Begin2.DVI
ben green
(Ben Green)
#1
Using the property that line integrals may be broken up into integration along
separate curves, one can write W=
∫B
O
F·dr =
∫A
O
F·dr +
∫B
A
F·dr
where F·dr = (x^2 +y)dx +xy dy.
Figure 6-21. Find the work done in moving particle from origin to point B.
The portion of the work done in moving along the parabola from 0 to A, where
y=^53 x^2 and dy =^53 (2 x dx ),is
∫A
O
F·dr =
∫ 1
0
[x^2 + (^5
3
x^2 )]dx +x(^5
3
x^2 )^5
3
(2 x dx ) = 2
The portion of the work done in moving along the straight-line from Ato B, where
y=− 518 (x−1) +^53 and dy =− 518 dx, is expressed as
∫B
A
F·dr =
∫ 2
1
[x^2 + ( −^18
5
(x−1) +^5
3
]dx +x(−^18
5
(x−1) +^5
3
)( −^18
5
dx ) = 4
The total work done is therefore given by the sum W= 2 + 4 = 6 ft-lbs. Here the unit
of work is the unit of force times unit of distance traveled.
Example 6-32. 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: Let the circular path be represented in the parametric form
x= cos t y = sin t,
