원기둥·구면좌표 삼중적분

$\frac{9\pi }{8}$

$\frac{1}{8}$

$V=\frac{\pi }{12}\approx 0.2618$

$180\pi \sqrt{10}$

$V=\frac{4\pi \sqrt{3}}{3}\approx 7.255$

$\displaystyle \int_{\theta_1}^{\theta_2}\int_{r_1}^{r_2}\int_{z_1}^{z_2}h(r,\theta,z)r\ dz\ dr\ d\theta$

$\displaystyle \int_{\varphi_1}^{\varphi_2}\int_{\theta_1}^{\theta_2}\int_{\rho_1}^{\rho_2}h(\rho,\theta,\varphi)\rho^2\sin(\varphi)\ d\rho\ d\theta\ d\varphi$

The sketch is below. To help build up this sketch, it is useful to recall the following facts.

• The cylindrical coordinate $r$ is the distance of the point from the $z$-axis. In particular all points with $r=0$ lie on the $z$-axis (for all values of $\theta$).
• The cylindrical coordinate $z$ is the distance of the point from the $xy$-plane. In particular all points with $z=0$ lie on the $xy$-plane.

The sketch is below. To help build up this sketch, it is useful to recall the following facts.

• The spherical coordinate $\rho$ is the distance of the point from the origin $(0,0,0)$. In particular if $\rho=0$, then the point is the origin (regardless of the values of $\theta$ and $\varphi$). If $\rho=1$ then the point lies on the sphere of radius $1$ centred on the origin.
• The spherical coordinate $\varphi$ is the angle between the positive $z$-axis and the radial line segment from the origin to $(x,y,z)$. In particular, all points with $\varphi=0$ lie on the positive $z$–axis (regardless of the value of $\theta$). All points with $\varphi=\frac{\pi}{2}$ lie in the $xy$-plane.

(a) Here is a sketch

(b) On $T$,

• the spherical coordinate $\varphi$ runs from $0$ (the positive $z$–xis) to ${\pi}/{2}$ (the $xy$–plane), and
• for each fixed $\varphi$ in that range, $\theta$ runs from $0$ to ${\pi}/{2}$, and
• for each fixed $\varphi$ and $\theta$, the spherical coordinate $\rho$ runs from $0$ to $1$.
• In spherical coordinates $\mathrm{d}V=\rho^2\,\sin\varphi\,\mathrm{d}\rho\,\mathrm{d}\theta\, \mathrm{d}\varphi$ and

$$xz=\big(\rho\sin\varphi\cos\theta\big)\big(\rho\cos\varphi\big) =\rho^2 \sin\varphi\ \cos\varphi\ \cos\theta$$

So

$$\begin{aligned} I=\int_0^{\pi/2}\mathrm{d}\varphi\int_0^{\pi/2}\mathrm{d}\theta\int_0^1\mathrm{d}\rho\ \rho^4\sin^2\varphi\cos\varphi\ \cos\theta \end{aligned}$$

(c) In spherical coordinates,

$$\begin{aligned} I&= \left[\int_0^{\pi/2}\mathrm{d}\varphi\ \sin^2\varphi\cos\varphi\right] \left[\int_0^{\pi/2}\mathrm{d}\theta\ \cos\theta\right] \left[\int_0^1\mathrm{d}\rho\ \rho^4\right] \\ &=\left[\frac{\sin^3\varphi}{3}\right]_0^{\pi/2} \left[\sin\theta\right]_0^{\pi/2} \left[\frac{\rho^5}{5}\right]_0^1 \\ &=\frac{1}{15} \end{aligned}$$

(a) Recall that, in spherical coordinates, $\varphi$ runs from $0$ (that's the positive $z$–axis) to $\pi$ (that's the negative $z$–axis), $\theta$ runs from $0$ to $2\pi$ ($\theta$ is the regular polar or cylindrical coordinate) and $\mathrm{d}V = \rho^2\ \sin\varphi\ \mathrm{d}\rho\,\mathrm{d}\theta\,\mathrm{d}\varphi$. So

$$\begin{aligned} \text{Volume} & = \int_0^\pi \mathrm{d}\varphi\int_0^{2\pi}\mathrm{d}\theta \int_0^{8\sin\varphi}\mathrm{d}\rho\ \rho^2\ \sin\varphi \\ & = \int_0^\pi \mathrm{d}\varphi\int_0^{2\pi}\mathrm{d}\theta\ \frac{(8\sin\varphi)^3}{3} \sin\varphi \\ & = \frac{2(8^3)\,\pi}{3} \int_0^\pi \mathrm{d}\varphi\ \sin^4\varphi \\ &= \frac{2(8^3)\,\pi}{3} \left[\frac{1}{32}\big(12\varphi -8\sin(2\varphi) +\sin(4\varphi)\big)\right]_0^\pi \\ &=\frac{2(8^3)\,\pi}{3}\ \frac{12\,\pi}{32} =128\pi^2 \end{aligned}$$

(b) Fix any $\varphi$ between $0$ and $\pi$. If $\rho=8\sin\varphi$, then as $\theta$ runs from $0$ to $2\pi$,

$$\begin{aligned} (x,y,z) &= \big(\rho\sin\varphi\cos\theta\,,\,\rho\sin\varphi\sin\theta\,,\, \rho\cos\varphi\big) \\ &= \big(8\sin^2\varphi\cos\theta\,,\,8\sin^2\varphi\sin\theta\,,\, 8\sin\varphi\cos\varphi\big) \\ &= \big(R\cos\theta\,,\,R\sin\theta\,,\,Z\big)\qquad \text{with }R=8\sin^2\varphi,\ Z= 8\sin\varphi\cos\varphi \end{aligned}$$

sweeps out a circle of radius $R=8\sin^2\varphi$ contained in the plane $z=Z=8\sin\varphi\cos\varphi$ and centred on $\big(0,0,Z=8\sin\varphi\cos\varphi\big)$. So the surface is a bunch of circles stacked one on top of the other. It is a surface of revolution. We can sketch it by

• first sketching the $\theta=0$ section of the surface (that's the part of the surface in the right half of the $xz$–plane)
• and then rotate the result about the $z$–axis.

The $\theta=0$ part of the surface is

$$\begin{aligned} &\big\{ \ (x,y,z)\ \big|\ x=8\sin^2\varphi,\ y=0,\ z=8\sin\varphi\cos\varphi,\ 0\le\varphi\le \pi\ \big\} \\ &=\big\{ \ (x,y,z)\ \big|\ x=4-4\cos(2\varphi),\ y=0,\ z=4\sin(2\varphi),\ 0\le\varphi\le \pi\ \big\} \end{aligned}$$

It's a circle of radius $4$, contained in the $xz$–plane (i.e. $y=0$) and centred on $(4,0,0)$! The figure on the left below is a sketch of the top half of the circle. When we rotate the circle about the $z$–axis we get a torus (a donut) but with the hole in the centre shrunk to a point. The figure on the right below is a sketch of the part of the torus in the first octant.