원기둥·구면좌표 삼중적분

We'll use cylindrical coordinates. In cylindrical coordinates

• the sphere $x^2+y^2+z^2=a^2$ becomes $r^2+z^2=a^2$ and
• the circular cylinder $x^2+y^2=ax$ (or equivalently $(x-a/2)^2+ y^2=a^2/4$) becomes $r^2=ar\cos\theta$ or $r=a\cos\theta$.

Here is a sketch of the top view of the solid.

The solid is

$$\big\{ \ (r\cos\theta\,,\,r\sin\theta\,,\,z)\ \big|\ -{\pi}/{2}\le\theta\le{\pi}/{2}\,,\, 0\le r\le a\cos\theta\,,\, -\sqrt{a^2-r^2}\le z\le \sqrt{a^2-r^2}\ \big\}$$

By symmetry, the volume of the specified solid is four times the volume of the solid

$$\big\{ \ (r\cos\theta\,,\,r\sin\theta\,,\,z)\ \big|\ 0\le\theta\le{\pi}/{2}\,,\, 0\le r\le a\cos\theta\,,\, 0\le z\le \sqrt{a^2-r^2}\ \big\}$$

Since $\mathrm{d}V = r\,\mathrm{d}r\,\mathrm{d}\theta\,\mathrm{d}z$, the volume of the solid is

$$\begin{aligned} 4\int_0^{\pi/2}\mathrm{d}\theta \int_0^{a\cos\theta} \mathrm{d}r \int_0^{\sqrt{a^2-r^2}}\mathrm{d}z\ r &=4\int_0^{\pi/2}\mathrm{d}\theta \int_0^{a\cos\theta} \mathrm{d}r\ r\sqrt{a^2-r^2} \\ &=-\frac{4}{3}\int_0^{\pi/2}\mathrm{d}\theta\ \big(a^2-r^2\big)^{3/2}\Big|_0^{a\cos\theta} \\ &=\frac{4}{3}\int_0^{\pi/2}\mathrm{d}\theta\ \Big[a^3-\big(a^2-a^2\cos^2\theta\big)^{3/2}\Big] \\ &=\frac{4a^3}{3}\int_0^{\pi/2}\mathrm{d}\theta\ \big[1-\sin^3\theta\big] \\ &=\frac{4a^3}{3}\left[\theta - \frac{1}{12}\cos(3\theta) + \frac{3}{4}\cos\theta \right]_0^{\pi/2} \\ &=\frac{4a^3}{3}\left[\frac{\pi}{2} + \frac{1}{12} - \frac{3}{4} \right] \\ &=\frac{4a^3}{3}\left[\frac{\pi}{2} - \frac{2}{3} \right] \end{aligned}$$