ans.<br />
n<br />
sin cos d0<br />
<br />
<br />
<br />
n<br />
cos sin d<br />
0
Vectors 445<br />
= (0 – 0) iˆ – (2 z – 0) ˆj (3 y – 1) kˆ – 2 zj ˆ (3 y – 1) kˆ<br />
<br />
ˆ ˆ ˆ 2 2 2<br />
<br />
i j k ( x y z 16)<br />
ˆn = | = x y z<br />
|<br />
| |<br />
2xiˆ<br />
2yj ˆ 2zkˆ<br />
xiˆ<br />
yj ˆ zkˆ<br />
xiˆ<br />
yj ˆ zkˆ<br />
=<br />
=<br />
=<br />
2 2 2 2 2 2<br />
4x 4y 4z<br />
x y z 4<br />
<br />
ˆ ˆ ˆ<br />
( F)<br />
nˆ<br />
= [– 2 ˆ (3 – 1) ˆ xi yj zk<br />
zj y k]<br />
<br />
2 yz (3y 1) z<br />
=<br />
4<br />
4<br />
ˆk n<br />
xiˆ<br />
yj ˆ zkˆ<br />
z<br />
ds = dx dy <br />
. k ds = dx dy ds = dx dy<br />
4<br />
4<br />
<br />
ds = 4 dx dy<br />
z<br />
2 (3 1) 4<br />
( F)<br />
nˆ<br />
ds yz y z dx dy<br />
<br />
<br />
4 z <br />
= [ 2 y (3 y 1)] dx dy = ( y 1) dxdy<br />
On putting x = r cos , y = r sin , dx dy = r d dr, we get<br />
=<br />
=<br />
= ( r sin 1) r d dr<br />
2<br />
3 2<br />
4<br />
<br />
<br />
2<br />
d ( r sin r)<br />
dr<br />
= r<br />
r <br />
d <br />
<br />
sin <br />
3 2 = 64 <br />
d <br />
sin 8 <br />
0 <br />
<br />
3<br />
0 0<br />
<br />
2<br />
64 64 64<br />
= cos 8 <br />
= 16<br />
= – 16 <br />
3 3 3<br />
0<br />
The line integral is equal to the surface integral, hence Stoke’s Theorem is verified. Proved.<br />
<br />
2 2<br />
Example 98. Verify Stoke’s theorem for a <strong>vector</strong> field defined by F ( x – y ) iˆ 2xy ˆj<br />
in<br />
the rectangular in xy-plane bounded by lines x = 0, x = a, y = 0, y = b.<br />
(Nagpur University, Summer 2000)<br />
Solution. Here we have to verify Stoke’s theorem F <br />
. dr<br />
<br />
= ( ). ˆ<br />
C F<br />
nds<br />
S<br />
Where ‘C’ be the boundary of rectangle (ABCD) and S be the surface enclosed by curve C.<br />
2<br />
F 2 2<br />
= ( x – y ) iˆ (2 xy)<br />
ˆj<br />
<br />
F . dr =<br />
2 2<br />
[( x – y ) iˆ 2 xyj ˆ].[ idx ˆ ˆj dy]<br />
<br />
F . dr = (x<br />
2<br />
+ y 2 ) dx + 2xy dy ...(1)<br />
Now, F <br />
. dr<br />
<br />
<br />
= F . dr F . dr F . dr F . dr<br />
C<br />
...(2)<br />
OA AB BC CO<br />
<br />
Along OA, put y = 0 so that k dy = 0 in (1) and F . dr = x 2 dx,<br />
Where x is from 0 to a.<br />
3<br />
a<br />
a<br />
2<br />
F <br />
. dr<br />
<br />
3<br />
x dx<br />
x a<br />
OA = <br />
0 3 3<br />
<br />
Along AB, put x = a so that dx = 0 in (1), we get F .<br />
Where y is from 0 to b.<br />
<br />
b<br />
0<br />
0<br />
<br />
d r<br />
= 2ay dy<br />
...(3)<br />
2 2<br />
F . dr 2aydy<br />
= [ ay ] b<br />
0<br />
ab<br />
...(4)<br />
AB
446 Vectors<br />
Along BC, put y = b and dy = 0 in (1) we get F . dr = (x 2 – b 2 ) dx,<br />
where x is from a to 0.<br />
F<br />
<br />
BC<br />
. dr =<br />
<br />
0<br />
3<br />
0<br />
3<br />
2 2 x 2<br />
– a 2<br />
Along CO, put x = 0 and dx = 0 in (1), we get F . dr 0<br />
( x – b ) dx – bx<br />
ba<br />
a<br />
3 3<br />
...(5)<br />
<br />
<br />
F <br />
. dr<br />
<br />
= 0 ...(6)<br />
CO<br />
Putting the values of integrals (3), (4), (5) and (6) in (2),<br />
we get<br />
<br />
<br />
F . dr =<br />
C<br />
3 3<br />
2 a 2 2<br />
a<br />
ab – ab 0 2ab<br />
...(7)<br />
3 3<br />
Now we have to evaluate R.H.S. of Stoke’s Theorem i.e.<br />
We have,<br />
F <br />
=<br />
iˆ<br />
ˆj kˆ<br />
<br />
x y z<br />
2 2<br />
x – y 2xy<br />
0<br />
<br />
S<br />
a<br />
<br />
( F).<br />
nds ˆ<br />
(2y 2 y) kˆ<br />
4ykˆ<br />
Also the unit <strong>vector</strong> normal to the surface S in outward direction is ˆn<br />
k<br />
(z-axis is normal to surface S)<br />
Also in xy-plane ds = dx dy<br />
<br />
( F). nˆ<br />
. ds = 4 ˆ. ˆ 4 .<br />
S yk kdxdy ydx dy<br />
R R<br />
Where R be the region of the surface S.<br />
Consider a strip parallel to y-axis. This strip starts on line y = 0 (i.e. x-axis) and end on the line<br />
y = b, We move this strip from x = 0 (y-axis) to x = a to cover complete region R.<br />
<br />
<br />
( F). nˆ<br />
. ds =<br />
S<br />
<br />
From (7) and (8), we get<br />
<br />
<br />
=<br />
<br />
<br />
ydy<br />
<br />
dx y dx<br />
<br />
a b 2<br />
4 <br />
a [2 ]<br />
b<br />
0 0 0<br />
0<br />
<br />
a 2 2 a 2<br />
b dx b x<br />
0<br />
ab<br />
...(8)<br />
0 2 2 [ ] 2<br />
F . dr = ( ). ˆ<br />
C F nds and hence the Stoke’s theorem is verified.<br />
S<br />
Example 99. Verify Stoke’s Theorem for the function<br />
<br />
F = xi<br />
2ˆ – xyj ˆ<br />
integrated round the square in the plane z = 0 and bounded by the lines<br />
x = 0, y = 0, x = a, y = a.<br />
Solution. We have, F = xi<br />
2ˆ – xyj ˆ<br />
<br />
F =<br />
iˆ<br />
ˆj kˆ<br />
<br />
x y z<br />
2<br />
x = 0<br />
Y<br />
y = b<br />
C<br />
B<br />
(a, b)<br />
x = a<br />
x xy 0<br />
= (0 –0) iˆ<br />
–(0–0) ˆj (– y –0) kˆ – ykˆ<br />
( ˆn to xy plane i.e. ˆk )<br />
O<br />
y = 0<br />
A<br />
X
Vectors 447<br />
( F)<br />
nˆ<br />
ds = ( yk)<br />
kdxdy<br />
S<br />
S<br />
<br />
=<br />
<br />
To obtain line integral<br />
a<br />
a<br />
dx<br />
ydy =<br />
0 0<br />
a<br />
a<br />
2<br />
y<br />
dx <br />
<br />
<br />
<br />
2 <br />
=<br />
0 0<br />
2<br />
a a<br />
( x) 0 =<br />
2<br />
<br />
F dr 2<br />
= ˆ ˆ ˆ ˆ<br />
2<br />
( x i xyj) ( idx j dy)<br />
= ( x dx xydy)<br />
C<br />
C<br />
C<br />
where c is the path OABCO as shown in the figure.<br />
Also,<br />
<br />
F dr =<br />
C<br />
<br />
Fdr =<br />
OABCO<br />
Along OA, y = 0, dy = 0<br />
F<br />
dr<br />
<br />
=<br />
OA<br />
<br />
2<br />
( xdx<br />
xydy )<br />
OA<br />
a <br />
3<br />
2 x <br />
= xdx = =<br />
0<br />
3 0<br />
Along AB, x = a, dx = 0<br />
Fdr<br />
<br />
2<br />
= ( x dx xyd y)<br />
AB AB<br />
=<br />
a<br />
aydy =<br />
Along BC, y = a, dy = 0<br />
<br />
<br />
0<br />
<br />
Fdr Fdr Fdr F<br />
dr<br />
OA AB BC CO<br />
3<br />
a<br />
...(1)<br />
2<br />
...(2)<br />
a<br />
3<br />
a<br />
3<br />
2<br />
y<br />
a <br />
<br />
<br />
2 <br />
F dr<br />
2<br />
= ( xdx xydy)<br />
=<br />
BC<br />
BC<br />
Along CO, x = 0, dx = 0<br />
<br />
a<br />
0<br />
=<br />
3<br />
a<br />
<br />
2<br />
0 2<br />
xdx<br />
a<br />
=<br />
Fdr <br />
2<br />
= ( )<br />
CO xdx xydy = 0<br />
CO<br />
Putting the values of these integrals in (2), we have<br />
Fdr<br />
<br />
=<br />
C<br />
From (1) and (3),<br />
3 3 3<br />
a a a<br />
0 =<br />
3 2 3<br />
( F)<br />
nˆ<br />
ds =<br />
Hence, Stoke’s Theorem is verified.<br />
S<br />
<br />
<br />
<br />
C<br />
<br />
3<br />
<br />
3<br />
x <br />
<br />
<br />
0<br />
3 a<br />
=<br />
3<br />
a<br />
<br />
3<br />
a<br />
...(3)<br />
2<br />
<br />
F dr<br />
Ans.<br />
Example 100. Verify Stoke’s Theorem for F<br />
<br />
= (x + y) î + (2x – z) ĵ + (y + z) ˆk for the<br />
surface of a triangular lamina with vertices (2, 0, 0), (0, 3, 0) and (0, 0, 6).<br />
(Nagpur University 2004, K. U. Dec. 2009, 2008, A.M.I.E.T.E., Summer 2000)<br />
Solution. Here the path of integration c consists of the straight lines AB, BC, CA where the<br />
co-ordinates of A, B, C and (2, 0, 0), (0, 3, 0) and (0, 0, 6) respectively. Let S be the plane<br />
surface of triangle ABC bounded by C. Let ˆn be unit normal <strong>vector</strong> to surface S. Then by<br />
Stoke’s Theorem, we must have<br />
Fdr<br />
<br />
=<br />
c<br />
curl Fnds ˆ<br />
...(1)<br />
s<br />
<br />
<br />
line Eq. of Lower Upper<br />
line limit limit<br />
OA y = 0 dy = 0 x = 0 x = a<br />
AB x = a dx = 0 y = 0 y = a<br />
BC y = a dy= 0 x = a x = 0<br />
CO x = 0 dx = 0 y = a y = 0
448 Vectors<br />
c<br />
L.H.S. of (1)= Fdr<br />
<br />
= Fdr F dr F dr<br />
ABC AB <br />
BC CA<br />
x y<br />
Along line AB, z = 0, equation of AB is = 1 2 3<br />
y = 3 3<br />
(2 x),<br />
dy =<br />
2<br />
2 dx<br />
At A, x = 2, At B, x = 0, r = xiˆ<br />
yj ˆ<br />
<br />
<br />
F dr =<br />
AB<br />
<br />
<br />
AB<br />
[( x y) iˆ 2 xj ˆ ykˆ<br />
]( idx ˆ ˆjdy)<br />
= ( x y) dx 2xdy<br />
=<br />
<br />
AB<br />
AB<br />
3x<br />
3 <br />
x 3 dx 2x<br />
dx<br />
2 2 <br />
2<br />
0<br />
7x<br />
7x<br />
<br />
= 3 dx<br />
3x<br />
2<br />
<br />
2 <br />
4 <br />
<br />
= (7 – 6) = + 1<br />
y z<br />
Along line BC, x = 0, Equation of BC is = 1 or z = 6 – 2y, dz = – 2dy<br />
3 6<br />
At B, y = 3, At C, y = 0, r =<br />
<br />
<br />
yj ˆ zkˆ<br />
F dr = [ ( ) ] ( )<br />
BC<br />
yi zj y z k jdy kdz = ( )<br />
BC zdy y z dz<br />
BC<br />
0<br />
= ( 6 2 y) dy ( y 6 2 y)( 2 dy)<br />
3<br />
0 2 0<br />
= (4y 18) dy (2y 18 y)<br />
3<br />
3 = 36<br />
x z<br />
Along line CA, y = 0, Eq. of CA, = 1 or z = 6 – 3x, dz = – 3dx<br />
2 6<br />
At C, x = 0, at A, x = 2, r = xiˆ<br />
zkˆ<br />
<br />
<br />
F dr = [ xiˆ (2 x z) ˆj zkˆ][ dxiˆ<br />
dzkˆ]<br />
CA = ( xdx zdz)<br />
CA<br />
=<br />
line Eq. of Lower Upper<br />
line limit limit<br />
AB<br />
x y 3 At A At B<br />
1<br />
dy – dx<br />
2 3 2 x 2 x 0<br />
z = 0<br />
BC<br />
y z<br />
At B At C<br />
1<br />
dz = – 2dy<br />
3 6<br />
y 3 y 0<br />
x = 0<br />
CA<br />
x z<br />
At C At A<br />
1<br />
dz = – 3dx<br />
2 6<br />
x 0 x 2<br />
y = 0<br />
CA<br />
2<br />
xdx (6 3 x )( 3 dx ) =<br />
0<br />
2<br />
0<br />
0<br />
2<br />
<br />
(10 x 18) dx 2 2<br />
= [5x<br />
18 x]<br />
0 = – 16
Vectors 449<br />
c<br />
L.H.S. of (1) =<br />
Fdr<br />
<br />
<br />
ABC<br />
=<br />
<br />
Fdr F dr F dr<br />
= 1 + 36 – 16 = 21 ...(2)<br />
AB BC CA<br />
<br />
Curl F = F =<br />
ˆ<br />
<br />
ˆ<br />
ˆ <br />
i j k [( x y) iˆ (2 x z) ˆj ( y z) k]<br />
x y z<br />
iˆ<br />
ˆj kˆ<br />
=<br />
<br />
x y z<br />
x y 2x z y z<br />
x y z<br />
Equation of the plane of ABC is = 1<br />
2 3 6<br />
Normal to the plane ABC is<br />
(1 1) iˆ(0 0) ˆj (2 1) kˆ<br />
2iˆ<br />
kˆ<br />
<br />
ˆ<br />
=<br />
ˆ<br />
<br />
ˆ<br />
x y z <br />
i j k 1<br />
=<br />
x y z2 3 6 <br />
iˆ<br />
ˆj kˆ<br />
<br />
Unit Normal Vector =<br />
2 3 6<br />
1 1 1<br />
<br />
4 9 36<br />
1<br />
ˆn = (3iˆ<br />
2 ˆj kˆ<br />
)<br />
14<br />
i ˆ ˆ ˆ<br />
<br />
j <br />
k<br />
2 3 6<br />
<br />
R.H.S. of (1) = curl Fnds <br />
= ˆ ˆ 1<br />
ˆ ˆ ˆ dx dy<br />
(2 ) (3 2 )<br />
s i k i j k<br />
s<br />
4 1<br />
(3iˆ<br />
2 ˆj kˆ).<br />
kˆ<br />
14<br />
(6 1) dx dy<br />
= = 7<br />
s<br />
14 1 dx dy = 7 Area of OAB<br />
14<br />
1<br />
<br />
= 7<br />
23<br />
= 21 ...(3)<br />
2<br />
<br />
with the help of (2) and (3) we find (1) is true and so Stoke’s Theorem is verified.<br />
Example 101. Verify Stoke’s Theorem for<br />
<br />
F<br />
= (y – z + 2) î + (yz + 4) ĵ – (xz) ˆk<br />
over the surface of a cube x = 0, y = 0, z = 0, x = 2, y = 2, z = 2 above the XOY plane<br />
(open the bottom).<br />
Solution. Consider the surface of the cube as shown in the figure. Bounding path is OABCO<br />
shown by arrows.<br />
<br />
Fdr = [( y z 2) iˆ ( yz 4) ˆj ( xz) kˆ] ( idx ˆ ˆjdy kdz ˆ )<br />
<br />
<br />
= ( y z 2) dx ( yz 4) dy xzdz<br />
c<br />
<br />
<br />
Fdr = Fdr Fdr Fdr Fdr<br />
c<br />
OA AB BC CO<br />
(1) Along OA, y = 0, dy = 0, z = 0, dz = 0<br />
...(1)
450 Vectors<br />
Line Equ. Lower Upper Fdr .<br />
1 OA<br />
2 AB<br />
3 BC<br />
4 CO<br />
of line limit limit<br />
y 0 dy 0<br />
z 0 dz 0<br />
x = 0 x = 2 2 dx<br />
x 2 dx 0<br />
z 0 dz 0<br />
y = 0 y = 2 4 dy<br />
y 2 dy 0<br />
z 0 dz 0<br />
x = 2 x = 0 4 dx<br />
x 0 dx 0<br />
z 0 dz 0<br />
y = 2 y = 0 4 dy<br />
Fdr =<br />
OA<br />
(2) Along AB, x = 2, dx = 0, z = 0, dz = 0<br />
AB<br />
<br />
<br />
<br />
<br />
Fdr =<br />
(3) Along BC, y = 2, dy = 0, z = 0, dz = 0<br />
F dr =<br />
BC<br />
(4) Along CO, x = 0, dx = 0, z = 0, dz = 0<br />
CO<br />
<br />
<br />
<br />
<br />
2<br />
2<br />
2 dx [2 x]<br />
0 = 4<br />
0<br />
2<br />
2<br />
4dy<br />
4( y)<br />
0 = 8<br />
0<br />
2<br />
0<br />
(2 0 2) dx (4 x)<br />
2 = – 8<br />
0<br />
Fdr = ( y 0 2) 0 (0 4) dy 0<br />
0<br />
= 4 dy 4( y) 2 = – 8<br />
On putting the values of these integrals in (1), we get<br />
Fdr = 4 + 8 – 8 – 8 = – 4<br />
c<br />
<br />
<br />
To obtain surface integral<br />
iˆ<br />
ˆj kˆ<br />
F <br />
=<br />
<br />
x y z<br />
y z 2 yz 4 xz<br />
= (0 – y) î – (– z + 1) ĵ + (0 – 1) ˆk = – y î + (z – 1) ĵ – ˆk<br />
Here we have to integrate over the five surfaces, ABDE, OCGF, BCGD, OAEF, DEFG.<br />
Over the surface ABDE (x = 2), ˆn = i, ds = dy dz<br />
<br />
( F)<br />
nds ˆ = [ yi ( z 1) j k]<br />
idx dz ydydz<br />
<br />
2<br />
= <br />
<br />
R<br />
( , )<br />
3 ( , , ) z f x y<br />
z f ( x, y)<br />
F x y z dx dy<br />
1
Vectors 451<br />
Surface Outward normal<br />
ds<br />
1 ABDE i dy dz x = 2<br />
2 OCGF – i dy dz x = 0<br />
3 BCGD j dx dz y = 2<br />
4 OAEF – j dx dz y = 0<br />
5 DEFG k dx dy z = 2<br />
=<br />
2 2 2<br />
y<br />
2<br />
ydydz [ z] 4<br />
0<br />
2 <br />
0 0 0<br />
2<br />
Over the surface OCGF (x = 0), ˆn = – i, ds = dy dz<br />
<br />
( F)<br />
nds ˆ = [ ˆ ( 1) ˆ ˆ] ( ˆ<br />
yi z j k i)<br />
dy dz<br />
=<br />
2 2 2<br />
y<br />
<br />
ydydz y dy dz 2<br />
4<br />
2 <br />
<br />
0 0 0<br />
2<br />
(3) Over the surface BCGD, (y = 2), ˆn = j, ds = dx dz<br />
<br />
( F)<br />
nds ˆ = [ yiˆ ( z 1) ˆj kˆ<br />
] ˆ<br />
j dx dz<br />
=<br />
= ( z 1)<br />
dxdz<br />
2 2<br />
dx ( z 1) dz =<br />
0 0<br />
( x)<br />
2<br />
0<br />
<br />
2<br />
z<br />
<br />
2<br />
2<br />
<br />
z<br />
<br />
<br />
0<br />
= 0<br />
(4) Over the surface OAEF, (y = 0), ˆn = – ĵ , ds = dx dz<br />
<br />
( F ) nˆ<br />
ds = ˆ<br />
<br />
[ yiˆ ( z 1) ˆj k]( ˆj)<br />
dx dz<br />
= ( z 1)<br />
dxdz = –<br />
2 2<br />
dx ( z 1) dz =<br />
0 0<br />
(5) Over the surface DEFG, (z = 2), ˆn = k, ds = dx dy<br />
( x)<br />
2<br />
0<br />
<br />
2<br />
z<br />
<br />
2<br />
2<br />
<br />
z<br />
<br />
<br />
0<br />
= 0<br />
<br />
( F)<br />
nds ˆ = [ yi ˆ ( z 1) ˆ j k ˆ]<br />
k ˆ<br />
dx dy = – dx dy<br />
= –<br />
2 2<br />
dx<br />
dy 2 2<br />
= [ x] [ y]<br />
= – 4<br />
0 0<br />
0 0<br />
Total surface integral = – 4 + 4 + 0 + 0 – 4 = – 4<br />
Thus<br />
curl F nds ˆ =<br />
S<br />
<br />
<br />
F dr = – 4<br />
c<br />
<br />
which verifies Stoke’s Theorem.<br />
<br />
Ans.
452 Vectors<br />
1. Use the Stoke’s Theorem to evaluate<br />
EXERCISE 5.14<br />
<br />
2<br />
ydx<br />
xydy xzdz,<br />
C<br />
where C is the bounding curve of the hemisphere x 2 + y 2 + z 2 = 1, z 0, oriented in the positive<br />
direction. Ans. 0<br />
2. Evaluate<br />
(curl F) ndA ˆ , using the Stoke’s Theorem, where F yiˆ<br />
zj ˆ xkˆ<br />
and s is the paraboloid<br />
s<br />
z = f (x, y) = 1 – x 2 – y 2 , z 0.<br />
Ans. <br />
<br />
2 2 2<br />
3. Evaluate the integral for ydx z dy xdz, where C is the triangular closed path joining the points<br />
C<br />
(0, 0, 0), (0, a, 0) and (0, 0, a) by transforming the integral to surface integral using Stoke’s Theorem.<br />
<br />
2<br />
4. Verify Stoke’s Theorem for A 3 yiˆ<br />
xzj ˆ yz kˆ<br />
, where S is the surface of the paraboloid 2z = x<br />
2<br />
+ y 2<br />
bounded by z = 2 and c is its boundary traversed in the clockwise direction.<br />
Ans. – 20 <br />
5. Evaluate<br />
F<br />
<br />
d R where<br />
C<br />
Ans.<br />
3<br />
a<br />
.<br />
3<br />
<br />
ˆ 3<br />
F yi xz ˆ j zy<br />
3ˆ k,<br />
C is the circl x 2 + y2 = 4, z = 1.5 Ans. 19 2 <br />
6. If S is the surface of the sphere x 2 + y 2 + z 2 <br />
= 9. Prove that curl F ds = 0.<br />
S<br />
7. Verify Stoke’s Theorem for the <strong>vector</strong> field<br />
<br />
F (2 y z) iˆ<br />
( x – z) ˆj ( y – x)<br />
kˆ<br />
over the portion of the plane x + y + z = 1 cut off by the co-ordinate planes.<br />
8. Evaluate <br />
dr by Stoke’s Theorem for F yziˆ zx ˆ j xyk and C is the curve of intersection of<br />
c<br />
x 2 + y 2 = 1 and y = z 2 . Ans. 0<br />
9.<br />
<br />
If ˆ 3 ˆ 2<br />
F ( x – z) i ( x yz) j 3xy kˆ<br />
and S is the surface of the cone z = a – 2 2<br />
( x y ) above the<br />
xy-plane, show that<br />
<br />
curl FdS = 3 a 4 / 4.<br />
s<br />
<br />
10. If F 3yiˆ<br />
xyj ˆ yz2kˆ<br />
and S is the surface of the paraboloid 2z = x 2 + y 2 bounded by z = 2, show by<br />
<br />
using Stoke’s Theorem that ( F)<br />
dS = 20 .<br />
s<br />
<br />
2 2 2 2 2 2 2 2 2<br />
11. If F ( y z – x ) iˆ<br />
( z x – y ) ˆj ( x y – z ) kˆ<br />
, evaluate curl F nds ˆ integrated over<br />
the portion of the surface x 2 + y 2 – 2ax + az = 0 above the plane z = 0 and verify Stoke’s Theorem; where<br />
ˆn is unit <strong>vector</strong> normal to the surface. (A.M.I.E.T.E., Winter 20002) Ans. 2 a 3<br />
where C is the boundary of<br />
C<br />
12. Evaluate by using Stoke’s Theorem sin zdx cos x dy sin<br />
ydz<br />
rectangle 0 x, 0 y1, z 3 . (AMIETE, June 2010)<br />
5.40 GAUSS’S THEOREM OF DIVERGENCE<br />
(Relation between surface integral and volume integral)<br />
(U.P., Ist Semester, Jan., 2011, Dec, 2006)<br />
Statement. The surface integral of the normal component of a <strong>vector</strong> function F taken around<br />
a closed surface S is equal to the integral of the divergence of F taken over the volume V enclosed<br />
by the surface S.<br />
Mathematically<br />
<br />
S<br />
<br />
F.<br />
nˆ<br />
ds <br />
<br />
V<br />
<br />
div Fdw
Vectors 453<br />
<br />
Proof. Let F Fi ˆ ˆ ˆ<br />
1 F2 j F3 k.<br />
<br />
Putting the values of F,<br />
nˆ<br />
in the statement of the divergence theorem, we have<br />
Fi ˆ ˆ<br />
1 F ˆ 2 j F3 k <br />
nds ˆ<br />
S<br />
= ˆ<br />
<br />
ˆ k ˆ <br />
( ˆ<br />
V F 1 F 2<br />
x y z<br />
ˆ F ˆ<br />
3 ) .<br />
1 2 3<br />
= F<br />
F<br />
F<br />
<br />
dx dy dz<br />
V<br />
<br />
x y z<br />
<br />
We require to prove (1).<br />
...(1)<br />
F3<br />
Let us first evaluate dx dy dz .<br />
V z<br />
V<br />
<br />
z<br />
F3<br />
dx dy dz<br />
=<br />
z f2 ( x, y)<br />
F3<br />
<br />
R<br />
dz dx dy<br />
z<br />
f1( x, y)<br />
z<br />
<br />
<br />
<br />
[ F ( x, y, f ) F ( x, y, f )] dx dy<br />
...(2)<br />
=<br />
R<br />
3 2 3 1<br />
For the upper part of the surface i.e. S 2<br />
, we have<br />
dx dy = ds 2<br />
cos r 2<br />
= ˆn 2<br />
. ˆk ds 2<br />
Again for the lower part of the surface i.e. S 1<br />
, we have,<br />
R<br />
dx dy = – cos r 1<br />
, ds 1<br />
= ˆn 1<br />
. ˆk ds 1<br />
F3 ( x, y, f2)<br />
dx dy ˆ ˆ<br />
F n kds<br />
S<br />
= 3 2 2<br />
= 3 1 1<br />
and F3 ( x, y, f1)<br />
dx dy ˆ ˆ<br />
R F n kds<br />
S1<br />
Putting these values in (2), we have<br />
F3<br />
dv F nˆ<br />
kds ˆ F nˆ<br />
kˆ<br />
ds<br />
V z<br />
S2 S1<br />
Similarly, it can be shown that<br />
F2<br />
dv<br />
ˆ<br />
V<br />
<br />
ˆ<br />
y<br />
F n<br />
jds<br />
S<br />
...(4)<br />
= 2<br />
2<br />
= 3 2 2 3 1 1 = 3<br />
F1<br />
dv =<br />
ˆ<br />
V<br />
1<br />
ˆ<br />
x<br />
F ni<br />
ds<br />
S<br />
...(5)<br />
Adding (3), (4) & (5), we have<br />
<br />
V<br />
F1 F2<br />
F3<br />
<br />
dv<br />
x y z<br />
<br />
= ( Fi ˆ ˆ ˆ<br />
1 F2 j F3 k)<br />
nˆ<br />
ds<br />
<br />
( F)<br />
dv<br />
V<br />
S<br />
= ˆ<br />
F nds<br />
Proved.<br />
S<br />
Example 102. State Gauss’s Divergence theorem<br />
<br />
ˆ ˆ<br />
F n<br />
kds ...(3)<br />
S<br />
F . nds ˆ Div Fdv<br />
S <br />
where S is the<br />
<br />
surface of the sphere x 2 + y 2 + z 2 <br />
= 16 and F 3xiˆ<br />
4y ˆj 5 zkˆ<br />
.<br />
(Nagpur University, Winter 2004)<br />
Solution. Statement of Gauss’s Divergence theorem is given in Art 24.8 on page 597.<br />
Thus by Gauss’s divergence theorem,<br />
<br />
= <br />
. Here F 3xiˆ<br />
4y ˆj 5zkˆ<br />
F . nds ˆ<br />
S<br />
v<br />
<br />
Fdv
454 Vectors<br />
. F <br />
=<br />
Putting the value of . F, we get<br />
Example 103. Evaluate<br />
<br />
<br />
ˆ<br />
<br />
ˆ<br />
ˆ <br />
i j k .(3xiˆ 4yj ˆ 5 zkˆ<br />
)<br />
x y z<br />
. F <br />
= 3 + 4 + 5 = 14<br />
= 14 .<br />
v<br />
F . nds ˆ<br />
S<br />
<br />
S<br />
dv<br />
where v is volume of a sphere<br />
= 14 v<br />
4 3 3584 <br />
= 14 (4)<br />
<br />
3 3<br />
Ans.<br />
<br />
ˆ 2<br />
F . nds ˆ where F 4 xzi – y ˆj yz kˆ<br />
and S is the surface of the<br />
cube bounded by x = 0, x = 1, y = 0, y = 1, z = 0, z = 1.<br />
(U.P., Ist Semester, 2009, Nagpur University, Winter 2003)<br />
Solution. By Divergence theorem,<br />
<br />
<br />
F . nds ˆ = ( . F)<br />
dv<br />
S<br />
<br />
v<br />
=<br />
=<br />
<br />
v<br />
v<br />
<br />
<br />
<br />
ˆ<br />
<br />
ˆ<br />
ˆ <br />
ˆ 2<br />
i j k .(4 xzi – y ˆj yz kˆ<br />
<br />
) dv<br />
x y z<br />
2 <br />
(4 xz) (– y ) ( yz)<br />
dxdydz<br />
x y z <br />
= (4 z –2 y y)<br />
dxdydz<br />
=<br />
=<br />
=<br />
v<br />
<br />
v<br />
2<br />
1 14z<br />
<br />
(4 z – y) dx dy dz – yz dx dy<br />
0<br />
0<br />
2 <br />
<br />
1 1 2 1<br />
1 1<br />
0 0<br />
0<br />
0 0<br />
<br />
(2 z – yz) dxdy (2 – y)<br />
dx dy<br />
1<br />
2<br />
1<br />
<br />
1<br />
y 3<br />
2 y – dx dx<br />
<br />
2 2<br />
=<br />
0 0<br />
0<br />
Note: This question is directly solved as on example 14 on Page 574.<br />
1<br />
0<br />
3 1 3 3<br />
[ x ] 0 (1) Ans.<br />
2 2 2<br />
<br />
2<br />
Example 104. Find Fn ˆ ds,<br />
where F (2x 3) z iˆ<br />
– ( xz y) ˆj ( y 2 z)<br />
kˆ<br />
and S is<br />
the surface of the sphere having centre (3, – 1, 2) and radius 3.<br />
(AMIETE, Dec. 2010, U.P., I Semester, Winter 2005, 2000)<br />
Solution. Let V be the volume enclosed by the surface S.<br />
By Divergence theorem, we’ve<br />
<br />
<br />
<br />
<br />
Fnˆ<br />
ds = div Fdv.<br />
S V<br />
Now, div F 2<br />
= (2x 3) z [ ( xz y)] ( y 2 z)<br />
= 2 – 1 + 2 = 3<br />
x y z<br />
<br />
<br />
= 3<br />
Fnˆ<br />
ds<br />
S dv = 3<br />
V dv = 3V.<br />
V<br />
Again V is the volume of a sphere of radius 3. Therefore<br />
4 3 4 3<br />
V = r = (3) = 36 .<br />
3 3<br />
<br />
Fn ˆ ds = 3V = 3 × 36 = 108 Ans.<br />
S
Vectors 455<br />
Example 105. Use Divergence Theorem to evaluate<br />
A <br />
S ds ,<br />
where 3 3<br />
A x iˆ<br />
y ˆj z<br />
3ˆ k and S is the surface of the sphere x2 + y2 + z2 = a2 .<br />
(AMIETE, Dec. 2009)<br />
Solution.<br />
=<br />
<br />
A <br />
<br />
S ds = div AdV<br />
v<br />
<br />
ˆ<br />
<br />
ˆ<br />
ˆ 3 ˆ 3ˆ<br />
3<br />
i j k ( x i y j z kˆ<br />
) dV<br />
x y z<br />
v<br />
2 2 2<br />
(3 x 3y 3 z ) dV =<br />
v<br />
2 2 2<br />
=<br />
3 ( x y z ) dV<br />
v<br />
On putting x = r sin cos , y = r sin sin , z = r cos , we get<br />
3 r ( r sin dr dd)<br />
= 3 × 8<br />
v<br />
=<br />
2 2<br />
<br />
<br />
2 2<br />
<br />
d sin d<br />
r dr<br />
0 0<br />
5<br />
a<br />
5 5<br />
<br />
= 24 ( )<br />
2<br />
( cos )<br />
2<br />
r a<br />
12a<br />
<br />
0 0<br />
<br />
= 24 ( 0 1)<br />
<br />
5 2 5 5<br />
Ans.<br />
<br />
<br />
<br />
0<br />
Example 106. Use divergence Theorem to show that<br />
2 2 2<br />
( x y z ) d <br />
s = 6 V<br />
S<br />
where S is any closed surface enclosing volume V. (U.P., I Semester, Winter 2002)<br />
<br />
Solution. Here (x 2 + y 2 + z 2 ˆ 2 2 2<br />
) =<br />
ˆ<br />
<br />
i ˆ<br />
<br />
j k ( x y z )<br />
x y z<br />
= 2 xiˆ 2 yj ˆ 2 zkˆ 2 ( xiˆ yj ˆ zkˆ)<br />
<br />
2 2 2<br />
( x y z ) ds<br />
=<br />
S<br />
<br />
S<br />
2 2 2<br />
ˆ<br />
( x y z ) nds<br />
ˆn being outward drawn unit normal <strong>vector</strong> to S<br />
= 2( xiˆ yj ˆ zkˆ<br />
) nds ˆ<br />
<br />
S<br />
a<br />
0<br />
4<br />
= 2 ( ˆ ˆ ˆ<br />
div xi yj zk)<br />
dv<br />
...(1)<br />
V<br />
(By Divergence Theorem)<br />
(V being volume enclosed by S)<br />
Now,<br />
<br />
div. ( xiˆ<br />
yj ˆ zkˆ<br />
ˆ<br />
ˆ<br />
) =<br />
ˆ<br />
<br />
ˆ<br />
<br />
i j k ( xiˆ y ˆj zk)<br />
x y z <br />
x y z<br />
=<br />
<br />
<br />
<br />
x y z<br />
= 3 ...(2)<br />
From (1) & (2), we have<br />
2 2 2<br />
( y z ) dS dv dv = 6 V<br />
V<br />
Proved.<br />
= 2 3 = 6<br />
V<br />
Example 107. Evaluate ˆ ˆ<br />
( y z i z x j z y k ) n ˆ dS , where S is the part of the sphere<br />
S<br />
x 2 + y 2 + z 2 = 1 above the xy-plane and bounded by this plane.<br />
Solution. Let V be the volume enclosed by the surface S. Then by divergence Theorem, we<br />
have<br />
2 2 2 2 2<br />
( y z i ˆ z x ˆ j z y 2ˆ k ) n ˆ dS<br />
2 2 2 2 2<br />
S<br />
= div ( y z iˆ<br />
z x ˆj z yk<br />
2ˆ)<br />
dV<br />
=<br />
<br />
x y z<br />
2 2 2 2 2 2ˆ<br />
<br />
<br />
2 2 2 2 2 2<br />
( y z ) ( z x ) ( z y ) dV<br />
V<br />
<br />
<br />
2 2 2<br />
2<br />
V V<br />
V<br />
<br />
<br />
zy dV zy dV
456 Vectors<br />
Changing to spherical polar coordinates by putting<br />
x = r sin cos , y= r sin sin , z = r cos , dV = r 2 sin dr d d<br />
To cover V, the limits of r will be 0 to 1, those of will be 0 to 2<br />
and those of will be 0 to<br />
2.<br />
<br />
2<br />
2<br />
dV =<br />
V zy<br />
=<br />
=<br />
=<br />
2 /2 1 2 2 2 2<br />
<br />
2 ( r cos )( r sin sin ) r sin drd d<br />
0 0 0<br />
2 /2 1 5 3 2<br />
<br />
2 r sin cos sin<br />
drd d<br />
<br />
0 0 0<br />
<br />
6<br />
1<br />
2 2<br />
3 2<br />
r<br />
<br />
0 <br />
<br />
0<br />
6 0<br />
2 sin cos sin<br />
d d<br />
2 2<br />
=<br />
2<br />
2<br />
sin <br />
6 0 4.2 d<br />
1<br />
12<br />
2<br />
2 <br />
sin d<br />
=<br />
0<br />
12<br />
Example 108. Use Divergence Theorem to evaluate FdS<br />
2<br />
where F 4 xiˆ<br />
–2y ˆj z<br />
2ˆ k<br />
S<br />
and S is the surface bounding the region x 2 + y 2 = 4, z = 0 and z = 3.<br />
(A.M.I.E.T.E., Summer 2003, 2001)<br />
Solution. By Divergence Theorem,<br />
FdS<br />
<br />
= div FdV<br />
S V =<br />
<br />
<br />
V<br />
<br />
ˆ<br />
<br />
ˆ<br />
ˆ <br />
ˆ 2ˆ<br />
2<br />
i j k (4 xi 2 y j z kˆ<br />
) dV<br />
x y z<br />
= (4 4y<br />
2 z)<br />
dx dy dz<br />
=<br />
V<br />
3<br />
dx dy (4 4y 2 z)<br />
dz =<br />
= (12 12y<br />
9)<br />
dxdy<br />
Let us put x = r cos , y = r sin <br />
= (21 12r sin )<br />
r d<br />
dr<br />
=<br />
0<br />
= (21 12 )<br />
=<br />
2<br />
2<br />
2<br />
2 2<br />
21r<br />
3<br />
<br />
d 4r<br />
sin <br />
2<br />
0 <br />
<br />
= <br />
0 0<br />
<br />
0<br />
<br />
<br />
2 2<br />
0<br />
<br />
<br />
2 3<br />
dx dy [4z 4 yz z ] 0<br />
y dxdy<br />
d (21r 12r sin )<br />
dr<br />
<br />
<br />
d (42 32 sin ) (42 32 cos )<br />
2<br />
0<br />
Ans.<br />
= 84 + 32 – 32 = 84 Ans.<br />
^<br />
Example 109. Apply the Divergence Theorem to compute u nds, where s is the surface of<br />
the cylinder x 2 + y 2 = a 2 bounded by the planes z = 0, z = b and where u ix ˆ – ˆjy kz ˆ .<br />
Solution. By Gauss’s Divergence Theorem<br />
u nds<br />
<br />
ˆ = ( u)<br />
dv<br />
V<br />
<br />
ˆ<br />
ˆ<br />
=<br />
ˆ<br />
<br />
ˆ<br />
<br />
i j k ( ix ˆ ˆ<br />
<br />
jy kzdv )<br />
V<br />
<br />
x y z<br />
= x y z<br />
dv<br />
V<br />
<br />
x y z<br />
= 111<br />
dv<br />
V<br />
= dv dx dy dz<br />
V = Volume of the cylinder = a 2 b Ans.<br />
=<br />
V
Vectors 457<br />
Example 110. Apply Divergence Theorem to evaluate<br />
F . ˆ , where<br />
V<br />
F<br />
4x 3 iˆ x 2 y ˆj x 2 zk ˆ and S is the surface of the cylinder x 2 + y 2 = a 2 bounded by the<br />
planes z = 0 and z = b.<br />
Solution. We have,<br />
(U.P. Ist Semester, Dec. 2006)<br />
F<br />
4x 3 iˆ<br />
x 2 y ˆj x 2 zk ˆ<br />
<br />
ˆ<br />
3 2 2 ˆ<br />
div F =<br />
ˆ<br />
<br />
ˆ<br />
<br />
i j k (4 x iˆ x yj ˆ x zk)<br />
x y z<br />
3 2 2<br />
= (4 x ) ( x y) ( x z)<br />
x y z<br />
= 12x 2 – x 2 + x 2 = 12 x 2<br />
Now,<br />
V<br />
<br />
div FdV<br />
=<br />
2 2<br />
12 a a <br />
x b<br />
xay a<br />
2<br />
x<br />
2<br />
z 0<br />
x<br />
2<br />
dzdydx<br />
a a<br />
2<br />
x<br />
2<br />
2 b<br />
xa y a<br />
2<br />
x<br />
2<br />
=<br />
= 12 x ( z)<br />
0 dydx<br />
a<br />
a<br />
<br />
a 2 2<br />
2 a <br />
x<br />
a a<br />
2<br />
x<br />
2<br />
12 b x ( y)<br />
dx<br />
2 2 2<br />
2 2 2<br />
= 12 b x .2 a x dx = 24 b<br />
x a x dx<br />
=<br />
=<br />
<br />
a<br />
2 2 2<br />
48 b x a x dx<br />
<br />
0<br />
/2 2 2<br />
48 b a sin a cos a cos d<br />
<br />
0<br />
a<br />
a<br />
[Put x = a sin , dx = a cos d]<br />
3 3<br />
4 /2<br />
2 2<br />
= 48 ba sin cos<br />
d<br />
4<br />
= 48 ba<br />
2 2<br />
0<br />
23<br />
1 1<br />
<br />
4 2 2<br />
= 48 ba<br />
= 3 b a 4 Ans.<br />
22 <br />
Example 111. Evaluate surface integral Fnds ˆ ,<br />
<br />
<br />
where F = (x 2 + y 2 + z 2 ) iˆ<br />
ˆj kˆ<br />
( ), S<br />
is the surface of the tetrahedron x = 0, y = 0, z = 0, x + y + z = 2 and n is the unit normal in<br />
the outward direction to the closed surface S.<br />
Solution. By Divergence theorem<br />
Fnds<br />
<br />
ˆ = div Fdv <br />
S V<br />
where S is the surface of tetrahedron x = 0, y = 0, z = 0, x + y + z = 2<br />
=<br />
<br />
<br />
<br />
i ˆ<br />
<br />
ˆ j k ˆ 2 2 2<br />
( x y z )( i ˆ ˆ j k ˆ)<br />
dv<br />
V<br />
<br />
x y z<br />
= (2x 2y 2 z)<br />
dv<br />
V<br />
<br />
= 2 ( x y z)<br />
dx dy dz<br />
=<br />
=<br />
V<br />
2 2x 2 x y<br />
<br />
2 dx dy ( x y z)<br />
dz<br />
2<br />
<br />
0 0 0<br />
2<br />
2 x<br />
y<br />
2 2 x <br />
z<br />
dx dy <br />
xz<br />
yz<br />
2 <br />
<br />
<br />
0 0<br />
0
458 Vectors<br />
=<br />
<br />
2<br />
2 2 x <br />
2 2 (2 x<br />
y)<br />
<br />
dx dy x x xy y xy y<br />
0 <br />
<br />
<br />
0 <br />
2 2 2<br />
<br />
3 3<br />
2 <br />
2 2 2 y (2 x<br />
y)<br />
<br />
= 2<br />
dx 2xy x y xy y <br />
<br />
0<br />
<br />
3 6 <br />
2<br />
2<br />
x<br />
3 3<br />
2 <br />
2 2 2 (2 x) (2 x)<br />
<br />
= 2<br />
dx 2 x(2 x) x (2 x) x (2 x) (2 x)<br />
<br />
0<br />
<br />
3 6 <br />
3 3<br />
2<br />
2 2 3 2 3 2 (2 x) (2 x)<br />
<br />
= 2<br />
4x2x 2x x 4x 4 x x (2 x)<br />
<br />
0 <br />
<br />
<br />
3 6 <br />
2<br />
<br />
3 4 3 4 3 4 4<br />
2 4 2 4 (2 ) (2 ) (2 )<br />
= 2 2 x x 2<br />
x x x x x <br />
x x <br />
3 4 3 4 3 12 24 <br />
<br />
3 4 4<br />
(2 x) (2 x) (2 x)<br />
8 16 16 <br />
= 2 = 2 <br />
3 12 24 3 12 24<br />
<br />
= 4 Ans.<br />
0<br />
Example 112. Use the Divergence Theorem to evaluate<br />
<br />
S<br />
( xdydz y dz dx zdxdy)<br />
where S is the portion of the plane x + 2 y + 3 z = 6 which lies in the first Octant.<br />
(U.P., I Semester, Winter 2003)<br />
<br />
Solution. ( f1 dydz f2 dx dz f3<br />
dxdy)<br />
S<br />
f1 f2<br />
f3<br />
<br />
= <br />
dxdydz<br />
V<br />
<br />
x y z<br />
<br />
where S is a closed surface bounding a volume V.<br />
<br />
( xdydz y dz dx zdxdy)<br />
S<br />
<br />
=<br />
V<br />
x y z<br />
dx dy dz<br />
x y z<br />
= 3 V 2<br />
= (1 11)<br />
dx dy dz dx dy dz<br />
V<br />
= 3 (Volume of tetrahedron OABC)<br />
1<br />
= 3[( Area of the base OAB) height OC]<br />
3<br />
1 1<br />
<br />
= 3<br />
63<br />
2<br />
3 2<br />
= 18 Ans.<br />
<br />
Example 113. Use Divergence Theorem to evaluate : ( xdydz y dz dx zdxdy)<br />
Solution.<br />
over the surface of a sphere radius a.<br />
Here, we have<br />
(K. University, Dec. 2009)<br />
xdydz y dx dz zdxdy<br />
<br />
S<br />
<br />
<br />
f1 f2<br />
f3<br />
<br />
x y z<br />
dx dy dz<br />
V<br />
<br />
dx dy dz<br />
x y z<br />
<br />
<br />
V <br />
x y z<br />
(1 + 1 + 1) dx dy dz = 3 (volume of the sphere)<br />
V<br />
4<br />
3 <br />
= 3 a<br />
<br />
3 = 4 a3 Ans.<br />
<br />
0<br />
<br />
0
Vectors 459<br />
Example 114. Using the divergence theorem, evaluate the surface integral<br />
( yz dy dz zxdz dx xy dy dx)<br />
where S : x 2 + y 2 + z 2 = 4.<br />
S<br />
<br />
Solution. ( f1 dydz f2 dx dz f3<br />
dxdy)<br />
S<br />
<br />
f1 f2<br />
f3<br />
<br />
= <br />
dx dy dz<br />
v<br />
<br />
x y z <br />
where S is closed surface bounding a volume V.<br />
<br />
( yz dy dz zx dx dz xy dx dy)<br />
S<br />
(AMIETE, Dec. 2010, UP, I Sem., Dec 2008)<br />
( ) ( ) ( )<br />
= yz zx xy <br />
dx dy dz<br />
v<br />
<br />
x y z <br />
= (0 0 0) dx dy dz<br />
v<br />
= 0 Ans.<br />
<br />
2 2 3 2<br />
Example 115. Evaluate xz dy dz ( x y z ) dzdx (2 xy y z)<br />
dxdy<br />
S<br />
where S is the surface of hemispherical region bounded by<br />
z =<br />
2 2 2<br />
a x y and z = 0.<br />
=<br />
Solution. ( f1 dydz f2 dz dx f3<br />
dxdy)<br />
S<br />
where S is a closed surface bounding a volume V.<br />
<br />
<br />
=<br />
S<br />
<br />
2 2 3 2<br />
xz dy dz ( x y z ) dzdx (2 xy y z)<br />
dxdy<br />
<br />
<br />
V<br />
V<br />
f1 f2<br />
f3<br />
<br />
dx dy dz<br />
x y z<br />
<br />
2 2 3 <br />
2 <br />
( xz ) ( x y z ) (2 xy y z)<br />
dxdydz<br />
x y z<br />
<br />
(Here V is the volume of hemisphere)<br />
2 2 2<br />
= ( z x y ) dx dy dz<br />
V<br />
Let x = r sin cos , y = r sin sin , z = r cos <br />
=<br />
=<br />
Example 116. Evaluate<br />
2 2<br />
r ( r sin dr d d )<br />
=<br />
5<br />
a<br />
2 /2<br />
r<br />
<br />
( ) 0 ( cos )<br />
0 <br />
<br />
5 <br />
<br />
0<br />
=<br />
<br />
2 a<br />
2 sin <br />
0 0 0<br />
<br />
d d r dr<br />
5<br />
2 ( 0 1) 5<br />
a<br />
=<br />
2a<br />
5<br />
4<br />
5<br />
Ans.<br />
F . n ˆ ds over the entire surface of the region above the xy-plane<br />
S<br />
bounded by the cone z 2 = x 2 + y 2 2<br />
and the plane z = 4, if F = 4xz iˆ<br />
xyz ˆj 3 zkˆ<br />
.<br />
Solution. If V is the volume enclosed by S, then V is bounded by the surfaces z = 0, z = 4,<br />
z 2 = x 2 + y 2 .<br />
By divergence theorem, we have<br />
=<br />
F . n ˆ ds<br />
S<br />
=<br />
=<br />
<br />
<br />
<br />
V<br />
V<br />
V<br />
<br />
div Fdxdydz<br />
2 <br />
(4 xz) ( xyz ) (3) z dxdydz<br />
x y z<br />
<br />
2<br />
(4z xz 3) dxdydz<br />
Limits of z are<br />
2 2<br />
x y and 4.
460 Vectors<br />
4 2<br />
x<br />
2<br />
y<br />
2 <br />
(4z xz 3) dzdydx =<br />
<br />
<br />
3<br />
2 xz <br />
2z 3z<br />
dy dx<br />
3 <br />
4<br />
x<br />
2<br />
y<br />
2<br />
<br />
64x<br />
2 2 2 2 3/2 2 2 <br />
= 32 12 {2( x y ) xx ( y ) 3 x y } dydx<br />
3<br />
<br />
<br />
<br />
<br />
64x<br />
2 2 2 2 3/2 2 2 <br />
= 44 2( x y ) xx ( y ) 3<br />
x y dydx<br />
3<br />
<br />
Putting x = r cos and y = r sin , we have<br />
64r<br />
cos 2 3 <br />
= 44 2r r cos r 3r<br />
rddr<br />
3<br />
<br />
Limits of r are 0 to 4.<br />
and limits of are 0 to 2<br />
=<br />
=<br />
=<br />
=<br />
=<br />
<br />
<br />
<br />
<br />
<br />
2<br />
2 4 64r<br />
cos 3 5 2<br />
44 2 cos 3<br />
<br />
0 <br />
0 <br />
<br />
<br />
r r r r ddr<br />
3<br />
<br />
3 4 6<br />
2 <br />
2 64 r cos r r<br />
3<br />
0<br />
2 <br />
0<br />
2 <br />
0<br />
2 <br />
0<br />
<br />
22r cos r d<br />
<br />
9 2 6 <br />
3 4 6<br />
2 64 (4) cos (4) (4)<br />
3<br />
22(4) cos (4)<br />
d<br />
<br />
9 2 6<br />
<br />
6<br />
64 64 (4) <br />
352 cos 128 cos 64<br />
d<br />
9 6<br />
<br />
<br />
6<br />
64 64 (4) <br />
160 cos d<br />
<br />
<br />
9 6 <br />
<br />
<br />
<br />
2<br />
<br />
6<br />
64 64 (4) <br />
<br />
6<br />
64 64 (4) <br />
= 160 <br />
sin <br />
9 6 = 160 (2 ) sin 2<br />
<br />
<br />
<br />
<br />
9 6 <br />
0<br />
<br />
<br />
= 320 Ans.<br />
Example 117. The <strong>vector</strong> field 2<br />
F x iˆ<br />
zj ˆ yzkˆ<br />
is defined over the volume of the cuboid<br />
given by 0 x a, 0 y b, 0 z c, enclosing the surface S. Evaluate the surface integral<br />
<br />
<br />
F . ds<br />
(U.P., I Semester, Winter 2001)<br />
S<br />
Solution. By Divergence Theorem, we have<br />
2 ˆ<br />
2<br />
( x iˆ zj ˆ yz k). ds div ( x iˆ zj ˆ yz kˆ) dv,<br />
<br />
S<br />
<br />
v<br />
where V is the volume of the cuboid enclosing the surface S.<br />
<br />
ˆ<br />
2<br />
ˆ<br />
=<br />
ˆ<br />
<br />
ˆ<br />
<br />
i j k .( x iˆ z ˆ<br />
<br />
j yzk)<br />
dv<br />
v<br />
<br />
x y z<br />
2 <br />
= ( x ) ( z) ( yz)<br />
dx dy dz<br />
v<br />
x y z <br />
=<br />
a b c<br />
=<br />
(2 x<br />
y ) dx dy dz<br />
0 0 0<br />
x y z <br />
a b a b<br />
c<br />
= 0<br />
0 0 0 0<br />
<br />
dx [2 xz yz] dy dx (2 xc yc)<br />
dy<br />
4<br />
0<br />
a b c<br />
<br />
dx dy (2 x y ) dz<br />
0 0 0
Vectors 461<br />
=<br />
=<br />
a b a b<br />
2 a<br />
2<br />
y b <br />
cdx(2 x y) dy c2xy dx c 2bx<br />
dx<br />
2<br />
2 <br />
<br />
0 0 0 0 0<br />
2 2<br />
a<br />
2<br />
2<br />
2<br />
bx b x ab b <br />
c c ab abc a<br />
<br />
2 2 2 2<br />
0<br />
Example 118. Verify the divergence Theorem for the function F = 2 x2 yi – y 2 j + 4 x z 2 k<br />
taken over the region in the first octant bounded by y 2 + z 2 = 9 and x = 2.<br />
Solution.<br />
<br />
FdV<br />
=<br />
V<br />
<br />
= (4 xy 2y 8 xz)<br />
dxdydz<br />
=<br />
=<br />
<br />
ˆ<br />
<br />
ˆ<br />
ˆ 2 ˆ 2ˆ<br />
2<br />
i j k (2x yi y j 4 xz k)<br />
dV<br />
x y z<br />
=<br />
<br />
<br />
2 3 9 y<br />
2<br />
<br />
dx dy (4 xy 2y 8 xz)<br />
dz<br />
0 0 0<br />
2 3 9<br />
2<br />
2 y<br />
dx (4 2 4 )<br />
0 dy xyz yz xz<br />
0<br />
0<br />
2 3 2 2 2<br />
<br />
dx [4 xy 9 y 2y 9 y 4 x(9 y )] dy<br />
0 0<br />
3<br />
2 4x<br />
2 2 3/2 2 2 3/2 4xy<br />
<br />
= dx (9 y ) (9 y ) 36<br />
xy <br />
0<br />
2 3 3 3 0<br />
2<br />
2<br />
= (0 0 108 x 36 x 36 x 18) dx<br />
2<br />
= (108 x 18) dx<br />
<br />
2<br />
x <br />
0 = 108 18 x<br />
0<br />
2 0<br />
= 216 – 36 = 180 ...(1)<br />
Here Fnds<br />
<br />
ˆ = Fnˆ ds Fnˆ ds Fnˆ ds Fnˆ ds F<br />
nˆ<br />
ds<br />
S<br />
OABC OCE OADE ABD BDEC<br />
F nˆ<br />
2 2 2<br />
ds (2 x yiˆ<br />
y ˆj 4 xzkˆ<br />
). nds ˆ<br />
=<br />
BDEC<br />
<br />
BDEC<br />
Normal <strong>vector</strong><br />
<br />
= = ˆ<br />
<br />
i ˆ<br />
<br />
j kˆ<br />
<br />
<br />
x y z (y2 + z 2 – 9)<br />
= 2 yj ˆ 2 zkˆ<br />
2 yj ˆ 2zkˆ<br />
yj ˆ zkˆ<br />
Unit normal <strong>vector</strong> = ˆn =<br />
=<br />
2 2 2 2<br />
4y<br />
4z<br />
y z<br />
yj ˆ zkˆ<br />
yj ˆ zkˆ<br />
= =<br />
9 3<br />
ˆ<br />
2 ˆ 2ˆ<br />
2 yj zk 1<br />
(2x yi y j 4 xzk)<br />
ds<br />
3 3<br />
= ( 4 )<br />
3<br />
BDEC<br />
3<br />
y xz ds<br />
BDEC<br />
<br />
<br />
<br />
yj ˆ zkˆ<br />
<br />
( ˆ ) ˆ z dxdy <br />
dx dy ds nk ds k ds or ds <br />
<br />
3 3 z<br />
<br />
<br />
<br />
3 <br />
3<br />
1<br />
3 3 dxdy<br />
2 3<br />
y 2<br />
<br />
= ( y 4 xz )<br />
3<br />
= dx 4 xz dy<br />
z<br />
0 <br />
<br />
0<br />
z <br />
BDEC<br />
<br />
<br />
3<br />
3<br />
2<br />
27 sin <br />
2<br />
2<br />
<br />
= dx<br />
4 x (9cos )<br />
0 <br />
0<br />
3cos <br />
<br />
3<br />
Ans.<br />
y<br />
3sin ,<br />
<br />
<br />
z<br />
3cos
462 Vectors<br />
=<br />
=<br />
F nˆ<br />
ds =<br />
OABC<br />
<br />
<br />
=<br />
2 2 2<br />
dx 27 108 x<br />
0<br />
<br />
3 3<br />
=<br />
2<br />
<br />
2<br />
0<br />
( 18 72 x ) dx<br />
<br />
2<br />
18x<br />
36 x <br />
<br />
<br />
<br />
= 108 ...(2)<br />
0<br />
2 2 2<br />
(2 x yiˆ y ˆj 4 xz kˆ) ( kˆ)<br />
ds<br />
<br />
OABC<br />
<br />
2<br />
4 xz ds = 0 ...(3) because in OABC xy-plane, z = 0<br />
OABC<br />
F nˆ<br />
ds =<br />
OADE<br />
<br />
<br />
Fnds ˆ<br />
=<br />
OCE<br />
<br />
<br />
Fnds ˆ =<br />
ABD<br />
<br />
<br />
=<br />
(2 ˆ ˆ 4 ) ( ˆ<br />
x yi y j xzkj)<br />
ds =<br />
OADE<br />
2 2 2 ˆ<br />
(2 ˆ ˆ 4 ) ( ˆ<br />
x yi y j xzki)<br />
ds =<br />
OCE<br />
2 2 2 ˆ<br />
(2 x yiˆ y ˆj 4 xzk) () iˆ<br />
ds =<br />
2<br />
2 x y ds<br />
ABD<br />
2<br />
2 x y dy dz =<br />
2 2 2 ˆ<br />
9 z<br />
2<br />
<br />
2<br />
y ds = 0 ...(4)<br />
OADE<br />
because in OADE xz-plane, y = 0<br />
2<br />
2 x y ds = 0 ...(5)<br />
OCE<br />
<br />
because in OCE yz-plane, x = 0<br />
ABD<br />
3 9<br />
z<br />
2<br />
2<br />
dz 2(2) ydy<br />
0 <br />
because in ABD plane, x = 2<br />
0<br />
2<br />
3<br />
= 8 y<br />
<br />
dz <br />
0<br />
2<br />
=<br />
3 <br />
3<br />
2<br />
z <br />
4 dz (9 z ) = 49z<br />
<br />
0<br />
<br />
3<br />
0<br />
<br />
On adding (2), (3), (4), (5) and (6), we get<br />
<br />
<br />
3<br />
0<br />
= 4 [27 – 9] = 72 ...(6)<br />
Fnds ˆ = 108 + 0 + 0 + 0 + 72 = 180 ...(7)<br />
S<br />
From (1) and (7), we have<br />
V<br />
<br />
FdV<br />
Fnds<br />
<br />
<br />
S<br />
<br />
= <br />
ˆ<br />
Hence the theorem is verified.<br />
Example 119. Verify the Gauss divergence Theorem for<br />
= (x 2 – yz) î + (y 2 – zx) ĵ + (z 2 – xy) ˆk taken over the rectangular parallelopiped<br />
0 x a, 0 y b, 0 z c. (U.P., I Semester, Compartment 2002)<br />
Solution. We have<br />
div F = F <br />
ˆ<br />
2 2 2 ˆ<br />
=<br />
ˆ<br />
<br />
ˆ<br />
<br />
i j k [( x yz) iˆ( y zx) ˆj ( z xy) k]<br />
x y z<br />
2 2 2<br />
= ( x yz) ( y zx) ( z xy)<br />
= 2x + 2y + 2z<br />
x y z<br />
F <br />
Volume integral =<br />
=<br />
=<br />
<br />
FdV<br />
= 2( x y z)<br />
dV<br />
V<br />
V<br />
a b c<br />
=<br />
2 ( x y z ) dx dy dz<br />
2<br />
x 0 y 0 z<br />
0<br />
<br />
dx dy <br />
xz yz <br />
<br />
2<br />
a b z<br />
0 0<br />
c<br />
<br />
2 <br />
<br />
=<br />
0<br />
b<br />
2 2<br />
= 2 a y c y <br />
dx cxyc<br />
<br />
0 <br />
2 2 <br />
<br />
<br />
= 2<br />
0<br />
2<br />
<br />
<br />
a b c<br />
<br />
2 dx dy ( x y z)<br />
dz<br />
0 0 0<br />
<br />
dx dy <br />
cx cy <br />
<br />
2<br />
a b c<br />
0 0<br />
a bc bc<br />
dx bcx<br />
0 <br />
<br />
2 2<br />
2 2<br />
<br />
2
Vectors 463<br />
a<br />
<br />
2 2 2<br />
bc x b cx bc x <br />
= 2 = [a<br />
2 2 2<br />
2 bc + ab 2 c + abc 2 ]<br />
0<br />
= abc (a + b + c) ...(A)<br />
<br />
<br />
To evaluate Fnds ˆ , where S consists of six plane surfaces.<br />
S<br />
<br />
<br />
= ˆ ˆ ˆ<br />
OABC <br />
DEFG OAFG<br />
Fnds ˆ<br />
S<br />
Fnds <br />
OABC<br />
<br />
ˆ =<br />
<br />
<br />
F nds ˆ<br />
DEFG<br />
<br />
Fnds Fnds Fnds<br />
<br />
<br />
Fnds ˆ Fnds ˆ Fnds ˆ<br />
=<br />
=<br />
=<br />
=<br />
=<br />
<br />
<br />
BCDE ABEF OCDG<br />
OABC<br />
ab<br />
<br />
00<br />
ab<br />
00<br />
2 2 2<br />
{( x yzi ) ˆ ( y xz) ˆj ( z xy) kˆ}( kˆ)<br />
dxdy<br />
2<br />
( z xy)<br />
dxdy<br />
(0 xy)<br />
dx dy =<br />
<br />
DEFG<br />
ab<br />
ab<br />
2<br />
( z xy)<br />
dxdy =<br />
<br />
00<br />
00<br />
a<br />
2<br />
b a<br />
<br />
2 xy <br />
c y dx<br />
2<br />
= <br />
0 0<br />
0<br />
<br />
2 2<br />
2 x b <br />
= cbx<br />
<br />
4 <br />
Fnds<br />
<br />
<br />
OAFG<br />
<br />
ˆ =<br />
=<br />
=<br />
<br />
<br />
a<br />
OAFG<br />
c<br />
0 0<br />
<br />
ˆ<br />
2<br />
<br />
a<br />
0<br />
=<br />
OAFG<br />
( y zx)<br />
dxdz<br />
dx<br />
(0 zx)<br />
dz =<br />
2 2<br />
a b<br />
4<br />
...(1)<br />
2 2 2<br />
{( x yz) iˆ<br />
( y xz) ˆj ( z xy) kˆ}( kˆ)<br />
dxdy<br />
2<br />
( c xy)<br />
dxdy<br />
<br />
2 xb<br />
cb<br />
<br />
2<br />
2 2<br />
2<br />
<br />
dx<br />
<br />
<br />
2 a b<br />
abc ...(2)<br />
4<br />
{( x yz) iˆ ( y zx) ˆj ( z xy) k}( ˆj)<br />
dxdz<br />
2 2 2 ˆ<br />
a<br />
2<br />
xz<br />
dx <br />
<br />
<br />
2 <br />
=<br />
<br />
c<br />
0 0<br />
a xc<br />
2<br />
dx =<br />
2<br />
0<br />
x c<br />
<br />
4<br />
F nds 2 2 2<br />
= {( x yz ) i ˆ ( y zx ) ˆ j ( z xy ) k ˆ}<br />
ˆ jdxdz<br />
BCDE = <br />
Fnds<br />
<br />
<br />
ABEF<br />
=<br />
<br />
a c<br />
a<br />
2<br />
c<br />
<br />
2<br />
2 xz <br />
dx ( b xz)<br />
dz =<br />
<br />
b z <br />
2 <br />
0 0<br />
0 0<br />
2 2<br />
a<br />
2 2<br />
2 x c 2 ac<br />
2 2<br />
dx =<br />
<br />
<br />
<br />
a<br />
0<br />
a<br />
<br />
0<br />
=<br />
2 2<br />
a c<br />
4<br />
BCDE<br />
<br />
2 xc<br />
<br />
b c <br />
2<br />
2<br />
...(3)<br />
( y xz)<br />
dxdz<br />
2<br />
<br />
<br />
dx<br />
<br />
<br />
= b cx = ab c ...(4)<br />
4 <br />
4<br />
0<br />
2 2 2<br />
<br />
ˆ<br />
ˆ<br />
= {( ) ˆ ( ) ˆ ( ) } ˆ<br />
x yz i y xz j z x y k idydz<br />
ABEF<br />
=<br />
2<br />
( x yz)<br />
dydz =<br />
ABEF<br />
S.No.<br />
Surface Outward normal<br />
ds<br />
1 OABC – k dx dy z = 0<br />
2 DEFG k dx dy z = c<br />
3 OAFG – j dx dz y = 0<br />
4 BCDE j dx dz y = b<br />
5 ABEF i dy dz x = a<br />
6 OCDG – i dy dz x = 0<br />
b c<br />
b<br />
2<br />
c<br />
2<br />
<br />
2 yz <br />
dy ( a yz)<br />
d z = dy<br />
<br />
a z <br />
2 <br />
0 0<br />
0 0
464 Vectors<br />
Fnds<br />
<br />
<br />
OCDG<br />
=<br />
<br />
ˆ =<br />
=<br />
b<br />
<br />
2<br />
2 yc <br />
<br />
ac<br />
dy<br />
2 <br />
0 <br />
=<br />
<br />
OCDG<br />
<br />
2 2<br />
2 yc <br />
acy<br />
<br />
4 <br />
2 2 2 ˆ<br />
b<br />
0<br />
=<br />
2 2<br />
2 b c<br />
a bc ...(5)<br />
4<br />
{( x yz) iˆ ( y zx) ˆj ( z x y) k} ( iˆ<br />
) dydz<br />
b c<br />
x 2 yz dydz<br />
=<br />
0 0 ( )<br />
2<br />
b yc <br />
2 2<br />
y c <br />
= dy = =<br />
0 2 4 0<br />
Adding (1), (2), (3), (4), (5) and (6), we get<br />
b<br />
<br />
b<br />
c<br />
<br />
=<br />
dy ( yz ) dz<br />
0 0<br />
2 2<br />
b c<br />
4<br />
2<br />
b yz<br />
dy <br />
<br />
<br />
0<br />
2 <br />
c<br />
0<br />
...(6)<br />
Fnds<br />
<br />
ˆ<br />
2 2 2 2 2 2 2 2<br />
2 2<br />
=<br />
a b abc a b a c ab c <br />
a c <br />
4 4 4 4 <br />
<br />
<br />
2 2 2 2<br />
b c <br />
2 b c <br />
a bc <br />
4 4 <br />
<br />
= abc 2 + ab 2 c + a 2 bc<br />
= abc (a + b + c) ...(B)<br />
From (A) and (B), Gauss divergence Theorem is verified.<br />
Verified.<br />
<br />
2<br />
Example 120. Verify Divergence Theorem, given that F 4 xziˆ<br />
– y ˆj yz kˆ<br />
and S is the<br />
surface of the cube bounded by the planes x = 0, x = 1, y = 0, y = 1, z = 0, z = 1.<br />
Solution.<br />
Volume Integral =<br />
F <br />
ˆ<br />
2 ˆ<br />
=<br />
ˆ<br />
<br />
ˆ<br />
<br />
i j k (4 zxiˆ y ˆj yzk)<br />
x y z<br />
= 4 z – 2 y + y<br />
= 4 z – y<br />
<br />
Fdv<br />
<br />
= (4 z y)<br />
dx dy dz<br />
=<br />
1 1 1<br />
<br />
dx dy (4 z y ) dz<br />
0 0 0<br />
=<br />
=<br />
1 dx 1 dy 2 1<br />
(2 z yz )<br />
0 0<br />
0<br />
=<br />
<br />
2<br />
1 y <br />
dx<br />
2 y <br />
0 2 =<br />
<br />
1<br />
0<br />
<br />
1 1<br />
<br />
dx dy (2 y)<br />
0 0<br />
<br />
1 1<br />
dx 2 <br />
<br />
0<br />
<br />
2 <br />
= 3 1 3 ( ) 1<br />
0<br />
0<br />
2 dx x = 3 2 2<br />
...(1)<br />
To evaluate Fnds ˆ , where S consists of six plane surfaces.<br />
S<br />
<br />
<br />
Over the face OABC , z = 0, dz = 0, ˆn = – ˆk , ds = dx dy<br />
<br />
<br />
1 1 2<br />
F. nds ˆ (– y ˆj) (– kˆ<br />
) dxdy 0<br />
<br />
0 0<br />
Over the face BCDE, y = 1, dy = 0
Vectors 465<br />
<br />
<br />
Fnds ˆ =<br />
1 1<br />
<br />
0 0<br />
(4 xziˆ ˆj zkˆ<br />
) ( ˆj)<br />
dx dz<br />
ˆn = ˆ, j ds dxdz =<br />
=<br />
1 1<br />
dx dz =<br />
0 0<br />
1 1<br />
<br />
0 0<br />
dxdz<br />
1 1<br />
0 z 0<br />
( x) ( ) = – (1) (1) = – 1<br />
Over the face DEFG, z = 1, dz = 0, ˆn = ˆk , ds = dx dy<br />
<br />
<br />
Fnds ˆ =<br />
=<br />
1 1 2<br />
<br />
0 0<br />
[4 x (1) y ˆj y(1) kˆ]()<br />
kˆ<br />
dx dy<br />
1 1<br />
ydxdy =<br />
0 0<br />
1 1<br />
dx ydy = ( x)<br />
0 0<br />
Over the face OCDG, x = 0, dx = 0, ˆn = – iˆ,<br />
<br />
<br />
1<br />
0<br />
ds = dy dz<br />
Fnds ˆ = 1 1 ˆ 2 ˆ ˆ ˆ<br />
(0 i y j yzk ) ( i ) dydz = 0<br />
0 0<br />
Over the face AOGF, y = 0, dy = 0, ˆn = – ĵ ,<br />
<br />
<br />
Fnds ˆ =<br />
1 1<br />
(4 ˆ) ( ˆ<br />
xzi j)<br />
dxdz = 0<br />
0 0<br />
Over the face ABEF, x = 1, dx = 0, ˆn = î ,<br />
<br />
<br />
Fnds ˆ =<br />
1 1 2<br />
0 0<br />
[(4 zi ˆ y ˆ j yzk ˆ)()]<br />
i ˆ dydz<br />
=<br />
1 dy<br />
1<br />
4<br />
0 0<br />
1 2 1<br />
dy (2 z )<br />
0<br />
0<br />
1<br />
<br />
2<br />
y <br />
<br />
2 <br />
<br />
0<br />
ds = dx dz<br />
ds = dy dz<br />
1<br />
0<br />
= 1 2<br />
1 1<br />
4 zdydz<br />
0 0<br />
= zdz = = 2 dy = 2( y) 2<br />
On adding we see that over the whole surface<br />
Fnds<br />
1 <br />
ˆ = 01 0 0<br />
2<br />
2 = 3 2<br />
From (1) and (2), we have<br />
1. Use Divergence Theorem to evaluate<br />
V<br />
<br />
Fdv<br />
Fnds<br />
<br />
<br />
S<br />
<br />
= <br />
ˆ<br />
EXERCISE 5.15<br />
<br />
2 2 2 2 2<br />
( y z i ˆ z x ˆ j x y 2ˆ k ). ds ,<br />
s<br />
1<br />
0<br />
...(2)<br />
Verified.<br />
where S is the upper part of the sphere x 2 + y 2 + z 2 = 9 above xy- plane. Ans. 243 <br />
8<br />
<br />
<br />
2. Evaluate ( F). ds,<br />
where S is the surface of the paraboloid x 2 + y 2 + z = 4 above the xy-plane and<br />
<br />
S<br />
2 ˆ ˆ<br />
2<br />
F ( x y 4) i 3 xyj (2 xz z ) kˆ<br />
.<br />
Ans. – 4 <br />
<br />
2 2 3 2<br />
3. Evaluate [ xz dy dz ( x y z ) dzdx (2 xy y z) dxdy],<br />
where S is the surface enclosing a<br />
s<br />
region bounded by hemisphere x 2 + y 2 + z 2 = 4 above XY-plane.<br />
4. Verify Divergence Theorem for 2<br />
F x iˆ zj ˆ yzkˆ<br />
, taken over the cube bounded by<br />
x = 0, x = 1, y = 0, y = 1, z = 0, z = 1.<br />
2<br />
5. Evaluate (2 ˆ ˆ ˆ<br />
xyi yz j xz k)<br />
ds <br />
over the surface of the region bounded by<br />
S<br />
x = 0, y = 0, y = 3, z = 0 and x + 2 z = 6 Ans. 351<br />
2
466 Vectors<br />
6.<br />
<br />
Verify Divergence Theorem for 2<br />
F ( x y ) iˆ<br />
– 2 xj ˆ 2 yzkˆ<br />
and the volume of a tetrahedron bounded<br />
by co-ordinate planes and the plane 2 x + y + 2 z = 6.<br />
(Nagpur, Winter 2000, A.M.I.E.T.E.. Winter 2000)<br />
7. Verify Divergence Theorem for the function<br />
x 2 + y 2 = 9, z = 0 and z = 2.<br />
8. Use the Divergence Theorem to evaluate<br />
<br />
s<br />
ˆ ˆ 2 ˆ over the region bounded by<br />
F yi xj z k<br />
3 2 2<br />
x dydz x ydzdx x zdxdy,<br />
where S is the surface of the region bounded by the closed cylinder<br />
x 2 + y 2 = a 2 , (0 z b) and z = 0, z = b.<br />
<br />
Ans.<br />
4<br />
5 ab<br />
4<br />
2<br />
9. Evaluate the integral ( z x) dy dz xy dx dz 3 zdxdy,<br />
where S is the surface of closed region<br />
s<br />
bounded by z = 4 – y 2 and planes x = 0, x = 3, z = 0 by transforming it with the help of Divergence<br />
Theorem to a triple integral. Ans. 16<br />
10. Evaluate<br />
<br />
s<br />
ds<br />
2 2 2 2 2 2<br />
a x b y c z<br />
over the closed surface of the ellipsoid ax 2 + by 2 + cz 2 = 1 by<br />
applying Divergence Theorem. Ans.<br />
11. Apply Divergence Theorem to evaluate 2 2 2<br />
( lx my nz ) ds<br />
<br />
4<br />
( abc)<br />
taken over the sphere (x – a) 2 + (y – b) 2 + (z – c) 2 = r 2 , l, m, n being the direction cosines of the external<br />
normal to the sphere. (AMIETE June 2010, 2009) Ans.<br />
8 ( )<br />
3<br />
3 a b c r<br />
12. Show that ( uV u<br />
V ) dv<br />
V<br />
<br />
<br />
= .<br />
s<br />
<br />
uV ds<br />
13. If E = grad and 2<br />
= 4 , prove that E <br />
n ds = 4 dv<br />
S<br />
V<br />
where n is the outward unit normal <strong>vector</strong>, while dS and dV are respectively surface and volume<br />
elements.<br />
Pick up the correct option from the following:<br />
14. If F <br />
is the velocity of a fluid particle then F.<br />
dr represents.<br />
(a) Work done (b) Circulation<br />
C<br />
(c) Flux (d) Conservative field.<br />
(U.P. Ist Semester, Dec 2009) Ans. (b)<br />
15. If f = ax i by j cz k , a, b, c, constants, then f.<br />
dS where S is the surface of a unit sphere is<br />
<br />
(a) ( )<br />
3 a b c (b) 4 ( a b c)<br />
(c) 2 ( a b c)<br />
(d) (a + b + c)<br />
3<br />
(U.P., Ist Semester, 2009) Ans. (b)<br />
16. A force field F is said to be conservative if<br />
<br />
(a) Curl F 0 (b) grad F 0 (c) Div F 0 (d) Curl (grad F ) = 0<br />
(AMIETE, Dec. 2006) Ans. (a)<br />
17. The line integral<br />
<br />
<br />
2 2<br />
x dx y dy, where C is the boundary of the region x 2 + y 2 < a 2 equals<br />
c<br />
(a) 0, (b) a (c) a 2 1 2<br />
(d) a<br />
2<br />
(AMIETE, Dec. 2006) Ans. (b)