1. Unit.2 Vector Calculus
UNIT II VECTOR CALCULUS
Part-A
     
        
                
   
 
    i  x  cy  z  bx y z j   x cy z           x  y  az
      
 y z x z
   
k bx y z x y az
          
 ic 1 j a  4 k b  2
1
Problem 1 Prove that div(grad )  2
Solution:
div(grad )  .
  
    
i j k
x y z
        
i j k i j k
x y z x y z
2 2 2
x2 y2 z2
     
  
  
2 2 2
x2 y2 z2

    
         
2 .
Problem 2 Find a, b, c, if F  x  2y  azi  bx 3y  z j  (4x  cy  2z)k
is
irrotational.
Solution:
F
 

is irrotational if  F  0
  
i j k
  
2 3 4 2
F
x y z
x y az bx y z x cy z
  
  
     

 4 2   3   4 2   2

 3 2   2

x y

  
ic 1 j a  4 k b  2
   
 F  0 0i  0 j  0k
  
c 1  0, a  4  0, b  2  0
c  1, a  4, b  2.
Problem 3 If S is any closed surface enclosing a volume V and r

is the position
vector of a point, prove  .  3
  r n ds  V
S


2. Unit.2 Vector Calculus
     
     
       
  
                                       
                                         
2
Solution:
Let r  xi  y j  zk
   
By Gauss divergence theorem
 

  F n ds     FdV
S V
 
Here F  .r
 

  r  n ds     rdV
S V
. 
            
V
    
i j k xi y j zk dV
x y z
   111dV
 3 .
  r n ds  V
S

  
Problem 4 If r  a cos nt  bsin nt,
 
where a,b, n
are constants show that
   
r  d r  na 
b
dt
Solution:
  
Given r  a cos nt  bsin nt
  
d r   na sin nt 
nb cos nt
dt
r  d r  a cos nt  bsin nt     na sin nt 
nb cos nt 
dt
   
 nabcos2 nt  basin2 nt
 nabcos2 nt  absin2 nt  
a b  ba
   
 n ab1 n ab

Problem 5 Prove that div curl A  0
Solution:
 
  
  
1 2 3
.
i j k
x y z
A A A
      
  
i 3 2 j 3 1 k 2 1
y z x z x y
3 2 3 1 2 1
x y z y x z z x y

3. Unit.2 Vector Calculus
  2   2     2   2     2   2
 
 3  2    1  3   2  1
  x  y  x  z      y  z  y  x    z  x  z  y

  
3

div curl A  0
Problem 6 Find the unit normal to surface xy3z2  4 at 1,1,2
Solution:
Let   xy3z2  4
  
  y3z2 i  3xy2z2 j  2xy3zk
   3  2   2  2   3  
1, 1,2  1 2 i 3 1 1 2 j 2 1 1 2 k           
  
  4i 12 j  4k
Unit normal to the surface is n





  
 i  j  k

4 12 4
16  144 
16
4 3 
 i  j  k
176
 
  
     
4 3   3 
.
 i  j  k  i  j  k
 
16 
11 11
Problem 7 Applying Green’s theorem in plane show that area enclosed by a simple
closed curve C is 1  
2
 xdy  ydx
Solution:
   
Pdx Qdy Q P dx dy
           
P  y, Q  x
P 1, Q 1
y x
C R
x y
 
  
 
  1 1 2
 x dy  ydx     dxdy   dx dy
R R
= 2 Area enclosed by C
Area enclosed by C = 1  
2
 xdy  ydx .

and B
Problem 8 If A

 
are irrotational show that A B
is solenoidal
Solution:
Given A

 
is irrotational i.e.,  A  0

4. Unit.2 Vector Calculus
  
 j 3z2 3xy 3x2 3yz
            
 
x dx x dx x dx x
2 2 2
   
      
3 3
4

 
B
is irrotational i.e.,  B  0
     
.A B  B. A A. B
   
 B0  A0  0
 
 A B
is solenoidal.

Problem 9 If F  grad x3  y3  z3  3xyz

find curl F
Solution:

F x3  y3  z3  3xyz
 3x2  3yzl  3y2 3xz  j  3z2 3xyk
  
i j k
  
3 2 3 3 2 3 3 2 3
F
x y z
x yz y xz z xy
  
  
  

  
 i  3z2  3xy   3y2  3xz
 y  z
 
x z

   
k 3y2 3xz 3x2 3yz
       x  y


  
i3x  3x j 3y  3y k 3z  3z
  
i 0  j0 k 0 0
.
  
Problem 10 If F  x2 i  y2 j,
 
evaluate  F  dr
along the straight line y  x from
0,0 to 1,1 .
Solution:
    
F.d r  x2 i  y2 jdxi  dy
 x2dx  y2dy
Given y  x
dy  dx
  1
 
 F d r   x 2 dx  y 2
dy
C 0
1 1 3 1
2 2 2
0 0 0
Problem 11 What is the unit normal to the surface   x, y, z  C at the point x, y, z?
Solution:

5. Unit.2 Vector Calculus
  
  
5
n





.

to be solenoidal
Problem 12 State the condition for a vector F
Solution:
 
.F  divF  0
Problem 13 If a 

?
is a constant vector what is  a
Solution:
Let 1 2 3 a  a i  a j  a k
   
  
 
  
   
1 2 3
0
i j k
a
 x  y 
z
a a a
Problem 14 Find grad  at 2,2,2 when   x2  y2  z2  2
Solution:
grad  
i x2 y2x2 2 j x2 y2 z2 2 k x2 y2 z2 2
x y z
          
  
  
 2x i  2y j  2zk
  
2,2,2   4i  4 j  4k
Problem 15 State Gauss Divergence Theorem
Solution:
The surface integral of the normal component of a vector function F over a closed

surface S enclosing volume V is equal to the volume integral of the divergence of F
  
  F nds     FdV
taken over V. i.e., . .
S V
Part –B
Problem 1 Find the directional derivative of   x2 yz  4xz2 at the point 1,2,1 in
  
the direction of the vector 2i  j  2k.
Solution:
  x2 yz  4xz2

6. Unit.2 Vector Calculus
  
                   
   
     
6
  
  2xyz  4z2 i  x2z j  x2 y 8xzk
       2  2    2     
1, 2, 1  2 1 2 1 4 1 i 1 1 j 1 2 8 1 1 k  
  
 4  4i  j  2 8k
  
 8i  j 10k
Directional derivative a 

is .a 




     
8 i  j  10 k .2 i  j  2 k

4 1 4

 
16  1 
20 37 .
 
3 3
Problem 2 Find the maximum directional derivative of   xyz2 at 1,0,3 .
Solution:
Given   xyz2
  
  yz2 i  xz2 j  2xyzk
   2   2    
1,0,3   0 3 i  1 3 j  2 1 0 3 k  9 j

Maximum directional directive of  is   9 j
Magnitude of maximum directional directive is   92  9.
Problem 3 Find the angle between the surfaces x2  y2  z2  9 and x2  y2  z  3at the
point 2,1, 2 .
Solution:
Let 2 2 2
1   x  y  z  9
  
1   2xi  2y j  2zk
      1 2, 1,2  2(2)i 2 1 j 2 2 k 4i 2 j 4k         
2 2
2   x  y  z  3
  
2   2xi  2y j  k
  
22, 1,2  4i 2 j 2k     
If  is the angle between the surfaces then
 
cos .
1 2
 
 

|  || 
|
1   2
 4 i  2 j  4 k   
.4 i  2 j  2 k

16 4 16 16 4 4

   

7. Unit.2 Vector Calculus
   
7
16  4 
8
36 24

12 1
6 2 6 6
 

cos 1 1 .
    
6
 
  
Problem 4 Find the work done, when a force F  x2  y2  xi  2xy  y j
moves a
particle from the origin to the point 1,1 along y2  x .
Solution:
 Given F   
x2  y2  xi  2xy  y j
   
F.d r  x2  y2  xdx  2xy  ydy
d r  dxi  dy j  dzk
 
Given y2  x
2ydy  dx
 
F.dr  x2  x  xdx  2y3  ydy
 x2dx  2y3  ydy
  1 1
 Fdr   x 2 dx   y 3
 y dy
0 0
2
C

3 1 4 2 1
 x   
   2
y  y  3   
 0  4 2

0
1 0 2 1 0 0
                3   4 2
  
1 1 1
3 2 2
      
 1 
 1 
2
3 3
 
Work done . 2
  F dr 
3 C
Problem 5 Prove that F   y2 cos x  z3 i  2y sin x  4 j  3xz2 k
is irrotational and
find its scalar potential.
Solution:
Given F   y2 cos x  z3 i  2y sin x  4 j  3xz2 k
   

8. Unit.2 Vector Calculus

      
y2 cos x z3 i 2y sin x 4j 3xz2 k i j
  
     
8
  
i j k
  
2 cos 3 2 sin 4 3 2
F
x y z
y x z y x xz
  
  
 

  
 i0  0 j 3z2 3z2   k2y cos x  2y cos
  
 0i  0 j  0k  0

 F  0

is irrotational
Hence F
F  
x y yz
 
  
Equating the coefficient i, j, k
  
y2 cos x z3 y2 cos x z3dx

x


     
2 3
1 1   y sin x  z x C
2y sin x 4 2y sin x 4dy

  
x


     
  x y  y C
 
2
2 2 2 sin 4
2
  
3xz2 3xz2dy

x


   
3
  3
x z C
3 3
3   y2 sin x  xz3  4y C
Problem 6   
If F  3xyi  y2 j
 
evaluate  F.dr
when C is curve in the xy plane
y  2x2 , from 0,0 to 1,2
Solution:
F  3xyi  y2 j
  
dr  dxi  dy j  dzk
   
F.dr  3xydx  y2dy
 
Given y  2x2
dy  4xdx
 
 F.d r  3x(2x2 )dx  2x2 2 4x dx
 6x3dx  4x4 (4x)dx
 6x3dx 16x5dx

9. Unit.2 Vector Calculus
  
 
C F d r   x dx   aydy   x  b dx  
 a 3   b 2    a 3
 
     a          ab
 
       
9
  1
 Fd r   x 3  x 5
dx
0
6 16
C
 
4 6 1
 x x 
   
 
6 16 7 .
4 6 6
0
6 16
4 6

   
Problem 7 Find .
 F dr
C
 
when F  x2  y2 i  2xy j
where the cure C is the
rectangle in the xy plane bounded by x  0, x  a, y  b, y  0.
Solution:
 Given F x2  y2  
i  2xy j
   
Fdr  (x2  y2 )dx  2xy dy
d r  dxi  dy j  dzk
 
C is the rectangle OABC and C consists of four different paths.
OA (y = 0)
AB (x = a)
BC (y = b)
CO (x = 0)
 
 F .
d r        
C OA AB BC CO
Along
OA, y  0, dy  0
AB, x  a, dx  0
BC, y  b, dy  0
CO, x  0, dx  0
. 2 2  2 2  0
C OA AB BC CO
  0
a b
  x 2 dx  2
a ydy  x 2  b 2
dx
0 0
a
a b o
 x 3   y 2   3

  a  x       b 2
x

     
0
2
3 2 3
o a
 
0 2 0 0 0 2
3 2 3
 2ab2.
  
Problem 8 If F  4xy 3x2 z2 i  2x2 j  2x3zk
 F dr
check whether the integral .
C
 
is
independent of the pathC .
Solution:
Given

10. Unit.2 Vector Calculus
10
  
F  4xy 3x2 z2 i  2x2 j  2x3zk
   
d r  dxi  dy j  dzk
 
. 4 3 2 2  2 2 2 3
C C C C
 F dr   xy  x z dx   x dy   x zdz
This  integral is independent 
of path of integration if
F    F  0
  
i j k
  
4 3 2 2 2 2 2 3
F
x y z
xy x z x x z
  
  
 

 
 i 0,0  j 6x2 z  6x2 z k 4x  4x
   
 0i  0i  0 j  0k  0.
Hence the line integral is independent of path.
Problem 9 Verify Green’s Theorem in a plane for  2 (1 ) ( 3 3 ) 
 x  y dx  y  x dy where
C
C is the square bounded x   a, y   a
Solution:
Let P  x2 (1 y)

P 
x2
y

Q  y3  x3

Q 
3x2
x

By green’s theorem in a plane
 
C C
   
Pdx Qdy Q P dxdy
    
    x  y
  Now
   
        
R
Q P dx dy
x y
3 2 2 
a a
   
a a
x x dx dy
 
2 2
a a
  
 
dy x dx
a a
 
2 3
3
a
a
a
a
y x 

 
  
 
  2  3 3 
 a  a a  a
3

11. Unit.2 Vector Calculus
      
Pdx Qdy x y dx x y dy
      
Pdx Qdy x y dx x y dy

      
Pdx Qdy x y dx x y dy
11
8 4 (1)
3
 a 
Now  
 Pdx Qdy        
C AB BC CD DA
Along AB, y  a, dy  0
X varies from a to a
a
   2 1   3 3  
AB 
a
   
 
2 (1 ) 0
a
a
x a dx

3
1
3
a
a
a x

 
   
 
  3 4
 1   a    a 3  a 3  2 a 
2
a  3 
3 3
Along BC
x  a, dx  0
Y varies from  a to a
a
   2 1   3 3  
BC 
a
  
( 3 3 )
a
a
a y dy

4
3
4
a
a
a y y

 
   
 
 4   4

        
   
a a a a a
4 4 2 4
4 4
Along CD
y  a, dy  0
X varies from a to a
a
   2 1   3 3  
CD a

  
 
2 (1 )
a
a
x a dx
 3

 1
a x dx
3
a
a
   
 
  
a a a
 
3 3
1
   
3
 

12. Unit.2 Vector Calculus

      
Pdx Qdy x y dx x y dy
3 4 3 4
 Pdx Qdy  2 a  2 a  2 a 4  2 a  2 a  2 a
4
3 3 3 3 C
            
Hence Green’s theorem verified.
Problem 10 Verify Green’s theorem in a plane for
3 2 8 2  4 6 
C
12
2 3 2 4
3 3
  a  a
Along DA,
x  a, dx  0
Y Varies from a to a
a
   2 1   3 3  
DA a
 2 1   3 3  
a
    
a
a y dx y a dy


 4

   3
 4


a
a
y a y
 4   4

        
   
a a a a a
4 4 2 4
4 4
 
4 4 4 4
 a  a
3
 8 a
4 ......(2)
3
From (1) and (2)
 
   
Pdx Qdy Q P dxdy 8 a
4 .
x y
3 C R
 x  y dx  y  xy dy where C is the boundary of the region defined by
x  y2 , y  x2.
Solution:
Green’s theorem states that
   
udx vdy v u dxdy
           
C R
x y
   
Pdx Qdy Q p dxdy
    
    x  y
 C R
 Given 3 2 8 2  4 6 
 x  y dx  y  xy dy
C
P  3x2 8y2

13. Unit.2 Vector Calculus
 Pdx Qdy   x  x dx  x  x xdx
 Pdx Qdy   y  y y dy  y  y dy
13
P 16y
y

 

Q  4y  6xy

Q  
6y
x

Evaluation of
 Pdx Qdy
C
(i) Along OA
y  x2 dy  2xdx
3 2 8 4  4 2 6 3 2
OA OA
  1
  3x 2 8x 4  8x 3 12x 4
dx
0
1
    20x 4  8x 3  3x 2
dx
0
5 4 3 1
0
 
 x x  20  8  3
x 
 5 4 3

20 8 3
5 5 3

  
  4  2 1  1
Along AO
y2  x2ydy  dx
3 4 8 2 2 4 6 3 
Ao Ao
  (6 y 5  16 y 3  4 y  6 y 3 )
dy
AO
0
    6y 5  22y 3
 4y dy
1
6 4 2 0
 y y y 
    
 
1
6 22 4
6 4 2
0
11 2 5
2 2
 y 6  y 4  y 2
    
1
1 5 3 (1)
 Pdx Qdy         
2 2 C OA AO
Evaluation of
   
        
R
Q P dx dy
x y
   
             
v u dxdy  6 y 16 y 
dxdy
x y
R R

14. Unit.2 Vector Calculus
14
 
y
1 1
2
10 10
0 0
x y
x y
y
y dx dy xy dy 
     
  1
 10y y  y 2
dy
0
1 3

2 3
  10 y y dy
0
 
   
1
 5

 2 4

0
y y
10 5 4
   
 
 2

10 2 1
     5 4

10 8 5
      20

  30  3 
(2)
20 2
For (1) and (2)
Hence Green’s theorem is verified.
Problem 11 Verify Gauss divergence theorem for F  yi  x j  z2 k
   
over the cylindrical
region bounded by x2  y2  9, Z  0 and Z  2 .
Solution:
Gauss divergence theorem is
  
 F .
n ds   divFdV
S V
  
   
  
div F  y  x z2  2z
x y z

2
3 9 
2
   
div F dV z dzdy dx
3 9 2
0
2
x
V   
x

3 9 2
2 2
   
3 9 2
0
2
2
x
x
z dydx

  
 
  
2
  
2
= 4 (Area of the circular region)
 4    3
2   36 .................(1)
3 9
3 9
4
x
x
dydx

  

15. Unit.2 Vector Calculus
15

 F n ds    
1 2 3
.
S S S S
S is the bottom of the circular region, S is the top of the circular region and S is the
1 2 3 cylindrical  region

On S , n   k, ds  dxdy, z  0
1  F n ds   z dxdy 
1
. 2 0
S
 
 
On 2 S , n  k
, ds  dxdy, z  2
 F n ds   z dx dy
2
. 2
s

 4 dxdy
 4 (Area of circular region)
  2   4  3  36
On 3 S ,   x2  y2  9

n xi y j
2 2
4
 
 
 
 x 2 y
2 


 
 
xi  y j

3
      
  
F n ds yi x j z k xi y j ds
 
3
. 2
    
3 S
 
 
yx 
yx ds 2
xy ds    
3 3 S
Let x  3cos , y  3sin
ds  3 d dy
 varies from 0 to 2
z varies from 0 to 2
2  9sin cos 
3
3
2 2
0 0
d dz

     
2 2
18 sin 2
2
0 0
d dz

    
2 2
 cos 2
    9   
2
0 0
dz

  2
 
9 1 1 0
2
    dz 
0
. 0 36 0 36 ...............(2)
 F n ds      
S
 

16. Unit.2 Vector Calculus
   
 
16
from (1) and (2)
  
 F .
n ds   div FdV
C V
Problem 12   Verify 
Stoke’s theorem for the vector field defined by
F  x2  y2 i  2xy j
in the rectangular region in the xy plane bounded by the lines
x  0, x  a, y  0, y  b.
Solution:
  
F  x2  y2 i  2xy j
   
 F dr  curl F n ds
By Stoke’s theorem . .
C S
  
i j k
  
2 2 2 0
curlF
x y z
x y xy

  
 

 i0  0 j 0  0 k 2y  2y  4yk
As the region is in the xy plane we can take n  k and ds  dxdy
curl F . n ds    4 yk .
k dx dy
S
  
0 0
4
b a
    y dx dy
 
 2

0
0
4
2
b
y x a
   
 
 2ab2..............(1)

 F .
dr        
C OA AB BC CO
Along OA
y  0dy  0,
x varies from 0 to a
 2 2 
0
    x  y dx  2
xy dy
a
OA
a  x 
a x dx
a 3 3

2
      
0 0 3 3
Along AB
x  adx  o, y varies from 0 to b

17. Unit.2 Vector Calculus
   
17
 2 2 
0
   a  y .0  2
ay dy
b
AB
 y 2

b a ab
2
     
0
2
2
 
Along BC
y  b, dy  0
x varies from a to 0
  0
   x  b dx 
2 2 0
BC a
3 0
 
 x   b 2
x

 
3 a
3
2
  a  ab
3
Along CO
x  0, dx  0,
y varies from b to 0
  0
    y  
0 2 0 0 0
CO b
3 3
 
 F dr  a  ab  a  ab 
. 2 2 0
3 3 c
 2ab2.........(2)
For (1) and (2)
   
 F . d r  curlF .
nds
C S
Here Stoke’s theorem is verified.
Problem 13 Find .
 F nds
S
 
if F  x  y2 i  2x j  2yzk
where S is the surface in
the plane 2x  y  2z  6 in the first octant.
Solution:
Let   2x  y  2z  6 be the given surface
  
Then   2i  j  2k
     
 i j k i j k
2   2 2  
2
4 1 4 3
  
 
 The unit outward normal n 
to the surface S is  1 2 2
n   i  j  k 
3
  
Let R be the projection of S on the xy plane

18. Unit.2 Vector Calculus
       
F n   x  y i  x j  yzk  i  j  k
  x  y dx  x  xy  where C is the boundary of the
18
   
F . n ds F .
n dx dy
S R .
1
n k
  
 
     
. 1 2 2 . 2
n k  i  j  k k 
3 3
.  2  2 2 .1 2 2 
3
2  2  2 4
3 3 3
 x  y  x  yz
2  2 2 
3
 y  yz
2  2 
3
 y y  z
2  6 2 
3
 y y   y  x
2 6 2 
3
 y  x
4 3 
3
 y  x
. .
   
F n ds F n dx dy
 
n k
  
S S .
  y  x dxdy
4 3 
3 2 / 3
 
1
   x dx dy
2 3
R
 
3 6 2
2 3
0 0
x
x dx dy
 
   
x x y dx
3 2 6 2
 
  
 
2 3
   
   
2
0 0
3
3
 4  3  x dx
0
4 3
0
  x 
4 (3 )
    4
 
 81units.
Problem 14 Evaluate   2 3 
C
triangle with vertices 2,0,0,0,3,0&0,0,6 using Stoke’s theorem.
Solution:

19. Unit.2 Vector Calculus
 

            
19
  
Stoke’s theorem is . .
 F dr  curlF nds
C S
where S is the surface of the triangle and n
is
the unit  vector 
normal to surface S .
Given F.dr   x  ydx  2x  z dy   y  z dz
   
dr i dx  jdy  k dz
F  x  yi  2x  z j   y  zk
   
  
i j k
  
2
curlF
x y z
x y x z y z

  
  

  
 i 11  j 0  0  k 2 1
  
curl F  2i  k
Equation of the plane ABC is
x  y  z 
1
2 3 6
3x  2y  z  6
Let   3x  2y  z  6
  
  3i  2 j  k
Unit normal vector to the surface ABCor  is
 3 2
n i j k 
  
 

14

  
     
    
curl F n i k i j k
.  2 . 3 2 6 1 7
     
14 14 14
 

Hence .  7
curlF nds   ds
S S 14

7
14 R
dxdy
nk
   where R is the projection of surface ABC on XOY plane
7 3 2 . 1 14 1 14 14
14
R
dxdy n k i j k k
    
 7
dxdy
R
 7 Area of le OAB
 73  21.

20. Unit.2 Vector Calculus
Problem 15 Verify Stoke’s theorem for F   y  zi  yz j  xzk
20
   
where S is the
surface bounded by the planes x  0, x 1, y  0, y 1, z  0 and z 1 above the
XOY plane.
Solution:
  
Stoke's theorem is . .
 F d r   F nds
C S
   
F   y  zi  yz j  xzk
  
i j k
F
  
x y z
y z yz xz
  
  
 

  
 yi  z 1 j  k


 F n ds      
1 2 3 4 5
.
S S S S S S
6 S  is not applicable, since the given condition is above the XOY plane.
 
   yi  z  j  k  idydz
1
1 .
S AEGD
   
  ydy dz
AEGD
1 1 1 2 1
 
y dy dz y dz
    
 1
    
0 0 0 0 2
1 1
2 2
  z  
0
   
   yi  z  j  k  i dydz
2
1 .
S OBFC
   
1 1 1 2 1
 
y dy dz y dz
    
    
0 0 0 0
1
2 2
 
   yi  z  j  k  jdxdz
3
1
S EBFG
   
1 1 1
   
1
0
   z 1 dx dz   xz  x dz
0 0 0
2 1
 
       
 
0
1 1 1
z z
2 2 2
   
   yi  z  j  k   j dxdz
4
1
S OADC
   

21. Unit.2 Vector Calculus
   y  z  dy  dz 
 
   x  z  dx  dz 
 
21
 
1 1
   z 1 dx dz
0 0
1 1
   
1
0
  xz  x   z 1 dz
0 0
2 1
   
      
 
0
   
1 1 1
z z
2 2 2
   yi  z  j  k kdxdy
5
1 .
S DGFC
   
1 1 1
   
1
0
   1 dxdy   x dy
0 0 0
1
   
1
0
  1 dy  y  1
0
      
S S1 S2 S3 S4 S5
1 1 1 1 1 1
2 2 2 2
       
 
L . H . S   F .
dr        
C OA AE EB BO
 
   y  z dx  yzdy  xzdz
OA OA
0 0  0, 0, 0, 0
OA
   y  z dx  yzdy  xzdz
AE AE
0 0  1, 0, 0, 0
AE
   y  z dx  yzdy  xzdz
EB EB
 
0
 1 dx y 1, z  0,
1
 0
1  x  0 1  1
 
   y  z dx  yzdy  xzdz
BO BO
0 0  0, 0
   x  z 
BO
 0
       
 0  0 1 0  1
C OA AE EB BO

22. Unit.2 Vector Calculus
22
L.HS = R.HS.
Hence Stoke’s theorem is verified.
z C (0,0,1) F
D G
O y
B(0,1,0)
x A (1,0,0) E (1,1,0)