원기둥·구면좌표 삼중적분

(a) The direction of maximum rate of increase is ${ \boldsymbol{\nabla} }\rho(1,0,-1)$. As

$$\begin{aligned} \frac{\partial \rho}{\partial x}(x,y,z) &= \frac{4x}{1+x^2+y^2} -\frac{2x(z+2x^2)}{{(1+x^2+y^2)}^2} & \frac{\partial \rho}{\partial x}(1,0,-1) &= \frac{4}{2} -\frac{2(-1+2)}{{(2)}^2} =\frac{3}{2} \\ \frac{\partial \rho}{\partial y}(x,y,z) &= -\frac{2y(z+2x^2)}{{(1+x^2+y^2)}^2} & \frac{\partial \rho}{\partial y}(1,0,-1) &= 0\\ \frac{\partial \rho}{\partial z}(x,y,z) &= \frac{1}{1+x^2+y^2} & \frac{\partial \rho}{\partial z}(1,0,-1) &= \frac{1}{2} \end{aligned}$$

So ${ \boldsymbol{\nabla} }\rho(1,0,-1) = \frac{1}{2}(3,0,1)$. The unit vector in this direction is $\frac{1}{\sqrt{10}}(3,0,1)$.

(b) The region swept by the space craft is, in cylindrical coordinates,

$$V=\big\{ \ (r\cos\theta\,,\,r\sin\theta\,,\,z)\ \big|\ 0\le\theta\le 2\pi,\ 0\le r\le 1,\ 0\le z\le 2\ \big\}$$

and the amount of hydrogen collected is

$$\begin{aligned} \iiint _V \rho\ \mathrm{d}V &=\iiint _V\frac{z+2r^2\cos^2\theta}{1+r^2}r\mathrm{d}r\,\mathrm{d}\theta\,\mathrm{d}z \\ &=\int_0^2\mathrm{d}z \int_0^{2\pi} \mathrm{d}\theta \int_0^1\mathrm{d}r\ \frac{zr+(2\cos^2\theta)r^3}{1+r^2} \\ &=\int_0^2\mathrm{d}z \int_0^{2\pi} \mathrm{d}\theta \int_0^1\mathrm{d}r\ \left[ z \frac{r}{1+r^2} +2r\cos^2\theta -\cos^2\theta\frac{2r}{1+r^2}\right]\\ &\text{since }\frac{r^3}{1+r^2} = \frac{r+r^3 - r}{1+r^2} =r-\frac{r}{1+r^2} \\ &=\int_0^2\mathrm{d}z \int_0^{2\pi} \mathrm{d}\theta\ \left[\frac{z}{2}\ln(1+r^2)+r^2\cos^2\theta -\ln(1+r^2)\cos^2\theta\right]_0^1 \\ &=\int_0^2\mathrm{d}z \int_0^{2\pi} \mathrm{d}\theta\ \left[\frac{\ln 2}{2} z+ \cos^2\theta -\ln(2)\cos^2\theta\right] \\ &=\int_0^2\mathrm{d}z \ \left[(\pi\ln 2) z+ \pi-\pi\ln 2\right] \\ &= 2\pi\ln 2 +2\pi -2\pi\ln 2 \\ &=2\pi \end{aligned}$$

In spherical coordinates, $x=\rho\sin\varphi\cos\theta$, $y=\rho\sin\varphi\sin\theta$, $z=\rho\cos\varphi$ so that

$$\begin{aligned} &\frac{\rho^2\sin^2\varphi\cos\theta\sin\theta +\rho^3\sin\varphi\sin\theta\,\cos^2\varphi +\rho^3\sin\varphi\,\cos^2\varphi} {\rho^2\sin^2\varphi+\rho^4\cos^4\varphi} \\ & =\frac{\sin^2\varphi\cos\theta\sin\theta +\rho\sin\varphi\sin\theta\,\cos^2\varphi +\rho\sin\varphi\cos\theta\,\cos^2\varphi} {\sin^2\varphi+\rho^2\cos^4\varphi} \end{aligned}$$

As $(x,y,z)\to (0,0,0)$, the radius $\rho\to 0$ and the second and third terms in the numerator and the second term in the denominator converge to $0$. But that leaves

$$\frac{\sin^2\varphi\cos\theta\sin\theta}{\sin^2\varphi} =\cos\theta\sin\theta$$

which takes many different values. In particular, if we send $(x,y,z)\to (0,0,0)$ along either the $x$– or $y$–axis, that is with $z=0$ and either $x=0$ or $y=0$, then

$$\frac{xy+yz^2+xz^2}{x^2+y^2+z^4}\bigg|_{\genfrac{}{}{0pt}{}{x=0\text{ or }y=0}{z=0}}=0$$

converges to $0$. But, if we send $(x,y,z)\to (0,0,0)$ along the line $y=x$, $z=0$

$$\frac{xy+yz^2+xz^2}{x^2+y^2+z^4}\bigg|_{\genfrac{}{}{0pt}{}{y=x}{z=0}}=\frac{x^2}{2x^2} =\frac{1}{2}$$

converges to $1/2$. So $\frac{xy+yz^2+xz^2}{x^2+y^2+z^4}$ does not approach a single value as $(x,y,z)\to(0,0,0)$ and the limit does not exist.

The disk of radius $2$ centred at the origin in the $xy$–plane is $x^2+y^2\le 4$. So

$$\begin{aligned} V&=\big\{ \ (x,y,z)\ \big|\ x^2+y^2\le 4,\ 0\le z\le 2\ \big\} \end{aligned}$$

The cone with vertex at the origin that contains the top edge, $x+y^2=4$, $z=2$, of $U$ is $x^2+y^2=z^2$. So

$$\begin{aligned} U&=\big\{ \ (x,y,z)\ \big|\ x^2+y^2\le 4,\ 0\le z\le 2,\ x^2+y^2\ge z^2\ \big\} \end{aligned}$$

Here are sketches of the $y=0$ cross–section of $V$, on the left, and $U$, on the right.

(a) In cylindrical coordinates, $x^2+y^2\le 4$ becomes $r\le 2$ and $x^2+y^2\ge z^2$ is $r\ge |z|$, and the density is $\sqrt{x^2+y^2}=r$. So

$$\begin{aligned} U&=\big\{ \ (r\cos\theta,r\sin\theta,z)\ \big|\ r\le 2,\ 0\le z\le 2,\ r\ge z\ \big\} \end{aligned}$$

Looking at the figure on the left below, we see that, on $U$

• $z$ runs from $0$ to $2$, and
• for each $z$ is that range, $r$ runs from $z$ to $2$ and $\theta$ runs from $0$ to $2\pi$.
• $\mathrm{d}V =r\,\mathrm{d}r\,\mathrm{d}\theta\,\mathrm{d}z$

So

$$\begin{aligned} \text{Mass}&= \int_0^2\mathrm{d}z\int_0^{2\pi}\mathrm{d}\theta\int_z^2\mathrm{d}r\ r\ \overbrace{r}^{\text{density}} = \int_0^2\mathrm{d}z\int_0^{2\pi}\mathrm{d}\theta\int_z^2\mathrm{d}r\ r^2 \end{aligned}$$

(b) 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 $x^2+y^2\le 4$ becomes $\rho\sin\varphi\le 2$, and $x^2+y^2\ge z^2$ becomes

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

and the density $\sqrt{x^2+y^2}=\rho\sin\varphi$. So

$$\begin{aligned} U&=\big\{ \ (\rho\sin\varphi\cos\theta,\rho\sin\varphi\sin\theta,\rho\cos\varphi)\ \big|\ {\pi}/{4}\le\varphi\le{\pi}/{2},\ 0\le \theta\le 2\pi,\ \rho\sin\varphi\le 2\ \big\} \end{aligned}$$

Looking at the figure on the right above, we see that, on $U$

• $\varphi$ runs from $\frac{\pi}{4}$ (on the cone) to $\frac{\pi}{2}$ (on the $xy$–plane), and
• for each $\varphi$ is that range, $\rho$ runs from $0$ to $\frac{2}{\sin\varphi}$ and $\theta$ runs from $0$ to $2\pi$.
• $\mathrm{d}V =\rho^2\sin\varphi\,\mathrm{d}\rho\,\mathrm{d}\theta\,\mathrm{d}\varphi$

So

$$\begin{aligned} \text{Mass}&= \int_{\pi/4}^{\pi/2}\mathrm{d}\varphi\int_0^{2\pi}\mathrm{d}\theta \int_0^{2/\sin\varphi}\mathrm{d}\rho\ \rho^2\sin\varphi\ \overbrace{\rho\sin\varphi}^{\text{density}} \\ &= \int_{\pi/4}^{\pi/2}\mathrm{d}\varphi\int_0^{2\pi}\mathrm{d}\theta \int_0^{2/\sin\varphi}\mathrm{d}\rho\ \rho^3\sin^2\varphi \end{aligned}$$

(c) We'll use the cylindrical form.

$$\begin{aligned} \text{Mass} &= \int_0^2\mathrm{d}z\int_0^{2\pi}\mathrm{d}\theta\int_z^2\mathrm{d}r\ r^2 \\ &=2\pi \int_0^2\mathrm{d}z\ \frac{8-z^3}{3} \\ &=\frac{2\pi}{3}\ \left[16-\frac{2^4}{4}\right] \\ &=8\pi \end{aligned}$$

(a) Call the solid $V$. In cylindrical coordinates

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

Observe that $r^2+z^2= 2$ and $r= z$ intersect when $2r^2=2$ so that $r=z=1$. Here is a sketch of the $y=0$ cross–section of $E$.

So

$$V =\big\{ \ (r\cos\theta\,,\,r\sin\theta\,,\,z)\ \big|\ 0\le r\le 1,\ 0\le\theta\le2\pi,\ r\le z\le \sqrt{2-r^2}\ \big\}$$

and

$$\begin{aligned} M &=\iiint _V \rho(x,y,z)\ \mathrm{d}V =\int_0^1\mathrm{d}r\int_0^{2\pi}\mathrm{d}\theta \int_r^{\sqrt{2-r^2}}\mathrm{d}z\ r\overbrace{(r^2)}^{\delta } =\int_0^1\mathrm{d}r\int_0^{2\pi}\mathrm{d}\theta \int_r^{\sqrt{2-r^2}}\mathrm{d}z\ r^3 \end{aligned}$$

(b)

In spherical coordinates

• $x^2+y^2+z^2\le 2$ is $\rho\le\sqrt{2}$, and
• $\sqrt{x^2+y^2}\le z$ is $\rho\sin\varphi\le \rho\cos\varphi$, or $\tan\varphi\le 1$ or $\varphi\le\frac{\pi}{4}$, and
• the density $x^2+y^2=\rho^2\sin^2\varphi$, and
• $\mathrm{d}V$ is $\rho^2\sin\varphi\,\mathrm{d}\rho\,\mathrm{d}\theta\,\mathrm{d}\varphi$

So

$$V =\big\{ \ (\rho\sin\varphi\cos\theta\,,\,\rho\sin\varphi\sin\theta\,,\, \rho\cos\varphi)\ \big|\ 0\le\rho\le \sqrt{2},\ 0\le\theta\le2\pi,\ 0\le \varphi\le {\pi}/{4}\ \big\}$$

and, since the integrand $x^2+y^2=\rho^2\sin^2\varphi$,

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

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

$$\begin{aligned} M &=\int_0^{\sqrt{2}}\mathrm{d}\rho\int_0^{2\pi}\mathrm{d}\theta \int_0^{\pi/4}\mathrm{d}\varphi\ \rho^4\sin^3\varphi \\ &=\int_0^{\sqrt{2}}\mathrm{d}\rho\int_0^{2\pi}\mathrm{d}\theta \int_0^{\pi/4}\mathrm{d}\varphi\ \rho^4\sin\varphi \big[1-\cos^2\varphi\big] \\ &=2\pi \int_0^{\sqrt{2}}\mathrm{d}\rho\ \rho^4 \left[-\cos\varphi +\frac{\cos^3\varphi}{3}\right]_0^{\pi/4} =2\pi \left[\frac{2}{3} - \frac{5}{6\sqrt{2}}\right] \int_0^{\sqrt{2}}\mathrm{d}\rho\ \rho^4 \\ &=2\pi \frac{4\sqrt{2}}{5}\left[\frac{2}{3} - \frac{5}{6\sqrt{2}}\right] =\pi \left[\frac{16\sqrt{2}}{15} - \frac{4}{3}\right] \approx 0.5503 \end{aligned}$$