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

Section 16.2 Line Integrals 1077
Example 2 Evaluate y C
2x ds, where C consists of the arc C 1 of the parabola y − x 2
from s0, 0d to s1, 1d followed by the vertical line segment C 2 from s1, 1d to s1, 2d.
y
C
(1, 2)
SOLUtion The curve C is shown in Figure 5. C 1 is the graph of a function of x, so we
can choose x as the parameter and the equations for C 1 become
x − x y − x 2 0 < x < 1
(0, 0)
C¡
(1, 1)
x
Therefore
y 2x ds −
C1
y 2xÎS 1
2
0 dxD 1S
dxD
dy 2
dx − y 1
2xs1 1 4x 2 dx
0
FIGURE 5
C − C 1 ø C 2
− 1 4 2 3 s1 1 4x 2 d 3y2 g 0
1
−
5s5 2 1
6
On C 2 we choose y as the parameter, so the equations of C 2 are
x − 1 y − y 1 < y < 2
and y 2x ds −
C2
y 2s1dÎS 2 dx 2
1 dyD 1S
dyD
2
dy − y 2
2 dy − 2
1
Thus y C
2x ds − y
C1
2x ds 1 y
C2
2x ds − 5s5 2 1
6
1 2 ■
Any physical interpretation of a line integral y C
f sx, yd ds depends on the physical
interpretation of the function f. Suppose that sx, yd represents the linear density at a
point sx, yd of a thin wire shaped like a curve C. Then the mass of the part of the wire
from P i21 to P i in Figure 1 is approximately sx i *, y i *d Ds i and so the total mass of the
wire is approximately o sx i *, y i *d Ds i . By taking more and more points on the curve, we
obtain the mass m of the wire as the limiting value of these approximations:
m − lim
n l ` o n
sx*, i y i *d Ds i − y sx, yd ds
i−1
C
[For example, if f sx, yd − 2 1 x 2 y represents the density of a semicircular wire, then
the integral in Example 1 would represent the mass of the wire.] The center of mass of
the wire with density function is located at the point sx, yd, where
4 x − 1 m y C x sx, yd ds y − 1 m y C
y sx, yd ds
Other physical interpretations of line integrals will be discussed later in this chapter.
Example 3 A wire takes the shape of the semicircle x 2 1 y 2 − 1, y > 0, and is thicker
near its base than near the top. Find the center of mass of the wire if the linear density at
any point is proportional to its distance from the line y − 1.
SOLUtion As in Example 1 we use the parametrization x − cos t, y − sin t,
0 < t < , and find that ds − dt. The linear density is
sx, yd − ks1 2 yd
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