第8章 8.22

(1)

$z = x^2 + y^2 \qquad z = y$

これから $y = x^2 + y^2$

$x = r\cos\theta \quad y = r\sin\theta$ とおくと $0 \leq r \leq \sin\theta \quad 0 \leq \theta \leq \pi$

$\int_0^\pi \int_0^{\sin\theta} (r\sin\theta - r^2) \cdot r\, dr\, d\theta = \int_0^\pi \left[ \frac{1}{3} r^3 \sin\theta - \frac{1}{4} r^4 \right]_0^{\sin\theta} d\theta = \int_0^\pi \frac{1}{3}\sin^4\theta - \frac{1}{4}\sin^4\theta\, d\theta$

$= \int_0^\pi \frac{1}{12}\sin^4\theta\, d\theta = 2\cdot\frac{1}{12}\int_0^{\frac{\pi}{2}} \sin^4\theta\, d\theta = \frac{\pi}{32}$

(2)

$x^2+y^2+z^2=a^2 \qquad x^2+y^2=\dfrac{a^2}{4}$

$x=r\cos\theta \quad y=r\sin\theta$ とおくと $0\leq r \leq \dfrac{a}{2} \quad 0\leq\theta\leq 2\pi$

$z=\sqrt{a^2-x^2-y^2}$ から

$2\int_0^{2\pi}\int_0^{\frac{a}{2}} \sqrt{a^2-r^2}\cdot r\,dr\,d\theta = 8\int_0^{\frac{\pi}{2}}\int_0^{\frac{a}{2}} r\sqrt{a^2-r^2}\,dr\,d\theta = 8\int_0^{\frac{\pi}{2}} \left[-\frac{1}{3}\sqrt{(a^2-r^2)^3}\right]_0^{\frac{a}{2}} d\theta$

$= 8\int_0^{\frac{\pi}{2}} -\frac{a^3\sqrt3}{8} + \frac{a^3}{3}\, d\theta = -\frac{\sqrt3}{2}\pi a^3 + \frac{4}{3}\pi a^3 = \frac{8-3\sqrt3}{6}\pi a^3$