Solution: The work done is determined by evaluating the line integral
∫
CF·dr where
F·dr = (x+z)dx + (y+z)dy + 2z dz
On the given curve C use x=t,y=t^2 and z=− 3 t+ 1 with dx =dt,dy = 2 t dt and
dz =− 3 dt and substitute these values into the line integral describing the work done.
This produces the result
∫
C
F·dr =
∫ 2
0
[
(t− 3 t+ 1)(dt) + (t^2 − 3 t+ 1)(2 t dt) + 2(− 3 t+ 1)(− 3 dt)
]
= 18
where work has the units of force times units of distance traveled.
Example 6-35.
For F=x(y+ 1)ˆe 1 +yˆe 2 +z(x+ 1) ˆe 3 evaluate the line integral
∫(1, 1 ,1)
(0, 0 ,0)
F·dr
(a) Along the line segments illustrated.
(b) Along the straight line path from (0 , 0 ,0) to (1 , 1 ,1).
Solution F·dr =x(y+ 1) dx +y dy +z(x+ 1) dz
(a) Along∫ (0, 0 ,0) to (1 , 0 ,0),x= 1, z = 0, 0 ≤x≤ 1
1
0
x dx =
1
2
Along∫ (1, 0 ,0) to (1 , 1 ,0),x= 1, z = 0, 0 ≤y≤ 1
1
0
y dy =
1
2
Along∫ (1, 1 ,0) to (1 , 1 ,1),x= 1, y = 1, 0 ≤z≤ 1
1
0
2 z dz = 1
therefore
∫(1, 1 ,1)
(0, 0 ,0)
F·dr =^1
2
+^1
2
+ 1 = 2
(b) The straight line path from (0, 0 ,0) to (1 , 1 ,1) is represented by the parametric
equation
x=t, y =t, z =t
for 0 ≤t≤ 1. Therefore
∫(1, 1 ,1)
(0, 0 ,0)
F·dr =
∫ 1
0
[t(t+ 1) + t+t(t+ 1)]dt =^13
6
The work done in moving from (0 , 0 ,0) to (1 , 1 ,1) is path dependent.
