Page 110 - Vector Analysis

106 CHAPTER 4. Vector Calculus

of γ). Since the work done by a constant force is the inner product of the displacement and

the force, we find the the work done by the force F along the small portion γ([ti, ti+1]), from

γ(ti) to γ(ti+1), of the curve, where |ti ´ ti+1| ! 1, is approximately

() ()
(F ¨ T)(γ(ti))ℓ γ([ti, ti+1]) ” F(γ(ti)) ¨ T(γ(ti))ℓ γ([ti, ti+1]) .

Summing over all the portions, we conclude that the work done by the force F along the
curve C, in the direction of the parametrization γ, is approximately

k´1 ¨ ( )
T)(γ(ti))ℓ γ([ti, ti+1])
ÿ (F

i=0

ż

which converges to the line integral (F ¨ T) ds. Therefore, the line integral of vector fields

C

F along C in the direction of the parametrization γ is simply the work done by the force F

in moving an object along the curve C from the starting point to the end point.

ż

Example 4.27. Let F(x, y) = (y2, 2xy). Evaluate the line integral F ¨ dr from (0, 0) to

(1, 1) along C

1. the straight line y = x,

2. the curve y = x2, and

3. the piecewise smooth path consisting of the straight line segments from (0, 0) to (0, 1)
and from (0, 1) to (1, 1).

For the straight line case, we parameterize the path by γ(t) = (t, t) for t P [0, 1]. Then

ż ż1 ż1

F ¨ dr = (t2, 2t2) ¨ (1, 1)dt = 3t2dt = 1 .

C0 0

For the case of parabola, we parameterize the path by γ(t) = (t, t2) for t P [0, 1]. Then

ż ż1 ż1

F ¨ dr = (t4, 2t3) ¨ (1, 2t)dt = 5t4dt = 1 .

C0 0

For the piecewise linear case, we let C1 denote the line segment joining (0, 0) and (0, 1),
and let C2 denote the line segment joining (0, 1) and (1, 1). Note that we can parameterize
C1 and C2 by

γ1(t) = (0, t) t P [0, 1] and γ2(t) = (t, 1) t P [0, 1] ,