원기둥·구면좌표 삼중적분

(a) In cylindrical coordinates, the paraboloid is $z=r^2$ and the cone is $z=r$. The two meet when $r^2=r$. That is, when $r=0$ and when $r=1$. So, in cylindrical coordinates

$$\begin{aligned} I = \int_0^1 \mathrm{d}r\ r \int_0^{2\pi}\mathrm{d}\theta \int_{r^2}^r\mathrm{d}z\ z(r^2+z^2) \end{aligned}$$

(b) In spherical coordinates, the paraboloid is

$$\rho\cos\varphi=\rho^2\sin^2\varphi\qquad\text{or}\qquad \rho=\frac{\cos\varphi}{\sin^2\varphi}$$

and the cone is

$$\rho\cos\varphi=\rho\sin\varphi\quad\text{or}\quad \tan\varphi=1\quad\text{or}\quad \varphi=\frac{\pi}{4}$$

The figure below shows a constant $\theta$ cross–section of $E$. Looking at that figure, we see that $\varphi$ runs from $\frac{\pi}{4}$ (i.e. the cone) to $\frac{\pi}{2}$ (i.e. the $xy$–plane).

So, is spherical coordinates,

$$\begin{aligned} I &= \int_{\pi/4}^{\pi/2} \mathrm{d}\varphi \int_0^{2\pi}\mathrm{d}\theta \int_0^{\cos\varphi/\sin^2\varphi} \mathrm{d}\rho\ \rho^2\sin\varphi \ \overbrace{\rho\cos\varphi}^{z} \overbrace{\rho^2}^{x^2+y^2+z^2} \\ &= \int_{\pi/4}^{\pi/2} \mathrm{d}\varphi \int_0^{2\pi}\mathrm{d}\theta \int_0^{\cos\varphi/\sin^2\varphi} \mathrm{d}\rho\ \rho^5\sin\varphi \cos\varphi \end{aligned}$$

(c) The cylindrical coordinates integral looks easier.

$$\begin{aligned} I &= \int_0^1 \mathrm{d}r\ r \int_0^{2\pi}\mathrm{d}\theta \int_{r^2}^r\mathrm{d}z\ z(r^2+z^2) \\ &= \int_0^1 \mathrm{d}r\ r \int_0^{2\pi}\mathrm{d}\theta\ \left[r^2\frac{z^2}{2}+\frac{z^4}{4}\right]_{r^2}^r \\ &= 2\pi \int_0^1 \mathrm{d}r\ r \left[r^2\frac{r^2}{2}+\frac{r^4}{4} -r^2\frac{r^4}{2}-\frac{r^8}{4}\right] \\ &=2\pi \left[\frac{1}{12}+\frac{1}{24} -\frac{1}{16}-\frac{1}{40}\right] =\frac{\pi}{2} \left[\frac{1}{3}+\frac{1}{6} -\frac{1}{4}-\frac{1}{10}\right] \\ &=\frac{3\pi}{40} \end{aligned}$$