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372 CHAPTER 12 Vector Integral Calculus

We often write a line integral in vector notation. Let F = f i + gj + hk and form the position
vector R(t) = x(t)i + y(t)j + z(t)k for C. Then
dR = dx i + dy j + dz k
and
F · dR = fdx + gdy + hdz

suggesting the notation

fdx + gdy + hdz = F · dR.
C C

EXAMPLE 12.6
2
2
A force F(x, y, z) = x i − zyj + x cos(z)k moves an object along the path C given by x = t ,
y = t, z = πt for 0 ≤ t ≤ 3. We want to calculate the work done by this force.
At any point on C the particle will be moving in the direction of the tangent to C at that point.
We may approximate the work done along a small segment of the curve starting at (x, y, z) by
F(x, y, z) · dR, with the dimensions of force times distance. The work done in moving the object
along the entire path is approximated by the sum of these approximations along segments of the
path. In the limit as the lengths of these segments tend to zero we obtain

2
work = F · dR = x dx − zy dy + x cos(z)dz
C C
3
4
2
= [t (2t) − (πt)(t) + t cos(πt)(π)]dt
0
3
2
2
5
= [2t − πt + πt cos(πt)]dt
0
6
= 243 − 9π − .
π
12.1.1 Line Integral With Respect to Arc Length
In some contexts it is useful to have a line integral with respect to arc length along C.If
ϕ(x, y, z) is a scalar field and C is a smooth curve with coordinate functions x = x(t), y =
y(t), z = z(t) for a ≤ t ≤ b, we define
b

2
2
2

ϕ(x, y, z)ds = ϕ(x(t), y(t), z(t)) x (t) + y (t) + z (t) dt.
C a
The rationale behind this definition is that

2
2
2

ds = x (t) + y (t) + z (t) dt
is the differential element of arc length along C.
To see how such a line integral arises, suppose C is a thin wire having density δ(x, y, z)
at (x, y, z), and we want to compute the mass. Partition [a,b] into n subintervals by inserting
points
a = t 0 < t 1 < t 2 < ··· < t n−1 < t n = b
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October 14, 2010 14:53 THM/NEIL Page-372 27410_12_ch12_p367-424