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

1012 Chapter 15 Multiple Integrals
2 Change to Polar Coodinates in a Double Integral If f is continuous on a
polar rectangle R given by 0 < a < r < b, < < , where 0 < 2 < 2,
then
y f sx, yd dA − y yb f sr cos , r sin d r dr d
a
R
O
FIGURE 5
r
d¨
r d¨
dA
dr
The formula in (2) says that we convert from rectangular to polar coordinates in a
double integral by writing x − r cos and y − r sin , using the appropriate limits of
integration for r and , and replacing dA by r dr d. Be careful not to forget the additional
factor r on the right side of Formula 2. A classical method for remembering this
is shown in Figure 5, where the “infinitesimal” polar rectangle can be thought of as an
ordinary rectangle with dimensions r d and dr and therefore has “area” dA − r dr d.
ExamplE 1 Evaluate yy R
s3x 1 4y 2 d dA, where R is the region in the upper half-plane
bounded by the circles x 2 1 y 2 − 1 and x 2 1 y 2 − 4.
SOLUTION The region R can be described as
R − hsx, yd | y > 0, 1 < x 2 1 y 2 < 4j
It is the half-ring shown in Figure 1(b), and in polar coordinates it is given by
1 < r < 2, 0 < < . Therefore, by Formula 2,
yy s3x 1 4y 2 d dA − y
0 y2 s3r cos 1 4r 2 sin 2 d r dr d
1
R
− y
0 y2 1
s3r 2 cos 1 4r 3 sin 2 d dr d
Here we use the trigonometric identity
sin 2 − 1 2s1 2 cos 2d
See Section 7.2 for advice on integrating
trigonometric functions.
− y
fr 3 cos 1 r 4 sin 2 r−2
g
0 r−1 d − y
s7 cos 1 15 0 sin2 d d
− y
0
15
f7 cos 1 2 s1 2 cos 2dg d
− 7 sin 1 15
2 2 15
4 sin 2 − 15
G0
2
■
ExamplE 2 Find the volume of the solid bounded by the plane z − 0 and the paraboloid
z − 1 2 x 2 2 y 2 .
x
z
(0, 0, 1)
0
D
y
SOLUTION If we put z − 0 in the equation of the paraboloid, we get x 2 1 y 2 − 1. This
means that the plane intersects the paraboloid in the circle x 2 1 y 2 − 1, so the solid
lies under the paraboloid and above the circular disk D given by x 2 1 y 2 < 1 [see Figures
6 and 1(a)]. In polar coordinates D is given by 0 < r < 1, 0 < < 2. Since
1 2 x 2 2 y 2 − 1 2 r 2 , the volume is
V − y s1 2 x 2 2 y 2 d dA − y 2
y 1
s1 2 r 2 d r dr d
0 0
D
FIGURE 6
− y 2
d y 1
sr 2 r 3 d dr − 2F r 2
0 0
2 2 r 4
1
4G0
− 2
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.