극좌표 이중적분

(a) Observe that

• the condition $x^2+y^2\le 1$ restricts $G$ to the interior of the circle of radius $1$ centred on the origin, and
• the conditions $0\le x\le 2y$ restricts $G$ to $x\ge 0$, $y\ge 0$, i.e. to the first quadrant, and
• the conditions $x\le 2y$ and $y\le 2x$ restrict $\frac{x}{2}\le y\le 2x$. So $G$ lies below the (steep) line $y=2x$ and lies above the (not steep) line $y=\frac{x}{2}$.

Here is a sketch of $G$

(b) Observe that the line $y=2x$ crosses the circle $x^2+y^2=1$ at a point $(x,y)$ obeying

$$\begin{aligned} x^2+(2x)^2=x^2+y^2=1 \implies 5x^2=1 \end{aligned}$$

and that the line $x=2y$ crosses the circle $x^2+y^2=1$ at a point $(x,y)$ obeying

$$\begin{aligned} (2y)^2+y^2=x^2+y^2=1 \implies 5y^2=1 \end{aligned}$$

So the intersection point of $y=2x$ and $x^2+y^2=1$ in the first octant is $\left(\frac{1}{\sqrt{5}}\,,\,\frac{2}{\sqrt{5}}\right)$ and the intersection point of $x=2y$ and $x^2+y^2=1$ in the first octant is $\left(\frac{2}{\sqrt{5}}\,,\,\frac{1}{\sqrt{5}}\right)$.

We'll set up the iterated integral using horizontal strips as in the sketch

Looking at that sketch, we see that, on $G$,

• $y$ runs from $0$ to $\frac{2}{\sqrt{5}}$, and
• for each fixed $y$ between $0$ and $\frac{1}{\sqrt{5}}$, $x$ runs from $\frac{y}{2}$ to $2y$, and
• for each fixed $y$ between $\frac{1}{\sqrt{5}}$ and $\frac{2}{\sqrt{5}}$ $x$ runs from $\frac{y}{2}$ to $\sqrt{1-y^2}$.

So

$$\begin{aligned} \iint _G f(x,y)\ \mathrm{d}A &=\int_0^{\frac{1}{\sqrt{5}}}\mathrm{d}y\int_{y/2}^{2y}\mathrm{d}x\ f(x,y) +\int_{\frac{1}{\sqrt{5}}}^{\frac{2}{\sqrt{5}}}\mathrm{d}y\int_{y/2}^{\sqrt{1-y^2}} \mathrm{d}x\ f(x,y) \end{aligned}$$

(b) In polar coordinates

• the equation $x^2+y^2=1$ becomes $r=1$, and
• the equation $y=x/2$ becomes $r\sin\theta = \frac{r}{2}\cos\theta$ or $\tan\theta =\frac{1}{2}$, and
• the equation $y=2x$ becomes $r\sin\theta = 2r\cos\theta$ or $\tan\theta =2$.

Looking at the sketch

we see that, on $G$,

• $\theta$ runs from $\arctan\frac{1}{2}$ to $\arctan 2$, and
• for each fixed $\theta$ in that range, $r$ runs from $0$ to $1$.

As $\mathrm{d}A=r\,\mathrm{d}r\,\mathrm{d}\theta$, and $x=r\cos\theta$, $y=r\sin\theta$,

$$\begin{aligned} \iint _G f(x,y)\ \mathrm{d}A &=\int_{\arctan\frac{1}{2}}^{\arctan 2}\mathrm{d}\theta \int_0^1\mathrm{d}r\ r\,f(r\cos\theta,r\sin\theta) \end{aligned}$$

(a) On the domain of integration

• $y$ runs from $0$ to $\sqrt{2}$ and
• for each $y$ in that range, $x$ runs from $y$ to $\sqrt{4-y^2}$. We can rewrite $x=\sqrt{4-y^2}$ in the more familiar form $x^2+y^2=4$, $x\ge 0$.

The figure on the left below provides a sketch of the domain of integration. It also shows the generic horizontal slice that was used to set up the given iterated integral.

(b) To reverse the order of integration observe, we use vertical, rather than horizontal slices. From the figure on the right above that, on the domain of integration,

• $x$ runs from $0$ to $2$ and
• for each $x$ in the range $0\le x\le\sqrt{2}$, $y$ runs from $0$ to $x$.
• for each $x$ in the range $\sqrt{2}\le x\le 2$, $y$ runs from $0$ to $\sqrt{4-x^2}$.

So the integral

$$J = \int_0^{\sqrt{2}}\int_0^x \frac{y}{x} e^{x^2+y^2}\ \mathrm{d}y\,\mathrm{d}x + \int_{\sqrt{2}}^2\int_0^{\sqrt{4-x^2}} \frac{y}{x} e^{x^2+y^2}\ \mathrm{d}y\,\mathrm{d}x$$

(c) In polar coordinates, the line $y=x$ is $\theta=\frac{\pi}{4}$, the circle $x^2+y^2=4$ is $r=2$, and $\mathrm{d}x\,\mathrm{d}y=r\,\mathrm{d}r\,\mathrm{d}\theta$. So

$$\begin{aligned} J&=\int_0^{\pi/4}\mathrm{d}\theta\int_0^2\mathrm{d}r\ r \overbrace{\frac{r\sin\theta}{r\cos\theta}}^{\frac{y}{x}}e^{r^2} \\ &=\int_0^{\pi/4}\mathrm{d}\theta\ \frac{\sin\theta}{\cos\theta} \left[\frac{1}{2} e^{r^2}\right]_0^2 \\ &=-\frac{1}{2}\left[e^4-1\right] \int_1^{1/\sqrt{2}}\mathrm{d}u\ \frac{1}{u} \qquad\text{with }u=\cos\theta,\ \mathrm{d}u=-\sin\theta\,\mathrm{d}\theta \\ &=-\frac{1}{2}\left[e^4-1\right]\ \Big[\ln|u|\Big]_1^{1/\sqrt{2}} \\ &= \frac{1}{4}\left[e^4-1\right]\ln 2 \end{aligned}$$

The paraboloid hits the $xy$–plane at $\frac{x^2}{a^2} +\frac{y^2}{b^2}=1$.

$$\begin{aligned} \text{Volume} &=\int_0^a \mathrm{d}x\int_0^{b\sqrt{1-{x^2\over a^2}}} \mathrm{d}y\ \left(1-\frac{x^2}{a^2}-\frac{y^2}{b^2}\right) \\ &=b\int_0^a \mathrm{d}x\int_0^{\sqrt{1-{x^2\over a^2}}} \mathrm{d}v\ \left(1-\frac{x^2}{a^2}-v^2\right) \qquad\text{ where $y=bv$}\\ \end{aligned}$$

Think of this integral as being of the form

$$\displaystyle b\int_0^a \mathrm{d}x\ g(x)\quad \text{with}\quad \displaystyle g(x)=\int_0^{\sqrt{1-{x^2\over a^2}}} \mathrm{d}v\ \left(1-\frac{x^2}{a^2}-v^2\right)$$

Then, substituting $x=au$,

$$\begin{aligned} \text{Volume} &=ab\int_0^1 \mathrm{d}u\int_0^{\sqrt{1-u^2}} \mathrm{d}v\ \big(1-u^2-v^2\big) \\ &=ab\iint _{u^2+v^2\le 1\atop u,v\ge 0} \mathrm{d}u \mathrm{d}v\ \big(1-u^2-v^2\big) \end{aligned}$$

Now switch to polar coordinates using $u=r\cos\theta$, $v=r\sin\theta$.

$$\begin{aligned} \text{Volume} &=ab\int_0^1 \mathrm{d}r\int_0^{\pi\over 2} \mathrm{d}\theta\ r\big(1-r^2\big) =ab\ \frac{\pi}{2}\ \left[\frac{r^2}{2}-\frac{r^4}{4}\right]_0^1 =\frac{\pi}{8}ab \end{aligned}$$

Let $r(z)$ be the radius of the urn at height $z$ above its middle. Because the bounding surface of the urn is parabolic, $r(z)$ must be a quadratic function of $z$ that varies between $3$ at $z=0$ and $2$ at $z=\pm 6$. That is, $r(z)$ must be of the form $r(z)=az^2+bz+c$. The condition that $r(0)=3$ tells us that $c=3$. The conditions that $r(\pm 6)=2$ tells us that

$$\begin{aligned} 6^2a + 6b + 3 &=2 \\ 6^2a - 6b + 3 &=2 \end{aligned}$$

So $b=0$ and $6^2a=-1$ so that $a=-\frac{1}{6^2}$. All together $r(z)=3-\big(\frac{z}{6}\big)^2$.

Slice the urn into horzontal slices, with the slice at height $z$ a disk of radius $r(z)$ and thickness $\mathrm{d}z$ and hence of volume $\pi r(z)^2\mathrm{d}z$. The volume to height $z_0$ is

$$\begin{aligned} V(z) &=\int_{-6}^{z_0} \mathrm{d}z\ \pi r(z)^2 =\int_{-6}^{z_0} \mathrm{d}z\ \pi\left[3-\frac{z^2}{36}\right]^2 =\pi\left[9z-\frac{z^3}{18}+\frac{z^5}{5\times36^2}\right]_{-6}^{z_0}\\ \end{aligned}$$

We are told that the mark is to be at the 6 cup level and that the urn holds 24 cups. So the mark is to be at the height $z_0$ for which the volume, $V(z_0)$, is one quarter of the total volume, $V(6)$. That is, we are to choose $z_0$ so that $V(z_0)=\frac{1}{4} V(6)$ or

$$\pi\left[9z-\frac{z^3}{18}+\frac{z^5}{5\times36^2}\right]_{-6}^{z_0} =\frac{\pi}{4}\left[9z-\frac{z^3}{18}+\frac{z^5}{5\times36^2}\right]_{-6}^{6} =\frac{\pi}{2}\left[9\times 6-\frac{6^3}{18}+\frac{6^5}{5\times36^2}\right]$$

or

$$9z_0-\frac{z_0^3}{18}+\frac{z_0^5}{5\times36^2} =-\frac{1}{2}\left[9\times 6-\frac{6^3}{18}+\frac{6^5}{5\times36^2}\right] =-21.60$$

Since $\Big[9z_0-\frac{z_0^3}{18}+\frac{z_0^5}{6480}\Big]_{z_0=-2.495}=-21.61$ and $\Big[9z_0-\frac{z_0^3}{18}+\frac{z_0^5}{6480}\Big]_{z_0=-2.490}=-21.57$, there is a solution $z_0=-2.49$ (to two decimal places). The mark should be about 3.5” above the bottom.

(a) In polar coordinates, the base region $x^2+y^2\le 9$ is $r\le 3$, $0\le\theta\le 2\pi$. So the

$$\begin{aligned} \text{Volume} &=\iint _{x^2+y^2\le 9} e^{x^2+y^2}\ \mathrm{d}x\mathrm{d}y =\int_0^3 \mathrm{d}r\int_0^{2\pi}\mathrm{d}\theta\ r e^{r^2} =2\pi\int_0^3\mathrm{d}r\ r e^{r^2} =\pi e^{r^2}\Big|_0^3 \\ &=\pi\big(e^9-1\big)\approx 25,453 \end{aligned}$$

(b) The two integrals have domains

$$\begin{aligned} \big\{ \ (x,y)\ \big|\ 0\le y\le 1,\ 0\le x\le y\ \big\}\qquad \big\{ \ (x,y)\ \big|\ 1\le y\le 2,\ 0\le x\le 2-y\ \big\} \end{aligned}$$

The union of those two domains (as well as horizontal strips that were used in setting up the two given integrals) is sketched in the figure on the left below.

To reverse the order of integration, we decompose the domain using vertical strips as in the figure on the right above. As

• $x$ runs from $0$ to $1$ and
• for each fixed $x$ between $0$ and $1$, $y$ runs from $x$ to $2-x$.

we have that the

$$\text{Volume}=\int_0^1 \mathrm{d}x\int_x^{2-x}\mathrm{d}y\ e^{x^2+y^2}$$