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

Section 15.3 Double Integrals in Polar Coordinates 1013
If we had used rectangular coordinates instead of polar coordinates, then we would
have obtained
V − y s1 2 x 2 2 y 2 d dA − y 1 2
21 ys12x s1 2 x 2 2 y 2 d dy dx
2s12x 2
D
which is not easy to evaluate because it involves finding y s1 2 x 2 d 3y2 dx.
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¨=∫
D
r=h(¨)
What we have done so far can be extended to the more complicated type of region
shown in Figure 7. It’s similar to the type II rectangular regions considered in Section
15.2. In fact, by combining Formula 2 in this section with Formula 15.2.5, we obtain
the following formula.
O
∫
å
¨=å
r=h¡(¨)
FIGURE 7
D=s(r, ¨) | 寨¯∫, h¡(¨)¯r¯h(¨)d
3 If f is continuous on a polar region of the form
then
y
D
D − hsr, d | < < , h 1sd < r < h 2 sdj
y f sx, yd dA − y yh2sd
h 1
sd
f sr cos , r sin d r dr d
In particular, taking f sx, yd − 1, h 1 sd − 0, and h 2 sd − hsd in this formula, we see
that the area of the region D bounded by − , − , and r − hsd is
AsDd − y
D
− y
and this agrees with Formula 10.4.3.
y 1 dA − y yhsd
F
0
r dr d
r 2 hsd
d − y 1
2 fhsdg2 d
2G0
¨= π 4
ExamplE 3 Use a double integral to find the area enclosed by one loop of the fourleaved
rose r − cos 2.
SOLUTION From the sketch of the curve in Figure 8, we see that a loop is given by the
region
D − hsr, d | 2y4 < < y4, 0 < r < cos 2j
So the area is
¨=_ π 4
AsDd − y dA − y y4 2
ycos r dr d
2y4 0
D
FIGURE 8
− y y4
2y4 f 1 2 r 2 g 0
cos 2
d −
1
2 y y4
2y4 cos2 2 d
− 1 4 y y4
s1 1 cos 4d d − 1 4 f 1 1 4 sin 4g −
2y4 2y4
8
y4
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