James Stewart-Calculus_ Early Transcendentals-Cengage Learning (2015)

Section 16.2 Line Integrals 1081
Line integrals along C with respect to x, y, and z can also be defined. For example,
y f sx, y, zd dz − lim
C n l ` o n
f sx*, i y i *, z i *d Dz i
i−1
− y b
f sxstd, ystd, zstdd z9std dt
a
Therefore, as with line integrals in the plane, we evaluate integrals of the form
10 y C
Psx, y, zd dx 1 Qsx, y, zd dy 1 Rsx, y, zd dz
by expressing everything sx, y, z, dx, dy, dzd in terms of the parameter t.
ExamplE 5 Evaluate y C
y sin z ds, where C is the circular helix given by the equa tions
x − cos t, y − sin t, z − t, 0 < t < 2. (See Figure 9.)
6
4
z
2
0
_1
C
_1
SOLUtion Formula 9 gives
y C
y sin z ds − y 2
0
0
ssin td sin tÎS
dtD
dx 2
1S
dtD
dy 2
1S
dtD
dz 2
dt
− y 2
sin 2 tssin 2 t 1 cos 2 t 1 1 dt − s2 y 2 1
2s1 2 cos 2td dt
− s2
2 ft 2 1 2 sin 2tg 2
0 − s2
0
■
0
y
FIGURE 9
1
1
0
x
ExamplE 6 Evaluate y C
y dx 1 z dy 1 x dz, where C consists of the line segment C 1
from s2, 0, 0d to s3, 4, 5d, followed by the vertical line segment C 2 from s3, 4, 5d to
s3, 4, 0d.
z
SOLUtion The curve C is shown in Figure 10. Using Equation 8, we write C 1 as
C¡
0
(2, 0, 0)
x
FIGURE 10
(3, 4, 5)
C
(3, 4, 0)
y
or, in parametric form, as
Thus
rstd − s1 2 td k2, 0, 0 l 1 t k3, 4, 5 l − k2 1 t, 4t, 5t l
x − 2 1 t y − 4t z − 5t 0 < t < 1
y y dx 1 z dy 1 x dz −
C1
y 1
s4td dt 1 s5td4 dt 1 s2 1 td5 dt
0
− y 1
s10 1 29td dt − 10t 1 29 t 2
0
1
2G0
− 24.5
Likewise, C 2 can be written in the form
rstd − s1 2 td k3, 4, 5 l 1 tk3, 4, 0 l − k3, 4, 5 2 5t l
or x − 3 y − 4 z − 5 2 5t 0 < t < 1
Copyright 2016 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.