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

Section 16.2 Line Integrals 1083
This integral is often abbreviated as y C
F dr and occurs in other areas of physics as
well. Therefore we make the following definition for the line integral of any continuous
vector field.
13 Definition Let F be a continuous vector field defined on a smooth curve C
given by a vector function rstd, a < t < b. Then the line integral of F along C
is
y C
F dr − y b
a
Fsrstdd r9std dt − y C
F T ds
Figure 12 shows the force field and the
curve in Example 7. The work done
is negative because the field impedes
movement along the curve.
y
1
When using Definition 13, bear in mind that Fsrstdd is just an abbreviation for the vector
field Fsxstd, ystd, zstdd, so we evaluate Fsrstdd simply by putting x − xstd, y − ystd,
and z − zstd in the expression for Fsx, y, zd. Notice also that we can formally write
dr − r9std dt.
ExamplE 7 Find the work done by the force field Fsx, yd − x 2 i 2 xy j in moving a
particle along the quarter-circle rstd − cos t i 1 sin t j, 0 < t < y2.
SOLUtion Since x − cos t and y − sin t, we have
Fsrstdd − cos 2 t i 2 cos t sin t j
and
r9std − 2sin t i 1 cos t j
Therefore the work done is
y C
F dr − y y2
0
Fsrstdd r9std dt − y y2
s22 cos 2 t sin td dt
0
0
FIGURE 12
1
x
y2
− 2 cos3 t
− 2 2 3
G0
3
■
Figure 13 shows the twisted cubic C in
Example 8 and some typical vectors
acting at three points on C.
2
Note Even though y C
F dr − y C
F T ds and integrals with respect to arc length
are unchanged when orientation is reversed, it is still true that
y 2C
F dr − 2y C
F dr
because the unit tangent vector T is replaced by its negative when C is replaced by 2C.
F{r(1)}
z 1
(1, 1, 1)
F{r(3/4)} C
0
F{r(1/2)}
0
y
1
2 2
1
x
FIGURE 13
0
ExamplE 8 Evaluate y C
F dr, where Fsx, y, zd − xy i 1 yz j 1 zx k and C is the
twisted cubic given by
SOLUtion We have
x − t y − t 2 z − t 3 0 < t < 1
rstd − t i 1 t 2 j 1 t 3 k
r9std − i 1 2t j 1 3t 2 k
Fsrstdd − t 3 i 1 t 5 j 1 t 4 k
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