다중적분 전체 문제

The top surface $x^2+y^2+z^2=2$ meets the bottom surface $z=x^2+y^2$ when $z$ obeys $x^2+y^2=z=2-z^2$. That is, when $0=z^2+z-2=(z-1)(z+2)$. The root $z=-2$ is inconsistent with $z=x^2+y^2\ge 0$. So the top and bottom surfaces meet at the circle $z=1$, $x^2+y^2=1$.

In polar coordinates, the top surface is $z^2=2-r^2$, or equivalently $z=\sqrt{2-r^2}$, and the bottom surface is $z=r^2$. So the height of the volume above the point with polar coordinates $(r,\theta)$ is $\sqrt{2-r^2}-r^2$ and

$$\begin{aligned} \text{Volume}&=\int_0^1\mathrm{d}r\int_0^{2\pi} \mathrm{d}\theta\ r\,\big[\sqrt{2-r^2}-r^2\big] =2\pi \int_0^1\mathrm{d}r\ r\,\big[\sqrt{2-r^2}-r^2\big]\\ &=2\pi \left[-\frac{1}{3}{(2-r^2)}^{3/2}-\frac{r^4}{4}\right]_0^1 =2\pi \left[-\frac{1}{3}-\frac{1}{4}+\frac{1}{3}2^{3/2}\right] \\ &=\pi \left[\frac{4}{3}\sqrt{2}-\frac{7}{6}\right] \approx 2.26 \end{aligned}$$

In Cartesian coordinates

$$\begin{aligned} \text{Volume}&=4\int_0^1\mathrm{d}x\int_0^{\sqrt{1-x^2}} \mathrm{d}y\ \big[\sqrt{2-x^2-y^2}-x^2-y^2\big] \end{aligned}$$

The $y$ integral can be done using the substitution $y=\sqrt{2-x^2}\cos t$, but it is easier to use polar coordinates.

For this region $x$ and $y$ run over the interior of the cylinder $x^2+(y-a)^2=a^2$. For each $(x,y)$ inside the cylinder, $z$ runs from $-\sqrt{x^2+y^2}$ to $\sqrt{x^2+y^2}$. As $x^2+(y-a)^2=a^2$ if and only if $x^2+y^2-2ay=0$, the cylinder has equation $r^2=2ar\sin\theta$, or equivalently, $r=2a\sin\theta$, in polar coordinates.

Thus $(r,\theta)$ runs over $0\le\theta\le\pi,\ 0\le r\le 2a\sin\theta$ and for each $(r,\theta)$ in this region $z$ runs from $-r$ to $r$. By symmetry under $x\rightarrow -x$, the volume is

$$\begin{aligned} \text{Volume} &=2\int_0^{{\pi\over 2}}\mathrm{d}\theta\int_0^{2a\sin\theta}\mathrm{d}r\ r\big[r-(-r)\big] =4\int_0^{{\pi\over 2}}\mathrm{d}\theta\int_0^{2a\sin\theta}\mathrm{d}r\ r^2 =\frac{4}{3}\int_0^{{\pi\over 2}}\!\!\mathrm{d}\theta\ (2a\sin\theta)^3 \\ &=\frac{32}{3}a^3\int_0^{{\pi\over 2}}\!\!\mathrm{d}\theta\ \sin\theta(1-\cos^2\theta) =-\frac{32}{3}a^3\int_1^0 dt\ (1-t^2)\text{ where $t=\cos\theta$ }\\ &=-\frac{32}{3}a^3\left[t-\frac{t^3}{3}\right]_1^0 =\frac{64}{9}a^3 \end{aligned}$$

The figure below shows the top view of the specified solid. $(x,y)$ runs over the interior of the circle $x^2+y^2=2ax$. For each fixed $(x,y)$ in this disk, $z$ runs from $-\sqrt{2ax}$ to $+\sqrt{2ax}$. In polar coordinates, the circle is $r^2=2ar\cos\theta$ or $r=2a\cos\theta$.

The solid is symmetric under $y\rightarrow -y$ and $z\rightarrow -z$, so we can restrict to $y\ge 0$, $z\ge 0$ and multiply by 4. The volume is

$$\begin{aligned} \text{Volume} &=4\int_0^{\pi\over 2} \mathrm{d}\theta\int_0^{2a\cos\theta} \mathrm{d}r\ r\sqrt{2ar\cos\theta} \\ &=4\int_0^{\pi\over 2} \mathrm{d}\theta\ \sqrt{2a\cos\theta}\ \frac{2}{5}\ r^{5/2}\bigg|_0^{2a\cos\theta} \\ &=\frac{8}{5}\int_0^{\pi\over 2} \mathrm{d}\theta\ \big(2a\cos\theta\big)^3 =\frac{64}{5}a^3\int_0^{\pi\over 2} \mathrm{d}\theta\ \cos\theta\big(1-\sin^2\theta\big)\\ &=\frac{64}{5}a^3\int_0^1 dt\ \big(1-t^2\big) \qquad\text{ where $t=\sin\theta$}\\ &=\frac{64}{5}a^3\left[t-\frac{t^3}{3}\right]_0^1 =\frac{128}{15}a^3 \end{aligned}$$

(a)

• The equation $x^2 + y^2 \le 2y$ is equivalent to the equation $x^2 + (y-1)^2=1$, which is the equation of the cylinder whose $z=z_0$ cross–section is the horizontal circle of radius $1$, centred on $x=0$, $y=1$, $z=z_0$. The part of this cylinder in the first octant is sketched in the figure on the left below.
• $z \le \sqrt{x^2 + y^2}$ is the equation of the cone with vertex $(0,0,0)$, and axis the positive $z$–axis, whose radius at height $z=2$ is $2$. The part of this cone in the first octant is sketched in the figure on the right below.

The region $E$ is the part of the cylinder that is above the $xy$–plane (since $z\ge 0$) outside the cone (since $z\le\sqrt{x^2+y^2}$). The part of $E$ that is in the first octant is outlined in red in the figure below. Both $x^2 + y^2 \le 2y$ and $0 \le z \le \sqrt{x^2 + y^2}$ are invariant under $x\rightarrow -x$. So $E$ is also invariant under $x\rightarrow -x$. That is, $E$ is symmetric about the $yz$–plane and contains, in the octant $x\le 0$, $y\ge 0$, $z\ge 0$, a mirror image of the first octant part of $E$.

(b) In polar coordinates, $x^2 + y^2 \le 2y$ becomes

$$r^2\le 2r\sin\theta \iff r\le 2\sin\theta$$

Let us denote by $D$ the base region of the part of $E$ in the first octant (i.e. the shaded region in the figure above). Think of $D$ as being part of the $xy$–plane. In polar coordinates, on $D$

• $\theta$ runs from $0$ to $\frac{\pi}{2}$. (Recall that $D$ is contained in the first quadrant.)
• For each $\theta$ in that range, $r$ runs from $0$ to $2\sin\theta$.

Because

• in polar coordinates $\mathrm{d}A = r\,\mathrm{d}r\,\mathrm{d}\theta$, and
• the height of $E$ above each point $(x,y)$ in $D$ is $\sqrt{x^2+y^2}$, or, in polar coordinates, $r$, and
• the volume of $E$ is twice the volume of the part of $E$ in the first octant,

we have

$$\begin{aligned} \text{Volume}(E) &= 2\int_0^{\frac{\pi}{2}}\mathrm{d}\theta \int_0^{2\sin\theta}\mathrm{d}r\ r^2 \\ &=\frac{16}{3} \int_0^{\frac{\pi}{2}}\mathrm{d}\theta\ \sin^3\theta =\frac{16}{3} \int_0^{\frac{\pi}{2}}\mathrm{d}\theta\ \sin\theta\ \big(1-\cos^2\theta\big) \\ &=-\frac{16}{3} \int_1^0\mathrm{d}u\ \big(1-u^2\big)\qquad \text{with }u=\cos\theta,\ \mathrm{d}u=-\sin\theta\ \mathrm{d}\theta \\ &=\frac{16}{3}\left[1-\frac{1}{3}\right] \\ &=\frac{32}{9} \end{aligned}$$

On the domain of integration

• $x$ runs for $0$ to $2$, and
• for each fixed $x$ in that range, $y$ runs from $0$ to $\sqrt{4-x^2}$. The equation $y=\sqrt{4-x^2}$ is equivalent to $x^2+y^2=4$, $y\ge 0$.

This domain is sketched in the figure on the left below.

Considering that

• the integrand, ${(x^2+y^2)}^{\frac{3}{2}}$, is invariant under rotations about the origin and
• the outer curve, $x^2+y^2=4$, is invariant under rotations about the origin

we'll use polar coordinates. In polar coordinates,

• the outer curve, $x^2+y^2=4$, is $r=2$, and
• the integrand, ${(x^2+y^2)}^{\frac{3}{2}}$ is $r^3$, and
• $\mathrm{d}A=r\,\mathrm{d}r\,\mathrm{d}\theta$

Looking at the figure on the right above, we see that the given integral is, in polar coordinates,

$$\begin{aligned} \int_0^{\pi/2}\mathrm{d}\theta\int_0^2\mathrm{d}r\ r (r^3) &=\frac{\pi}{2}\ \frac{2^5}{5} =\frac{16\pi}{5} \end{aligned}$$

(a) The region $\mathcal{L}$ is sketched in the figure on the leflt below.

(b) In polar coordinates

• the circle $x^2+y^2=2$ is $r^2=2$ or $r=\sqrt{2}$, and
• the circle $x^2+y^2=4$ is $r^2=4$ or $r=2$, and
• the line $y=x$ is $r\sin\theta = r\cos\theta$, or $\tan\theta =1$, or (for the part in the first quadrant) $\theta=\frac{\pi}{4}$, and
• the positive $x$–axis ($y=0$, $x\ge 0$) is $\theta=0$

Looking at the figure on the right above, we see that, in $\mathcal{L}$,

• $\theta$ runs from $0$ to $\frac{\pi}{4}$, and
• for each fixed $\theta$ in that range, $r$ runs from $\sqrt{2}$ to $2$.
• $\mathrm{d}A$ is $r\,\mathrm{d}r\,\mathrm{d}\theta$

So

$$\begin{aligned} M = \int_0^{\pi/4}\mathrm{d}\theta\int_{\sqrt{2}}^2\mathrm{d}r\ r\, \rho(r\cos\theta\,,\,r\sin\theta) \end{aligned}$$

(c) When

$$\begin{aligned} \rho = \frac{2xy}{x^2+y^2} = \frac{2r^2\cos\theta\,\sin\theta}{r^2} =\sin(2\theta) \end{aligned}$$

we have

$$\begin{aligned} M &= \int_0^{\pi/4}\mathrm{d}\theta\int_{\sqrt{2}}^2\mathrm{d}r\ r\, \sin(2\theta) \\ &=\left[\int_0^{\pi/4}\sin(2\theta)\ \mathrm{d}\theta\right] \left[\int_{\sqrt{2}}^2 r\ \mathrm{d}r\right] \\ &=\left[-\frac{1}{2}\cos(2\theta)\right]_0^{\pi/4} \left[ \frac{r^2}{2}\right]_{\sqrt{2}}^2 =\frac{1}{2}\ \frac{4-2}{2} \\ &=\frac{1}{2} \end{aligned}$$

We'll use polar coordinates. The domain of integration is

$$\mathbb{R}^2=\big\{ \ (r\cos\theta\,,\,r\sin\theta)\ \big|\ 0\le r<\infty,\ 0\le\theta\le 2\pi\ \big\}$$

The given integral is improper, so we'll start by integrating $r$ from $0$ to an arbitrary $R>0$, and then we'll take the limit $R\rightarrow\infty$. In polar coordinates, the integrand $\frac{1}{{(1+x^2+y^2)}^2}=\frac{1}{{(1+r^2)}^2}$, and $\mathrm{d}A=r\,\mathrm{d}r\,\mathrm{d}\theta$, so

$$\begin{aligned} \iint _{\mathbb{R}^2} \frac{1}{{(1+x^2+y^2)}^2}\ \mathrm{d}A &=\lim_{R\rightarrow\infty} \int_0^{2\pi}\mathrm{d}\theta\int_0^R\mathrm{d}r \frac{r}{{(1+r^2)}^2} \\ &=\lim_{R\rightarrow\infty} \int_0^{2\pi}\mathrm{d}\theta\ \left[-\frac{1}{2(1+r^2)}\right]_0^R \\ &=\lim_{R\rightarrow\infty} 2\pi \left[\frac{1}{2}-\frac{1}{2(1+R^2)}\right] \\ &=\pi \end{aligned}$$

Let's switch to polar coordinates. In polar coordinates, the circle $x^2+y^2=2$ is $r=\sqrt{2}$ and the line $y=x$ is $\theta=\frac{\pi}{4}$.

In polar coordinates $\mathrm{d}A = r\,\mathrm{d}r\,\mathrm{d}\theta$, so the integral

$$\begin{aligned} \iint _D y\sqrt{x^2+y^2}\,\mathrm{d}A &=\int_0^{\pi/4}\mathrm{d}\theta\int_0^{\sqrt{2}}\mathrm{d}r\ r\ \overbrace{r\sin\theta}^{y}\ \overbrace{r}^{\sqrt{x^2+y^2}} \\ &=\int_0^{\pi/4}\mathrm{d}\theta\ \sin\theta\ \left[\frac{r^4}{4}\right]_0^{\sqrt{2}} \\ &=\Big[-\cos\theta\Big]_0^{\pi/4} \\ &= 1-\frac{1}{\sqrt{2}} \end{aligned}$$