원기둥·구면좌표 삼중적분

(a) Recall that in spherical coordinates,

$$\begin{aligned} x&=\rho\sin\varphi\cos\theta \\ y&=\rho\sin\varphi\sin\theta \\ z&=\rho\cos\varphi \\ x^2+y^2 &=\rho^2\sin^2\varphi \end{aligned}$$

so that $x^2+y^2+z^2= a^2$ becomes $\rho = a$, and $\sqrt{x^2+y^2}= z$ becomes

$$\begin{aligned} \rho\sin\varphi = \rho\cos\varphi \iff \tan\varphi= 1 \iff \varphi=\frac{\pi}{4} \end{aligned}$$

Here is a sketch of the $y=0$ cross–section of the specified region.

Looking at the figure above, we see that, on that region,

• $\varphi$ runs from $0$ (the positive $z$–axis) to $\frac{\pi}{4}$ (on the cone), and
• for each $\varphi$ is that range, $\rho$ runs from $0$ to $a$ and $\theta$ runs from $0$ to $2\pi$.

so that

$$\begin{aligned} \text{Volume} &= \int_{0}^{a}\mathrm{d}\rho \int_{0}^{2\pi} \mathrm{d}\theta \int_0^{{\pi\over 4}}\mathrm{d}\varphi\ \rho^2\sin\varphi = \Big\{\int_{0}^{a}\mathrm{d}\rho\ \rho^2 \Big\}\Big\{\int_{0}^{2\pi} \mathrm{d}\theta\Big\}\Big\{ \int_0^{{\pi\over 4}}\mathrm{d}\varphi\ \sin\varphi\Big\} \\ &= \frac{a^3}{3}\ 2\pi\ \Big[-\cos\varphi\Big]_0^{{\pi\over 4}} =2\pi\frac{a^3}{3}\left(1-\frac{1}{\sqrt{2}}\right) \end{aligned}$$

(b) The part of the sphere in question is

$$\begin{aligned} R&=\big\{ \ (x,y,z)\ \big|\ x^2+y^2+z^2\le a^2,\ x\ge0,\ y\ge 0,\ z\ge 0\ \big\} \\ &=\big\{ \ (\rho\sin\varphi\cos\theta\,,\, \rho\sin\varphi\sin\theta\,,\, \rho\cos\varphi)\ \big|\ \rho\le a,\ 0\le\varphi\le\tfrac{\pi}{2}, \ 0\le\theta\le\tfrac{\pi}{2}\ \big\} \end{aligned}$$

By symmetry, the two specified integrals are equal, and are

$$\begin{aligned} \int_{0}^{a}\mathrm{d}\rho\ \rho^2 \int_0^{{\pi\over 2}}\mathrm{d}\varphi\ \sin\varphi \int_{0}^{{\pi\over 2}} \mathrm{d}\theta\ \overbrace{\rho\cos\varphi}^{z} &=\frac{a^4}{4}\frac{\pi}{2} \int_0^{{\pi\over 2}}\mathrm{d}\varphi\ \sin\varphi\cos\varphi \\ &=\frac{\pi a^4}{8} \int_0^{1}\mathrm{d}t\ t\qquad\text{ where $t=\sin\varphi$, $\mathrm{d}t=\cos\varphi\,\mathrm{d}\varphi$} \\ &=\frac{\pi a^4}{16} \end{aligned}$$

(c) The planet in question is

$$\begin{aligned} P&=\big\{ \ (x,y,z)\ \big|\ x^2+y^2+z^2\le a^2\ \big\} \\ &=\big\{ \ (\rho\sin\varphi\cos\theta\,,\, \rho\sin\varphi\sin\theta\,,\, \rho\cos\varphi)\ \big|\ \rho\le a,\ 0\le\varphi\le\pi, \ 0\le\theta\le2\pi\ \big\} \end{aligned}$$

So the

$$\begin{aligned} \text{mass} &= \int_{0}^{a}\mathrm{d}\rho\ \rho^2 \int_0^{\pi}\mathrm{d}\varphi\ \sin\varphi \int_{0}^{2\pi} \mathrm{d}\theta\ \overbrace{\frac{A}{B+\rho^2}}^{\rm density} =2\pi A\Big\{ \int_0^{\pi}\mathrm{d}\varphi\ \sin\varphi\Big\} \Big\{\int_{0}^{a}\mathrm{d}\rho\ \frac{\rho^2}{B+\rho^2} \Big\} \\ &=4\pi A\int_{0}^{a}\mathrm{d}\rho\ \left(1-\frac{B}{B+\rho^2} \right) \\ &=4\pi Aa-4\pi A\sqrt{B}\int_{0}^{a/\sqrt{B}}\mathrm{d}s\ \frac{1}{1+s^2} \qquad\text{ where $\rho=\sqrt{B}\,s$, $\mathrm{d}\rho=\sqrt{B}\,\mathrm{d}s$} \\ &=4\pi A\left(a-\sqrt{B}\tan^{-1}\frac{a}{\sqrt{B}}\right) \end{aligned}$$

(d) Observe that

• when $\varphi=0$ (i.e. on the positive $z$–axis), $\cos\varphi=1$ so that $\rho=a(1-\cos\varphi)=0$ and
• as $\varphi$ increases from $0$ to $\frac{\pi}{2}$, $\cos\varphi$ decreases so that $\rho=a(1-\cos\varphi)$ increases and
• when $\varphi=\frac{\pi}{2}$ (i.e. on the $xy$–plane), $\cos\varphi=0$ so that $\rho=a(1-\cos\varphi)=a$ and
• as $\varphi$ increases from $\frac{\pi}{2}$ to $\pi$, $\cos\varphi$ continues to decrease so that $\rho=a(1-\cos\varphi)$ increases still more and
• when $\varphi=\pi$ (i.e. on the negative $z$–axis), $\cos\varphi=-1$ so that $\rho=a(1-\cos\varphi)=2a$

So we have the following sketch of the intersection of the specified volume with the right half of the $yz$–plane.

The volume in question is invariant under rotations about the $z$–axis so that

$$\begin{aligned} \text{Volume} &= \int_{0}^{2\pi} \mathrm{d}\theta\int_0^{\pi}\mathrm{d}\varphi\ \sin\varphi\int_{0}^{a(1-\cos\varphi)}\mathrm{d}\rho\ \rho^2 \\ &=2\pi\frac{a^3}{3} \int_0^{\pi}\mathrm{d}\varphi\ \sin\varphi(1-\cos\varphi)^3 \\ &=2\pi\frac{a^3}{3} \int_0^2\mathrm{d}t\ t^3 \qquad\text{ where $t=1-\cos\varphi$, $\mathrm{d}t=\sin\varphi\,\mathrm{d}\varphi$} \\ &=2\pi\ \frac{a^3}{3}\ \frac{2^4}{4} =\frac{8}{3}\pi a^3 \end{aligned}$$