Calculus- Early Transcendentals, 2021a

540 Vector Calculus
Definition 16.4: Line Integral of a Function
Let f (x,y,z) be a function defined along a curve C. If the curve C is parametrized by r(t) with
t ∈ [a,b], then the integral of f along C is given by
∫
∫ b
f (x,y)ds = f (r(t))
dr
∣ dt ∣ dt
C
a
Example 16.5: Compute Integral Over Line Segment
∫
Compute ye x ds where C is the line segment from (1,2) to (4,7).
C
Solution. We write the line segment as a vector function: v = 〈1,2〉 + t〈3,5〉, 0≤ t ≤ 1, or in parametric
form x = 1 + 3t, y = 2 + 5t. Then
∫
C
ye x ds =
∫ 1
0
(2 + 5t)e 1+3t√ 3 2 + 5 2 dt = 16 √
34e 4 − 1 √
34e.
9 9
All of these ideas extend to three dimensions in the obvious way.
Example 16.6: Compute Integral Over Line Segment
∫
Compute x 2 zds where C is the line segment from (0,6,−1) to (4,1,5).
C
♣
Solution. We write the line segment as a vector function: v = 〈0,6,−1〉 + t〈4,−5,6〉, 0≤ t ≤ 1, or in
parametric form x = 4t, y = 6 − 5t, z = −1 + 6t. Then
∫ ∫ 1
x 2 zds= (4t) 2 (−1 + 6t) √ 16 + 25 + 36dt = 16 √ ∫ 1
77 −t 2 + 6t 3 dt = 56 √
77.
C
0
0
3
♣
16.3.2 Line Integrals of a Vector Field
Now we turn to a perhaps more interesting example. Recall that in the simplest case, the work done by a
force on an object is equal to the magnitude of the force times the distance the object moves; this assumes
that the force is constant and in the direction of motion. We have already dealt with examples in which the
force is not constant; now we are prepared to examine what happens when the force is not parallel to the
direction of motion.
We have already examined the idea of components of force, in Example 12.5: the component of a
force F in the direction of a vector v is
F · v
|v| 2 v,