then the above line integral can be written
∫
C
©F·dr =
∫
C
(xˆe 1 +yeˆ 2 )·(dx ˆe 1 +dy ˆe 2 )
=
∫
C
x dx +y dy
=
∫ 2 π
0
(cos t)(−sin t)dt + (sin t)(cos t)dt = 0.
Here the direction of integration is in the positive sense as the parameter tvaries
from 0 to 2 π.
Example 6-33. Compute the value of the line integral
∫
C
©F×dr, where
F=xˆe 1 +yˆe 2 and Cis the circle x^2 +y^2 = 1
Solution: Write
F×dr =
∣∣
∣∣
∣∣
eˆ 1 ˆe 2 ˆe 3
x y 0
dx dy 0
∣∣
∣∣
∣∣=ˆe^3 (x dy −y dx )
and therefore ∫
C
©F×dr =
∫
C
©ˆe 3 (x dy −y dx ).
If the circular path of integration is represented in the parametric form
x= cos t y = sin t
one finds
∫
C
©F×dr =ˆe 3
∫ 2 π
0
(cos t)(cos t)dt −(sin t)(−sin t)dt =eˆ 3
∫ 2 π
0
dt = 2πeˆ 3.
Example 6-34.
Examine the work done in moving a particle through the force field
F= (x+z)ˆe 1 + (y+z)ˆe 2 + 2 zˆe 3
as the particle moves along the curve Cdescribed by the position vector
r =tˆe 1 +t^2 eˆ 2 + (− 3 t+ 1) ˆe 3
as the parameter tranges from 0 to 2.
