第8章 8.24

$(x-a)^2+(y-a)^2 \le a^2 \qquad z=ax \qquad z=-ax$

$x=a+r\cos\theta,\quad y=a+r\sin\theta$ とおくと $r^2\cos^2\theta+r^2\sin^2\theta\le a^2,\quad 0\le r\le a,\quad 0\le\theta\le2\pi$

$2\int_0^{2\pi}\int_0^a a(a+r\cos\theta)r\,dr\,d\theta = \int_0^{2\pi} 2\left[\frac{1}{2}a^2r^2+\frac{1}{3}ar^3\cos\theta\right]_0^a d\theta = \int_0^{2\pi}2\left(\frac{1}{2}a^4+\frac{1}{3}a^4\cos\theta\right)d\theta$

$= 2\left[\frac{1}{2}a^4\theta+\frac{1}{3}a^4\sin\theta\right]_0^{2\pi} = 2\pi a^4$