第8章 8.5

(1)

$x = r\cos\theta$ $y = r\sin\theta$ として

$x^2 + y^2 \leq a^2$ から $0 \leq r \leq a$ $0 \leq \theta \leq 2\pi$ これから

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

$= \int_0^{2\pi}\frac{1}{4}a^4\frac{1+\cos 2\theta}{2}d\theta = \frac{1}{4}a^4\left[\frac{1}{2}\theta + \frac{1}{4}\sin 2\theta\right]_0^{2\pi} = \frac{\pi}{4}a^4$

(1)

と同様に

$\iint_D x^4\,dxdy = \int_0^{2\pi}\int_0^a r^4\cos^4\theta \cdot r\,dr\,d\theta = \int_0^{2\pi}\left[\frac{1}{6}r^6\cos^4\theta\right]_0^a d\theta = \int_0^{2\pi}\frac{1}{6}a^6\cos^4\theta\,d\theta$

$= \frac{2}{3}a^6\left(\frac{3}{4}\cdot\frac{1}{2}\cdot\frac{\pi}{2}\right) = \frac{\pi}{8}a^6$

(1)

と同様に

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

(1)

と同様に

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

$= \frac{2}{3}\pi a^3$

(1)

と同様に

$\iint_D e^{-x^2-y^2}\,dxdy = \int_0^{2\pi}\int_0^a e^{-r^2(\cos^2\theta+\sin^2\theta)} \cdot r\,dr\,d\theta = \int_0^{2\pi}\int_0^a e^{-r^2} \cdot r\,dr\,d\theta = \int_0^{2\pi}\left[-\frac{1}{2}e^{-r^2}\right]_0^a d\theta$

$= \int_0^{2\pi}\frac{1}{2} - \frac{1}{2}e^{-a^2}\,d\theta = \pi(1 - e^{-a^2})$