Vector Integration

Vector Line Integrals

Consider a curve C in space which consist of infinitesimally small line elements of length dr.
Then the line integral of a vector 𝐴(𝑥, 𝑦, 𝑧) along the curve C is defined to be the sum of

the scalar products of 𝐴 and 𝑑𝑟 and is represented by 𝐴 𝑑𝑟 .

If C is a closed curve which do not intersect anywhere, the line integral around C is denoted

by 𝐹 . 𝑑𝑟

If 𝐹 is the force acted upon by a particle in displacing it along the curve C then represents

the 𝐹 𝑑𝑟

represents the total work done by the force. It also represents the circulation of 𝐹 about
C where 𝐹 represents the velocity of the fluid.

𝐹 is said to be irrotational if 𝐹 . 𝑑𝑟 = 0
Problems:
  
1. If F  x y i  y z j  z x k , Evaluate  F.d r where C is the curve represented by
c
x  t , y  t 2 , z  t 3 ,  1  t  1.
 
Soln: we have F  x y i  y z j  z x k and r  x i  y j  z k will give

d r  d xi  d y j  d z k
2 3
Since, x  t , y  t , z  t by data, we obtain dx  dt, dy  2t dt, dz  3t 2 dt

 
 
 F . d r  t 3 dt  t 5 2 t  dt  t 4 3t 2 dt

 
  
F . d r  t 3  2 t 6  3 t 6 dt  t 3  5 t 6 dt 
1 1
 
  1 t 4  t 7   1 1   1 1  10
 F . d r   t  5 t dt   4   5 7    4  4   5 7  7   7
3 6

C t  1   1   1
  
2. Evaluate  F . d r where
C


F  x y i  x 2  y 2 j along 
(i) The path of the straight line from 0,0to 1, 0 and then to 1,1

(ii) The straight line joining the origin and 1, 2

Soln: 

F . d

r   x y d x  x 2
 y 2
 
d y.......... .......... ....1
C C
     
 F .d r   F . d r   F . d r .......... .......... .......... .2
C OA AB

Along O A : y  0  d y  0 and 0  x  1
 
From (1),  F .d r  0.......... .......... .......... .......... ......... 3
OA

Along A B : x  1  d x  0 and 0  y  1
3 1
  
 
1
 1   .......... .....4
y 1 4
From (1),  F . d r   0  1  y 2 d y   y  
AB y 0  3 
y 0
3 3
  4 4
Using 3 and 4 in 2 we obtain  F .d r  0  
3 3
C
(ii) C is the straight line joining 0, 0 and 1, 2

The equation of the line is given by y  0  2  0

x0 1 0

y  2 x  dy  2 d x and x var ies from 0 to1

 F . d r   x y d x  x 
 
Hence from (1),
2
 y 2 d y.......... .......... ....1
C C

 
  1
 F .d r  
2 2
x .2 x d x  x  4 x 2d x
C x 0

 1
 1  x3 
Thus  F . d r   12 x dx  12   4
2
x 0
C  3  0
  
   
3. If F  x i  x y j evaluate  F . d r
2
where F  x y i  x 2  y 2 j along
C
(i) The line 𝑦 = 𝑥
(ii) The parabola 𝑦 = 𝑥
 

2
Soln: F . d r  x d x  x yd y
C
(i) Along y  x : we have 0  x  1 and dy  dx
1
  1 1 1  2 x3  2
   
2 2 2
F . d r  x d x  x d x  2 x d x    
C x 0 x 0 x 0  3  x  0 3

(ii) Along y  x : y 2  x and 2 y d y  d x, 0  y  1

1 1
  1 1  y6   y4  1 1 7
  
5 3
F . d r  2 t d y  y d y         
C y 0 y 0   y  0   y  0
3 4 3 4 12
 
4. Find the total work done by the force represented by F  3 x y i  y j  2 z x k
in moving a particle round the circle x 2  y 2  4
 
Soln: Total work done, W   F . d r
C
x 2  y 2  4 can be represented in the parametric form

x  2 cos , y  2 sin  and z  0. 0    2

 
W   F . d r   3 x y d x  y d y  2z x d z
C
2 2
W  3 4 cos sin  2 sin  d   4 sin cos d
0 0

2 2  sin 3     cos 2 
 24  sin  cos d  2  sin 2 d  24
2
  2 0
0 0  3   2 

Thus the total work done is 0.