第5章 5.22

(1)

$r = 2\sin\theta \qquad r = 6\sin\theta$

$S = \frac{1}{2}\cdot 2\int_0^{\pi/2}(6\sin\theta)^2\,d\theta - \frac{1}{2}\cdot 2\int_0^{\pi/2}(2\sin\theta)^2\,d\theta = 36\left[\frac{1}{2}\theta-\frac{1}{4}\sin2\theta\right]_0^{\frac{\pi}{2}} - 4\left[\frac{1}{2}\theta-\frac{1}{4}\sin2\theta\right]_0^{\frac{\pi}{2}}$

$=9\pi-\pi=8\pi$

(2)

$r=a(1+\cos\theta) \qquad r=2a\cos\theta$

$S=\frac{1}{2}\cdot 2\int_0^\pi (2a\cos\theta)^2\,d\theta-\frac{1}{2}\cdot 2\int_0^\pi \{a(1+\cos\theta)\}^2\,d\theta$

$=\int_0^\pi 4a^2\,\frac{1+\cos2\theta}{2}\,d\theta-\int_0^\pi a^2\left(1+2\cos\theta+\frac{1+\cos2\theta}{2}\right)d\theta$

$=\left[4a^2\left(\frac{1}{2}+\frac{1}{4}\sin2\theta\right)\right]_0^\pi-\left[a^2\left(\theta+2\sin\theta+\frac{1}{2}\theta+\frac{1}{4}\sin2\theta\right)\right]_0^\pi=2\pi a^2-\pi a^2-\frac{\pi}{2}a^2=\frac{\pi}{2}a^2$