第8章 8.6

(1)

$0 \leq r \leq 2a\cos\theta$ $x = r\cos\theta$ $y = r\sin\theta$ から

$\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \int_0^{2a\cos\theta} r^2 \cdot r \, dr\, d\theta = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \left[\frac{1}{4}r^4\right]_0^{2a\cos\theta} d\theta = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} 4a^4 \cos^4\theta \, d\theta = 2\int_0^{\frac{\pi}{2}} 4a^4 \cos^4\theta \, d\theta$

$= 8a^4 \cdot \frac{3}{4} \cdot \frac{1}{2} \cdot \frac{\pi}{2} = \frac{3}{2}\pi a^4$

(2)

$0 \leq r \leq a(1+\cos\theta)$ $x = r\cos\theta$ $y = r\sin\theta$ から

$\int_0^{2\pi} \int_0^{a(1+\cos\theta)} r^2 \cdot r \, dr\, d\theta = \int_0^{2\pi} \left[\frac{1}{4}r^4\right]_0^{a(1+\cos\theta)} d\theta = \int_0^{2\pi} \frac{1}{4}a^4(1+\cos\theta)^4 \, d\theta$

$= \frac{1}{4}a^4 \int_0^{2\pi} (1 + 4\cos\theta + 6\cos^2\theta + 4\cos^3\theta + \cos^4\theta)\, d\theta$

$0 \sim 2\pi$ の範囲において、$\cos\theta$、$\cos^3\theta$ は期間数であるため、積分は $0$ となる

$= \frac{1}{4}a^4 \int_0^{2\pi} (1 + 6\cos^2\theta + \cos^4\theta)\, d\theta = a^4 \int_0^{\frac{\pi}{2}} (1 + 6\cos^2\theta + \cos^4\theta)\, d\theta$

$= a^4 \left\{ [\theta]_0^{\frac{\pi}{2}} + 6 \cdot \frac{1}{2} \cdot \frac{\pi}{2} + \frac{3}{4} \cdot \frac{1}{2} \cdot \frac{\pi}{2} \right\} = a^4 \left(\frac{\pi}{2} + \frac{3}{2}\pi + \frac{3}{16}\pi\right) = a^4 \cdot \frac{35}{16}\pi = \frac{35}{32}\pi a^4$