181202(MA 2161) – MATHEMATICS - II UNIT 2 notes




Unit.2 Vector Calculus
1
UNIT II VECTOR CALCULUS
Part-A
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
 
2 3 4 2
i j k
F
x y z
x y az bx y z x cy z
  
  
  
     
  

       
   
4 2 3 4 2 2
3 2 2
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
x y
                             
   
          
 

ic 1 j a  4 k b  2
  
 F  0 0i  0 j  0k
   
 ic 1 j a  4 k b  2
  
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
S
  r n ds  V

Unit.2 Vector Calculus
2
Solution:
Let r  xi  y j  zk
   
By Gauss divergence theorem

S V
  F n ds     FdV
 
Here F  .r
 

S V
  r  n ds     rdV
 
. 
V
i j k xi y j zk dV
x y z
    
            
     
   111dV
 3 .
S
  r n ds  V

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
                                         
Unit.2 Vector Calculus
3
2 2 2 2 2 2
3 2 1 3 2 1
x y x z y z y x z x z y
                 
                            
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 




4 12 4
16 144 16
 i  j  k

 
  
4 3 
176
 i  j  k
 
  
4 3   3 
.
16 11 11
 i  j  k  i  j  k
 

     
Problem 7 Applying Green’s theorem in plane show that area enclosed by a simple
closed curve C is 1  
2
 xdy  ydx
Solution:
C R
Pdx Qdy Q P dx dy
x y
   
           
P  y, Q  x
P 1, Q 1
y x
 
  
 
  1 1 2
R R
 x dy  ydx     dxdy   dx dy
= 2 Area enclosed by C
Area enclosed by C = 1  
2
 xdy  ydx .
Problem 8 If A

and B

are irrotational show that A B
 
is solenoidal
Solution:
Given A

is irrotational i.e.,  A  0
 
Unit.2 Vector Calculus
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
  
3 2 3 3 2 3 3 2 3
i j k
F
x y z
x yz y xz z xy
  
  
  
  
  

i 3z2 3xy 3y2 3xz
y z
   
        
 j 3z2 3xy 3x2 3yz
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
2 2
C 0
 F d r   x dx  y dy
 
1 1 3 1
2 2 2
0 0 0
2 2 2
3 3
x dx x dx x dx x
 
      
   
Problem 11 What is the unit normal to the surface   x, y, z  C at the point x, y, z?
Solution:
Unit.2 Vector Calculus
5
n





.
Problem 12 State the condition for a vector F

to be solenoidal
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

taken over V. i.e., . .
S V
  F nds     FdV
  
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
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 10 .2 2 
4 1 4
i  j  k i  j  k

 
     
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 2 4 .4 2 2 
16 4 16 16 4 4
i  j  k i  j  k

   
     
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
  
d r  dxi  dy j  dzk
   
F.d r  x2  y2  xdx  2xy  ydy
 
Given y2  x
2ydy  dx
F.dr  x2  x  xdx  2y3  ydy
 
 x2dx  2y3  ydy
  1 1
2 3
0 0
2
C
 Fdr   x dx   y  y dy

3 1 4 2 1
0 0
2
3 4 2
 x   y y 
      
   
1 0 2 1 0 0
3 4 2
                    
1 1 1
3 2 2
      
1 1 2
3 3

  
Work done . 2
3 C
  F dr 
 
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
   
Unit.2 Vector Calculus
8
2 cos 3 2 sin 4 3 2
i j k
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

Hence F

is irrotational
F  

 y2 cos x z3 i 2y sin x 4 j 3xz2 k i j
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



     
  
 
2
2 2 2 sin 4
2
  x y  y C
3xz2 3xz2dy
x



   
  
3
3 3 3
3
  x z C
  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
Unit.2 Vector Calculus
9
  1
3 5
0
6 16
C
 Fd r   x  x dx
 
4 6 1
0
6 16
4 6
 x x 
   
 
6 16 7 .
4 6 6

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

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

      
2 (1 ) 0
a
a
x a dx

   
 
3
1
3
a
a
a x

 
   
 
  3 4
1 3 3 2 2
3 3 3
a a a a a         
 
Along BC
x  a, dx  0
Y varies from  a to a
   2 1   3 3  
a
BC a
Pdx Qdy x y dx x y dy

      
( 3 3 )
a
a
a y dy

  
4
3
4
a
a
a y y

 
   
 
4 4
4 4 2 4
4 4
a a a a a
   
        
   
Along CD
y  a, dy  0
X varies from a to a
   2 1   3 3  
a
CD a
Pdx Qdy x y dx x y dy

      
2 (1 )
a
a
x a dx

  
 
3
1
3
a
a
a x dx
  
   
 
 
3 3
1
3
a a a
  
   
 
Unit.2 Vector Calculus
12
2 3 2 4
3 3
  a  a
Along DA,
x  a, dx  0
Y Varies from a to a
   2 1   3 3  
a
DA a
Pdx Qdy x y dx x y dy

      
 2 1   3 3  
a
a
a y dx y a dy


    
4
3
4
a
a
y a y
 
   
 
4 4
4 4 2 4
4 4
a a a a a
   
        
   
 
3 4 3 4
2 2 2 4 2 2 2 4
3 3 3 3 C
 Pdx Qdy  a  a  a  a  a  a
4 4 4 4
3
 a  a
8 4 ......(2)
3
 a
From (1) and (2)
 
8 4 .
3 C R
Pdx Qdy Q P dxdy a
x y
   
            
Hence Green’s theorem verified.
Problem 10 Verify Green’s theorem in a plane for
3 2 8 2  4 6 
C
 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
C R
udx vdy v u dxdy
x y
   
           
C R
Pdx Qdy Q p dxdy
x y
   
           
Given 3 2 8 2  4 6 
C
 x  y dx  y  xy dy
P  3x2 8y2
Unit.2 Vector Calculus
13
P 16y
y

 

Q  4y  6xy
Q 6y
x

 

Evaluation of
C
 Pdx Qdy
(i) Along OA
y  x2 dy  2xdx
3 2 8 4  4 2 6 3 2
OA OA
 Pdx Qdy   x  x dx  x  x xdx
  1
2 4 3 4
0
  3x 8x  8x 12x dx
  1
4 3 2
0
  20x  8x  3x dx
5 4 3 1
0
20 8 3
5 4 3
 x x x 
    
 
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
 Pdx Qdy   y  y y dy  y  y dy
(6 5 16 3 4 6 3 )
AO
  y  y  y  y dy
  0
5 3
1
  6y  22y  4y dy
6 4 2 0
1
6 22 4
6 4 2
 y y y 
    
 
0
6 4 2
1
11 2 5
2 2
  y  y  y    
1 5 3 (1)
2 2 C OA AO
 Pdx Qdy         
Evaluation of
R
Q P dx dy
x y
   
        
 6 16 
R R
v u dxdy y y dxdy
x y
   
             
Unit.2 Vector Calculus
14
 
2
1 1
0 0
10 10
y
x y
x y
y
y dx dy xy dy 
     
  1
2
0
 10y y  y dy
1 3
2 3
0
10 y y dy
 
   
  
1
5
2 4
0
10 5 4
2
y y
 
 
   
 
 
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
.
S V
 F n ds   divFdV
  
div F  y  x z2  2z
x y z
  
   
  

2
2
3 9 2
3 9 0
2
x
V x
div F dV z dzdy dx

  
   

2
2
3 9 2 2
3 9 0
2
2
x
x
z dydx

  
 
  
   
2
2
3 9
3 9
4
x
x
dydx

  
  
= 4 (Area of the circular region)
  2   4  3
 36 .................(1)
Unit.2 Vector Calculus
15
1 2 3
.
S S S S
 F n ds    

1 S is the bottom of the circular region, 2 S is the top of the circular region and 3 S is the
cylindrical region
On 1 S , n   k, ds  dxdy, z  0
 
1
. 2 0
S
 F n ds   z dxdy 
 
On 2 S , n  k
 
, ds  dxdy, z  2
2
. 2
s
 F n ds   z dx dy

 4 dxdy
 4 (Area of circular region)
  2   4  3  36
On 3 S ,   x2  y2  9

 2 2 
2 2
4
n xi y j
x y


 
 
 
 
3
xi  y j

 
 
3
. 2
3 S
F n ds yi x j z k xi y j ds
  
    
 
 
      
2
3 3 S
yx yx ds xy ds 
   
Let x  3cos , y  3sin
ds  3 d dy
 varies from 0 to 2
z varies from 0 to 2
 
2 2
0 0
2 9sin cos 3
3
d dz

     
2 2
0 0
18 sin 2
2
d dz

    
2 2
0 0
9 cos 2
2
dz
       
  
 
2
0
9 1 1 0
2
    dz 
. 0 36 0 36 ...............(2)
S
 F n ds      
 
Unit.2 Vector Calculus
16
from (1) and (2)
.
C V
 F n ds   div FdV
  
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
  
By Stoke’s theorem . .
C S
 F dr  curl F n ds
   
2 2 2 0
i j k
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
 
. 4 .
S
curl F n ds    yk k dx dy
  
0 0
4
b a
    y dx dy
 
2
0
0
4
2
b
y x a
 
   
 
 2ab2..............(1)
.
C OA AB BC CO
 F dr        

Along OA
y  0dy  0,
x varies from 0 to a
 2 2 
0
2
a
OA
    x  y dx  xy dy
3 3
2
0 0 3 3
a x a a x dx
 
    
  
Along AB
x  adx  o, y varies from 0 to b
Unit.2 Vector Calculus
17
 2 2 
0
.0 2
b
AB
   a  y  ay dy
2
2
0
2
2
b a y ab
 
     
 
Along BC
y  b, dy  0
x varies from a to 0
  0
2 2 0
BC a
   x  b dx 
3 0
2
3 a
x b x
 
   
 
3
2
3
  a  ab
Along CO
x  0, dx  0,
y varies from b to 0
  0
0 2 0 0 0
CO b
    y  
3 3
. 2 2 0
3 3 c
 F dr  a  ab  a  ab 
 
 2ab2.........(2)
For (1) and (2)
. .
C S
 F d r  curlF nds
   
Here Stoke’s theorem is verified.
Problem 13 Find .
S
 F nds
 
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
  
2 2 2 2
4 1 4 3
 i j k i j k
   
  
 
     
 The unit outward normal n

to the surface S is  1 2 2
3
n   i  j  k 
  
Let R be the projection of S on the xy plane
Unit.2 Vector Calculus
18
1
. .
S R .
F n ds F n dx dy
n k
  
   
 
. 1 2 2 . 2
3 3
n k  i  j  k k 
     
.  2  2 2 .1 2 2 
3
F n   x  y i  x j  yzk  i  j  k
       
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
. .
S S .
F n ds F n dx dy
n k
  
   
 
4 3 
3 2 / 3
  y  x dxdy
 
1
2 3
R
   x dx dy
 
3 6 2
0 0
2 3
x
x dx dy
 
   
 
3 2 6 2
0 0
2 3
2
x x y dx
   
   
  
 
3
3
0
 4  3  x dx
4 3
0
4 (3 )
4
  x 
     
 81units.
Problem 14 Evaluate   2 3 
C
  x  y dx  x  xy  where C is the boundary of the
triangle with vertices 2,0,0,0,3,0&0,0,6 using Stoke’s theorem.
Solution:
Unit.2 Vector Calculus
19
Stoke’s theorem is . .
C S
 F dr  curlF nds
  
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
 
F  x  yi  2x  z j   y  zk
   
dr i dx  jdy  k dz
   
2
i j k
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
1
2 3 6
x  y  z 
3x  2y  z  6
Let   3x  2y  z  6
  3i  2 j  k
  
Unit normal vector to the surface ABCor  is
 3 2
14
n i j k 

  
 

  
.  2 . 3 2 6 1 7
14 14 14
curl F n i k i j k
    
     
 
     
Hence .  7
S S 14
curlF nds   ds


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
R
 dxdy
 7 Area of le OAB
 73  21.
Unit.2 Vector Calculus
20
Problem 15 Verify Stoke’s theorem for F   y  zi  yz j  xzk
   
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 . .
C S
 F d r   F nds
  
F   y  zi  yz j  xzk
   
i j k
F
x y z
y z yz xz
  
  
  
 
  

 yi  z 1 j  k
  

1 2 3 4 5
.
S S S S S S
 F n ds      

6 S
 is not applicable, since the given condition is above the XOY plane.
 
1
1 .
S AEGD
   yi  z  j  k  idydz
   
AEGD
  ydy dz
1 1 1 2 1
0 0 0 0 2
y dy dz y dz
 
    
    
 1
0
1 1
2 2
  z  
   
2
1 .
S OBFC
   yi  z  j  k  i dydz
   
1 1 1 2 1
0 0 0 0
1
2 2
y dy dz y dz
 
    
    
 
3
1
S EBFG
   yi  z  j  k  jdxdz
   
   
1 1 1
1
0
0 0 0
   z 1 dx dz   xz  x dz
2 1
0
1 1 1
2 2 2
z z
 
       
 
   
4
1
S OADC
   yi  z  j  k   j dxdz
   
Unit.2 Vector Calculus
21
 
1 1
0 0
   z 1 dx dz
   
1 1
1
0
0 0
  xz  x   z 1 dz
2 1
0
1 1 1
2 2 2
z z
   
      
 
   
5
1 .
S DGFC
   yi  z  j  k kdxdy
   
   
1 1 1
1
0
0 0 0
   1 dxdy   x dy
   
1
1
0
0
  1 dy  y  1
S S1 S2 S3 S4 S5
      
1 1 1 1 1 1
2 2 2 2
       
. . .
C OA AE EB BO
L H S   F dr        
 
 
OA OA
   y  z dx  yzdy  xzdz
0 0  0, 0, 0, 0
OA
   y  z  dy  dz 
 
AE AE
   y  z dx  yzdy  xzdz
0 0  1, 0, 0, 0
AE
   x  z  dx  dz 
 
EB EB
   y  z dx  yzdy  xzdz
 
0
1
 1 dx y 1, z  0,
 0
1  x  0 1  1
 
BO BO
   y  z dx  yzdy  xzdz
0 0  0, 0
BO
   x  z 
 0
C OA AE EB BO
       
 0  0 1 0  1
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)
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