Vector Field
Definition:
1. A vector field in R2 is a vector-valued function with domain being a subset of R2 and
the images being vectors in R2
2. A vector field in R3 is a vector-valued function with domain being a subset of R3 and
the images being vectors in R3
Example (1)
Let F (x,y) = < x + 1 , x + y >
Then F is a vector field in R2
Examples of values of F at different points of the domain
A Point (x0 , y0 ) of the
domain is a point in R2
The image F (x0 , y0 )
of (x0 , y0 ) is a vector
< a , b > in R2
(0.0)
< 0+1,0+0 >
= < 1,0 >
< 0+1,0+1 >
= < 1,1 >
< 1+1,1+0 >
= < 2,1>
< 1+1,1+1 >
= < 2,2 >
< -1 + 1 , -1 + 0 >
= < 0 , -1 >
< 0+1,0-1 >
= < 1 , -1 >
(0,1)
(1,0)
(1,1)
(-1,0)
(0,-1)
Considering (x0 , y0 ) to be the
initial point of the vector < a , b >
The terminal point ( x1 , y1 ) will
be:
(0+1 , 0+0)
=(1,0)
(0+1 , 1+1)
=(1,2)
(1+2 , 0+1)
=(3,1)
(1+2 , 1+2)
=(3,3)
( -1 + 0 , 0 - 1 )
= ( -1 , -1 )
( 0 + 1 , -1 - 1 )
= ( 1 , -2 )
Representing the vector field F
From each point (x0 , y0 ) of selected points of the domain of F, we draw the image
vector F (x0 , y0 ), such that (x0 , y0 ) is the initial point of the this vector.
Exercise (1)
Consider the vector field Let F (x,y) = < x , y >
Find the image of F at each of the following points
(0 , 0) , (1 , 0) , (0 , 1) , (-1 , 0) , (0 , -1) , (1 , 1) , (1 , 2) , (-1 , 1) , (-2 , -1) , (-1 , -2)
(1 , -2) , (2 , -1) , (-1 , 2). Represent this vector field geometrically.
Line Integrals
Example (1)
Find I =  C 8xy3 ds , where C is the curve having the parameterization:
x = cost , y = sint ; 0  t  π/2 .
Solution:
ds =  [ (dx/dt)2 + ( dy/dt) 2 ] dt =  [ (-sint)2 + ( cost) 2 ] = dt
I =  C 8xy3 ds
= 0π/2 8cost (sint)3 dt
= 8 0π/2 (sinx)3 cos .dt
= (8/4) (sinx)40 π/2 = 2 [ (1)4 – (0 ) 4 ] = 2
Example (2)
Find I =  C xy dx + x2 dy , where C is the curve consisting from the line
segment C1 from (2,1) to ( 4 ,1 ) and the line segment C2 from (4,1) to ( 4 ,5 )
Solution:
Parameterization of C1 : x = t , y = 1 : 2  t  4
dx = dt , dy = 0
I1 = C1 xy dx + x2 dy
I1 = 24 t(1) dt + t2 (0)
= 24 dt = (t2 / 2 )2 4 = ( 16 – 4 ) / 2 = 6
Parameterization of C2 : x = 4 , y = t : 1  t  5
dx = 0 , dy = dt
I1 = C2 xy dx + x2 dy
= 15 4(0) dt + 16 dt
= 16 15 dt = 16 (t )1 5 = 16 ( 5 – 1 ) = 64
C = C1 U C2
I =  C xy dx + x2 dy = C1 xy dx + x2 dy + C2 xy dx + x2 dy
= 6 + 64 = 70
Example (3)
Find I =  C xy dx + x2 dy , where C is the line segment from (2,1) to ( 4 ,5 )
Solution:
The equation of the line through the points (2,1) and ( 4 ,5 ) is :
y – 1 =[ (5-1)/(4-2)] ( x – 2)
→ y = 2x - 3
Parameterization of C1 : x = t , y = 2t-3 : 2  t  5
dx = dt , dy = 2dt
I = C xy dx + x2 dy
= 24 t(2t-3) dt + t2 . 2dt
= 24 (4 t2 – 3t )dt = [ (4t3 / 3 ) - (3t2 / 2 ) ] 2 4
= [ (4(4)3 / 3 ) - (3(4)2 / 2 ) ] - [ (4(2)3 / 3 ) - (3(2)2 / 2 ) ] = 170 / 3
Another way:
y = 2x – 3
→ dy = 2dx
I = C xy dx + x2 dy
= 24 x(2x-3) dx + x2 . 2dx
= 24 (4 x2 – 3x )dx = [ (4x3 / 3 ) - (3x2 / 2 ) ] 2 4
= [ (4(4)3 / 3 ) - (3(4)2 / 2 ) ] - [ (4(2)3 / 3 ) - (3(2)2 / 2 ) ] = 170 / 3
Example (4)
Find I =  C xy dx + x2 dy , where C is the curve with the parameterization:
x = 3t – 1 , y = 3 t2 – 2t
; 1  t  5/3
Solution:
We have:
dx = 3dt , dy = (6t – 2 )dt
I = C xy dx + x2 dy
= 15/3 (3t – 1 ) ( 3 t2 – 2t ) 3dt + (3t – 1 )2 . ( 6t – 2 )dt = 58
Another way:
t=(x+1)/3→
y = 3 [( x + 1 ) / 3 ] 2 – 2 [( x + 1 ) / 3 ] = (1/3) [ (x+1) 2 – 2x – 2 ]
= (1/3) [ x2 + 2x + 1 – 2x – 2 ]
= (1/3) [ x2– 1]
→ dy = (2/3) x dx
When t = 1 → x = 3(1) – 1 = 2
When t = 5/3 → x = 3(5/3) – 1 = 4
Thus,
I = C xy dx + x2 dy
= 24 x(1/3) ( x2– 1)dx + x2 . (2/3) x dx
= 24 [ x3– (1/3)x ]dx = [ (x4 / 4 ) - (x2 / 6 ) ] 2 4
= [ (44 / 4 ) - (42 / 6 ) ] - [ (24 / 4 ) - (22 / 6 ) ]
= [64 – (8/3) ] – [ 4 – (2/3) ] = 60 – 6 = 58
Example (5)
Find I =  C yz dx + xz dy + xy dz , where C is the curve with the
parameterization:
x = t , y = t2 , z= t3
; 0t2
Solution:
We have:
dx = dt , dy = 2tdt , dz = 3t2 dt
I =  C yz dx + xz dy + xy dz
= 02 t2 t3 dt + t t3 2tdt + t t2 3t2 dt
= 02 6 t5 dt = t6 0 2 = 26 - 06 = 64
Exercises
Exercise(1)
Find I =  C (2x + y3 ) dx + (3xy2 + 4 ) dy , where C is the curve consisting from the
line segment C1 from (2,1) to ( 4 ,1 ) and the line segment C2 from (4,1) to ( 4 ,5 )
Exercise(2)
Find I =  C (2x + y3 ) dx + (3xy2 + 4 ) dy , where C is the line segment from (2,1) to
( 4 ,5 ).
Exercise(3)
Find I =  C (2x + y3 ) dx + (3xy2 + 4 ) dy , where C is is the curve with the
parameterization:
x = 3t – 1 , y = 3 t2 – 2t
; 1  t  5/3