원기둥·구면좌표 삼중적분

(a) Recall that in spherical coordinates

$$\begin{aligned} x=\rho\,\sin\varphi\,\cos\theta\qquad y=\rho\,\sin\varphi\,\sin\theta\qquad z=\rho\,\cos\varphi \end{aligned}$$

so that

• $x^2+y^2+z^2\le 9$ is $\rho\le 3$, and
• $\sqrt{3x^2+3y^2}\le z$ is $\sqrt{3}\rho\sin\varphi\le \rho\cos\varphi$, or $\tan\varphi\le \frac{1}{\sqrt{3}}$ or $\varphi\le\frac{\pi}{6}$, and
• the integrand $x^2+y^2=\rho^2\sin^2\varphi$, and
• $\mathrm{d}V$ is $\rho^2\sin\varphi\,\mathrm{d}\rho\,\mathrm{d}\theta\,\mathrm{d}\varphi$

So

$$T =\big\{ \ (\rho\sin\varphi\cos\theta\,,\,\rho\sin\varphi\sin\theta\,,\, \rho\cos\varphi)\ \big|\ 0\le\rho\le 3,\ 0\le\theta\le2\pi,\ 0\le \varphi\le {\pi}/{6}\ \big\}$$

and,

$$\begin{aligned} I &=\iiint _T \big(x^2+y^2\big)\ \mathrm{d}V =\int_0^{3}\mathrm{d}\rho\int_0^{2\pi}\mathrm{d}\theta \int_0^{\pi/6}\mathrm{d}\varphi\ \rho^2\sin\varphi\ \overbrace{\rho^2\sin^2\varphi}^{x^2+y^2} \\ &=\int_0^{3}\mathrm{d}\rho\int_0^{2\pi}\mathrm{d}\theta \int_0^{\pi/6}\mathrm{d}\varphi\ \rho^4\sin^3\varphi \end{aligned}$$

(b) In cylindrical coordinates

• $x^2+y^2+z^2\le 9$ is $r^2+z^2\le 9$ and
• $\sqrt{3 x^2+3 y^2}\le z$ is $\sqrt{3}\,r\le z$ and
• the integand $x^2+y^2=r^2$, and
• $\mathrm{d}V$ is $r\,\mathrm{d}r\,\mathrm{d}\theta\,\mathrm{d}z$

Observe that $r^2+z^2= 9$ and $\sqrt{3}\,r= z$ intersect when $r^2+3r^2=9$ so that $r=\frac{3}{2}$ and $z=\frac{3\sqrt{3}}{2}$. Here is a sketch of the $y=0$ cross–section of $T$.

So

$$T =\big\{ \ (r\cos\theta\,,\,r\sin\theta\,,\,z)\ \big|\ 0\le r\le \tfrac{3}{2},\ 0\le\theta\le2\pi,\ \sqrt{3}\,r\le z\le \sqrt{9-r^2}\ \big\}$$

and

$$\begin{aligned} I &=\iiint _V (x^2+y^2)\ \mathrm{d}V =\int_0^{3/2}\mathrm{d}r\int_0^{2\pi}\mathrm{d}\theta \int_{\sqrt{3}\,r}^{\sqrt{9-r^2}}\mathrm{d}z\ r\overbrace{(r^2)}^{x^2+y^2} =\int_0^{3/2}\mathrm{d}r\int_0^{2\pi}\mathrm{d}\theta \int_{\sqrt{3}\,r}^{\sqrt{9-r^2}}\mathrm{d}z\ \ r^3 \end{aligned}$$

(c) We'll use the spherical coordinate form.

$$\begin{aligned} I &=\int_0^{3}\mathrm{d}\rho\int_0^{2\pi}\mathrm{d}\theta \int_0^{\pi/6}\mathrm{d}\varphi\ \rho^4\sin^3\varphi \\ &=\int_0^{3}\mathrm{d}\rho\int_0^{2\pi}\mathrm{d}\theta \int_0^{\pi/6}\mathrm{d}\varphi\ \rho^4\sin\varphi \big[1-\cos^2\varphi\big] \\ &=2\pi \int_0^3\mathrm{d}\rho\ \rho^4 \left[-\cos\varphi +\frac{\cos^3\varphi}{3}\right]_0^{\pi/6} =2\pi \left[-\frac{\sqrt{3}}{2} + \frac{\sqrt{3}}{8}+1-\frac{1}{3}\right] \int_0^3\mathrm{d}\rho\ \rho^4 \\ &=2\pi \frac{3^5}{5}\left[\frac{2}{3} - \frac{3\sqrt{3}}{8}\right] =\overbrace{81}^{3^4}\pi \left[\frac{4}{5} - \frac{9\sqrt{3}}{20}\right] \approx 5.24 \end{aligned}$$