We have used iterated integrals to evaluate double integrals, which give the signed volume under a surface, $z=f(x,y)$, over a region $R$ of the $x$-$y$ plane. The integrand is simply $f(x,y)$, and the bounds of the integrals are determined by the region $R$.
Some regions $R$ are easy to describe using rectangular coordinates — that is, with equations of the form $y=f(x)$, $x=a$, etc. However, some regions are easier to handle if we represent their boundaries with polar equations of the form $r=f(\theta )$, $\theta =\alpha $, etc.
The basic form of the double integral is ${\iint}_{R}f(x,y)\mathrm{d}A$. We interpret this integral as follows: over the region $R$, sum up lots of products of heights (given by $f({x}_{i},{y}_{i})$) and areas (given by $\mathrm{\Delta}{A}_{i}$). That is, $dA$ represents “a little bit of area.” In rectangular coordinates, we can describe a small rectangle as having area $\mathrm{d}x\mathrm{d}y$ or $\mathrm{d}y\mathrm{d}x$ — the area of a rectangle is simply length$\times $width — a small change in $x$ times a small change in $y$. Thus we replace $\mathrm{d}A$ in the double integral with $\mathrm{d}x\mathrm{d}y$ or $\mathrm{d}y\mathrm{d}x$.
Now consider representing a region $R$ with polar coordinates. Consider Figure 14.3.1(a). Let $R$ be the region in the first quadrant bounded by the curve. We can approximate this region using the natural shape of polar coordinates: portions of sectors of circles. In the figure, one such region is shaded, shown again in part (b) of the figure.
As the area of a sector of a circle with radius $r$, subtended by an angle $\theta $, is $A=\frac{1}{2}{r}^{2}\theta $, we can find the area of the shaded region. The whole sector has area $\frac{1}{2}{r}_{2}^{2}\mathrm{\Delta}\theta $, whereas the smaller, unshaded sector has area $\frac{1}{2}{r}_{1}^{2}\mathrm{\Delta}\theta $. The area of the shaded region is the difference of these areas:
$$\mathrm{\Delta}{A}_{i}=\frac{1}{2}{r}_{2}^{2}\mathrm{\Delta}\theta -\frac{1}{2}{r}_{1}^{2}\mathrm{\Delta}\theta =\frac{1}{2}\left({r}_{2}^{2}-{r}_{1}^{2}\right)\left(\mathrm{\Delta}\theta \right)=\frac{{r}_{2}+{r}_{1}}{2}\left({r}_{2}-{r}_{1}\right)\mathrm{\Delta}\theta .$$ |
Note that $({r}_{2}+{r}_{1})/2$ is just the average of the two radii.
To approximate the region $R$, we use many such subregions; doing so shrinks the difference ${r}_{2}-{r}_{1}$ between radii to 0 and shrinks the change in angle $\mathrm{\Delta}\theta $ also to 0. We represent these infinitesimal changes in radius and angle as $dr$ and $d\theta $, respectively. Finally, as $dr$ is small, ${r}_{2}\approx {r}_{1}$, and so $({r}_{2}+{r}_{1})/2\approx {r}_{1}$. Thus, when $\mathrm{d}r$ and $\mathrm{d}\theta $ are small,
$$\mathrm{\Delta}{A}_{i}\approx {r}_{i}\mathrm{d}r\mathrm{d}\theta .$$ |
Taking a limit, where the number of subregions goes to infinity and both ${r}_{2}-{r}_{1}$ and $\mathrm{\Delta}\theta $ go to 0, we get
$$\mathrm{d}A=r\mathrm{d}r\mathrm{d}\theta .$$ |
So to evaluate ${\iint}_{R}f(x,y)\mathrm{d}A$, replace $\mathrm{d}A$ with $r\mathrm{d}r\mathrm{d}\theta $. Convert the function $z=f(x,y)$ to a function with polar coordinates with the substitutions $x=r\mathrm{cos}\theta $, $y=r\mathrm{sin}\theta $. Finally, find bounds ${g}_{1}(\theta )\le r\le {g}_{2}(\theta )$ and $\alpha \le \theta \le \beta $ that describe $R$. This is the key principle of this section, so we restate it here as a Key Idea.
Let $R$ be a plane region bounded by the polar equations $\alpha \le \theta \le \beta $ and ${g}_{1}(\theta )\le r\le {g}_{2}(\theta )$. Then
$${\iint}_{R}f(x,y)\mathrm{d}A={\int}_{\alpha}^{\beta}{\int}_{{g}_{1}(\theta )}^{{g}_{2}(\theta )}f(r\mathrm{cos}\theta ,r\mathrm{sin}\theta )r\mathrm{d}r\mathrm{d}\theta .$$ |
Watch the video:
Double Integral Using Polar Coordinates — Part 1 of 3 from https://youtu.be/sQM-8Oj4Ecg
Examples will help us understand this Key Idea.
Find the signed volume under the plane $z=4-x-2y$ over the disk with equation ${x}^{2}+{y}^{2}\le 1$.
Solution^{†}^{†}margin:
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The bounds of the integral are determined solely by the region $R$ over which we are integrating. The surface and boundary of the region are shown in Figure 14.3.2. In this case the boundary of the region is the circle with equation ${x}^{2}+{y}^{2}=1$. We need to find polar bounds for this region. It may help to review Section 10.4; bounds for this circle are $0\le r\le 1$ and $0\le \theta \le 2\pi $.
We replace $f(x,y)$ with $f(r\mathrm{cos}\theta ,r\mathrm{sin}\theta )$. That means we make the following substitutions:
$$4-x-2y\mathit{\hspace{1em}}\Rightarrow \mathit{\hspace{1em}}4-r\mathrm{cos}\theta -2r\mathrm{sin}\theta .$$ |
Finally, we replace $dA$ in the double integral with $r\mathrm{d}r\mathrm{d}\theta $. This gives the final iterated integral, which we evaluate:
${\iint}_{R}}f(x,y)\mathrm{d}A$ | $={\displaystyle {\int}_{0}^{2\pi}}{\displaystyle {\int}_{0}^{1}}\left(4-r\mathrm{cos}\theta -2r\mathrm{sin}\theta \right)r\mathrm{d}r\mathrm{d}\theta $ | ||
$={\displaystyle {\int}_{0}^{2\pi}}{\displaystyle {\int}_{0}^{1}}\left(4r-{r}^{2}(\mathrm{cos}\theta -2\mathrm{sin}\theta )\right)\mathrm{d}r\mathrm{d}\theta $ | |||
$={{\displaystyle {\int}_{0}^{2\pi}}\left(2{r}^{2}-{\displaystyle \frac{1}{3}}{r}^{3}(\mathrm{cos}\theta -2\mathrm{sin}\theta )\right)|}_{0}^{1}\mathrm{d}\theta $ | |||
$={\displaystyle {\int}_{0}^{2\pi}}\left(2-{\displaystyle \frac{1}{3}}\left(\mathrm{cos}\theta -2\mathrm{sin}\theta \right)\right)\mathrm{d}\theta $ | |||
$={\left(2\theta -{\displaystyle \frac{1}{3}}\left(\mathrm{sin}\theta +2\mathrm{cos}\theta \right)\right)|}_{0}^{2\pi}$ | |||
$=4\pi .$ |
Find the volume under the paraboloid $z=4-{(x-2)}^{2}-{y}^{2}$ over the region bounded by the circles ${(x-1)}^{2}+{y}^{2}=1$ and ${(x-2)}^{2}+{y}^{2}=4$.
SolutionAt first glance, this seems like a very hard volume to compute as the region $R$ (shown in Figure 14.3.3(a)) has a hole in it, cutting out a strange portion of the surface, as shown in part (b) of the figure. However, by describing $R$ in terms of polar equations, the volume is not very difficult to compute. We can use techniques from Section 10.4 to show that the circle ${(x-1)}^{2}+{y}^{2}=1$ has polar equation $r=2\mathrm{cos}\theta $, and that the circle ${(x-2)}^{2}+{y}^{2}=4$ has polar equation $r=4\mathrm{cos}\theta $. Each of these circles is traced out on the interval $0\le \theta \le \pi $. The bounds on $r$ are $2\mathrm{cos}\theta \le r\le 4\mathrm{cos}\theta .$
Replacing $x$ with $r\mathrm{cos}\theta $ in the integrand, along with replacing $y$ with $r\mathrm{sin}\theta $, prepares us to evaluate the double integral ${\iint}_{R}f(x,y)\mathrm{d}A$:
${\iint}_{R}}f(x,y)\mathrm{d}A$ | $={\displaystyle {\int}_{0}^{\pi}}{\displaystyle {\int}_{2\mathrm{cos}\theta}^{4\mathrm{cos}\theta}}\left(4-{\left(r\mathrm{cos}\theta -2\right)}^{2}-{\left(r\mathrm{sin}\theta \right)}^{2}\right)r\mathrm{d}r\mathrm{d}\theta $ | ||
$={\displaystyle {\int}_{0}^{\pi}}{\displaystyle {\int}_{2\mathrm{cos}\theta}^{4\mathrm{cos}\theta}}\left(-{r}^{3}+4{r}^{2}\mathrm{cos}\theta \right)\mathrm{d}r\mathrm{d}\theta $ | |||
$={{\displaystyle {\int}_{0}^{\pi}}\left(-{\displaystyle \frac{1}{4}}{r}^{4}+{\displaystyle \frac{4}{3}}{r}^{3}\mathrm{cos}\theta \right)|}_{2\mathrm{cos}\theta}^{4\mathrm{cos}\theta}\mathrm{d}\theta $ | |||
$={\displaystyle {\int}_{0}^{\pi}}([-{\displaystyle \frac{1}{4}}(256{\mathrm{cos}}^{4}\theta )+{\displaystyle \frac{4}{3}}(64{\mathrm{cos}}^{4}\theta )]-$ | |||
$[-{\displaystyle \frac{1}{4}}(16{\mathrm{cos}}^{4}\theta )+{\displaystyle \frac{4}{3}}(8{\mathrm{cos}}^{4}\theta )])\mathrm{d}\theta $ | |||
$={\displaystyle {\int}_{0}^{\pi}}{\displaystyle \frac{44}{3}}{\mathrm{cos}}^{4}\theta \mathrm{d}\theta .$ |
To integrate ${\mathrm{cos}}^{4}\theta $, rewrite it as ${\mathrm{cos}}^{2}\theta {\mathrm{cos}}^{2}\theta $ and employ the half-angle formula twice:
${\mathrm{cos}}^{4}\theta $ | $={\mathrm{cos}}^{2}\theta {\mathrm{cos}}^{2}\theta $ | ||
$={\displaystyle \frac{1}{2}}\left(1+\mathrm{cos}(2\theta )\right){\displaystyle \frac{1}{2}}\left(1+\mathrm{cos}(2\theta )\right)$ | |||
$={\displaystyle \frac{1}{4}}\left(1+2\mathrm{cos}(2\theta )+{\mathrm{cos}}^{2}(2\theta )\right)$ | |||
$={\displaystyle \frac{1}{4}}\left(1+2\mathrm{cos}(2\theta )+{\displaystyle \frac{1}{2}}\left(1+\mathrm{cos}(4\theta )\right)\right)$ | |||
$={\displaystyle \frac{3}{8}}+{\displaystyle \frac{1}{2}}\mathrm{cos}(2\theta )+{\displaystyle \frac{1}{8}}\mathrm{cos}(4\theta ).$ |
Picking up from where we left off above, we have
${\iint}_{R}}f(x,y)\mathrm{d}A$ | $={\displaystyle {\int}_{0}^{\pi}}{\displaystyle \frac{44}{3}}{\mathrm{cos}}^{4}\theta \mathrm{d}\theta $ | ||
$={\displaystyle {\int}_{0}^{\pi}}{\displaystyle \frac{44}{3}}\left({\displaystyle \frac{3}{8}}+{\displaystyle \frac{1}{2}}\mathrm{cos}(2\theta )+{\displaystyle \frac{1}{8}}\mathrm{cos}(4\theta )\right)\mathrm{d}\theta $ | |||
$={{\displaystyle \frac{44}{3}}\left({\displaystyle \frac{3}{8}}\theta +{\displaystyle \frac{1}{4}}\mathrm{sin}(2\theta )+{\displaystyle \frac{1}{32}}\mathrm{sin}(4\theta )\right)|}_{0}^{\pi}$ | |||
$={\displaystyle \frac{11}{2}}\pi .$ |
While this example was not trivial, the double integral would have been much harder to evaluate had we used rectangular coordinates.
Find the volume under the surface $f(x,y)={\displaystyle \frac{1}{{x}^{2}+{y}^{2}+1}}$ over the sector of the circle with radius $a$ centered at the origin in the first quadrant, as shown in Figure 14.3.4.
Solution^{†}^{†}margin:
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The region $R$ we are integrating over is a circle with radius $a$, restricted to the first quadrant. Thus, in polar, the bounds on $R$ are $0\le r\le a$, $0\le \theta \le \pi /2$. The integrand is rewritten in polar as
$$\frac{1}{{x}^{2}+{y}^{2}+1}\Rightarrow \frac{1}{{r}^{2}{\mathrm{cos}}^{2}\theta +{r}^{2}{\mathrm{sin}}^{2}\theta +1}=\frac{1}{{r}^{2}+1}.$$ |
We find the volume as follows:
${\iint}_{R}}f(x,y)\mathrm{d}A$ | $={\displaystyle {\int}_{0}^{\pi /2}}{\displaystyle {\int}_{0}^{a}}{\displaystyle \frac{r}{{r}^{2}+1}}\mathrm{d}r\mathrm{d}\theta $ | ||
$={{\displaystyle {\int}_{0}^{\pi /2}}{\displaystyle \frac{1}{2}}\left(\mathrm{ln}|{r}^{2}+1|\right)|}_{0}^{a}\mathrm{d}\theta $ | |||
$={\displaystyle {\int}_{0}^{\pi /2}}{\displaystyle \frac{1}{2}}\mathrm{ln}({a}^{2}+1)\mathrm{d}\theta $ | |||
$={\left({\displaystyle \frac{1}{2}}\mathrm{ln}({a}^{2}+1)\theta \right)|}_{0}^{\pi /2}$ | |||
$={\displaystyle \frac{\pi}{4}}\mathrm{ln}({a}^{2}+1).$ |
Figure 14.3.4 shows that $f$ shrinks to near 0 very quickly. Regardless, as $a$ grows, so does the volume, without bound. ^{†}^{†}margin: Note: Previous work has shown that there is finite area under $\frac{1}{{x}^{2}+1}$ over the entire $x$-axis. However, Example 14.3.3 shows that there is infinite volume under $\frac{1}{{x}^{2}+{y}^{2}+1}$ over the entire $x$-$y$ plane. Λ
Find the volume of a sphere with radius $a$.
SolutionThe sphere of radius $a$, centered at the origin, has equation ${x}^{2}+{y}^{2}+{z}^{2}={a}^{2}$; solving for $z$, we have $z=\sqrt{{a}^{2}-{x}^{2}-{y}^{2}}$. This gives the upper half of a sphere. We wish to find the volume under this top half, then double it to find the total volume.
The region we need to integrate over is the circle of radius $a$, centered at the origin. Polar bounds for this equation are $0\le r\le a$, $0\le \theta \le 2\pi $.
All together, the volume of a sphere with radius $a$ is:
$2{\displaystyle {\iint}_{R}}\sqrt{{a}^{2}-{x}^{2}-{y}^{2}}\mathrm{d}A$ | $=2{\displaystyle {\int}_{0}^{2\pi}}{\displaystyle {\int}_{0}^{a}}\sqrt{{a}^{2}-{(r\mathrm{cos}\theta )}^{2}-{(r\mathrm{sin}\theta )}^{2}}r\mathrm{d}r\mathrm{d}\theta $ | |||
$=2{\displaystyle {\int}_{0}^{2\pi}}{\displaystyle {\int}_{0}^{a}}r\sqrt{{a}^{2}-{r}^{2}}\mathrm{d}r\mathrm{d}\theta .$ | ||||
We can evaluate this inner integral with substitution. With $u={a}^{2}-{r}^{2}$, $du=-2r\mathrm{d}r$. The new bounds of integration are $u(0)={a}^{2}$ to $u(a)=0$. Thus we have: | ||||
$={\displaystyle {\int}_{0}^{2\pi}}{\displaystyle {\int}_{{a}^{2}}^{0}}\left(-{u}^{1/2}\right)\mathrm{d}u\mathrm{d}\theta $ | ||||
$={{\displaystyle {\int}_{0}^{2\pi}}\left(-{\displaystyle \frac{2}{3}}{u}^{3/2}\right)|}_{{a}^{2}}^{0}\mathrm{d}\theta $ | ||||
$={\displaystyle {\int}_{0}^{2\pi}}\left({\displaystyle \frac{2}{3}}{a}^{3}\right)\mathrm{d}\theta $ | ||||
$={\left({\displaystyle \frac{2}{3}}{a}^{3}\theta \right)|}_{0}^{2\pi}$ | ||||
$={\displaystyle \frac{4}{3}}\pi {a}^{3}.$ |
Generally, the formula for the volume of a sphere with radius $r$ is given as $4/3\pi {r}^{3}$; we have justified this formula with our calculation.
A sculptor wants to make a solid bronze cast of the solid shown in Figure 14.3.5, where the base of the solid has boundary, in polar coordinates, $r=\mathrm{cos}(3\theta )$, and the top is defined by the plane $z=1-x+0.1y$. Find the volume of the solid.
SolutionFrom the outset, we should recognize that knowing how to set up this problem is probably more important than knowing how to compute the integrals. The iterated integral to come is not “hard” to evaluate, though it is long, requiring lots of algebra. Once the proper iterated integral is determined, one can use readily-available technology to help compute the final answer.
The region $R$ that we are integrating over is bound by $0\le r\le \mathrm{cos}(3\theta )$, for $0\le \theta \le \pi $ (note that this rose curve is traced out on the interval $[0,\pi ]$, not $[0,2\pi ]$). This gives us our bounds of integration. The integrand is $z=1-x+0.1y$; converting to polar, we have that the volume $V$ is:
$$V={\iint}_{R}f(x,y)\mathrm{d}A={\int}_{0}^{\pi}{\int}_{0}^{\mathrm{cos}(3\theta )}\left(1-r\mathrm{cos}\theta +0.1r\mathrm{sin}\theta \right)r\mathrm{d}r\mathrm{d}\theta .$$ |
Distributing the $r$, the inner integral is easy to evaluate, leading to
$${\int}_{0}^{\pi}\left(\frac{1}{2}{\mathrm{cos}}^{2}(3\theta )-\frac{1}{3}{\mathrm{cos}}^{3}(3\theta )\mathrm{cos}\theta +\frac{0.1}{3}{\mathrm{cos}}^{3}(3\theta )\mathrm{sin}\theta \right)\mathrm{d}\theta .$$ |
This integral takes time to compute by hand; it is rather long and cumbersome. The powers of cosine need to be reduced, and products like $\mathrm{cos}(3\theta )\mathrm{cos}\theta $ need to be turned to sums using the Product To Sum formulas in the back cover of this text.
We rewrite $\frac{1}{2}{\mathrm{cos}}^{2}(3\theta )$ as $\frac{1}{4}(1+\mathrm{cos}(6\theta ))$. We can also rewrite the second term as:
$$\begin{array}{c}\frac{1}{3}{\mathrm{cos}}^{3}(3\theta )\mathrm{cos}\theta =\frac{1}{3}{\mathrm{cos}}^{2}(3\theta )\mathrm{cos}(3\theta )\mathrm{cos}\theta \hfill \\ \hfill =\frac{1}{3}\frac{1+\mathrm{cos}(6\theta )}{2}\left(\mathrm{cos}(4\theta )+\mathrm{cos}(2\theta )\right).\end{array}$$ |
This last expression still needs simplification, but eventually all terms can be reduced to the form $a\mathrm{cos}(m\theta )$ or $a\mathrm{sin}(m\theta )$ for various values of $a$ and $m$.
We forgo the algebra and recommend the reader employ technology, such as WolframAlpha®, to compute the numeric answer. Such technology gives:
$${\int}_{0}^{\pi}{\int}_{0}^{\mathrm{cos}(3\theta )}\left(1-r\mathrm{cos}\theta +0.1r\mathrm{sin}\theta \right)r\mathrm{d}r\mathrm{d}\theta =\frac{\pi}{4}{\text{units}}^{3}.$$ |
Since the units were not specified, we leave the result as almost $0.8$ cubic units (meters, feet, etc.).
We have used iterated integrals to find areas of plane regions and volumes under surfaces. Just as a single integral can be used to compute much more than “area under the curve,” iterated integrals can be used to compute much more than we have thus far seen. The next two sections show two, among many, applications of iterated integrals.
When evaluating ${\iint}_{R}}f(x,y)\mathrm{d}A$ using polar coordinates, $f(x,y)$ is replaced with and $\mathrm{d}A$ is replaced with .
Why would one be interested in evaluating a double integral with polar coordinates?
In Exercises 3–10., a function $f(x,y)$ is given and a region $R$ of the $x$-$y$ plane is described. Set up and evaluate ${\iint}_{R}f(x,y)\mathrm{d}A$ using polar coordinates.
$f(x,y)=3x-y+4$; $R$ is the region enclosed by the circle ${x}^{2}+{y}^{2}=1$.
$f(x,y)=4x+4y$; $R$ is the region enclosed by the circle ${x}^{2}+{y}^{2}=4$.
$f(x,y)=8-y$; $R$ is the region enclosed by the circles with polar equations $r=\mathrm{cos}\theta $ and $r=3\mathrm{cos}\theta $.
$f(x,y)=4$; $R$ is the region enclosed by the petal of the rose curve $r=\mathrm{sin}\left(2\theta \right)$ in the first quadrant.
$f(x,y)=\mathrm{ln}\left({x}^{2}+{y}^{2}\right)$; $R$ is the annulus enclosed by the circles ${x}^{2}+{y}^{2}=1$ and ${x}^{2}+{y}^{2}=4$.
$f(x,y)=1-{x}^{2}-{y}^{2}$; $R$ is the region enclosed by the circle ${x}^{2}+{y}^{2}=1$.
$f(x,y)={x}^{2}-{y}^{2}$; $R$ is the region enclosed by the circle ${x}^{2}+{y}^{2}=36$ in the first and fourth quadrants.
$f(x,y)=\left(x-y\right)/\left(x+y\right)$; $R$ is the region enclosed by the lines $y=x$, $y=0$ and the circle ${x}^{2}+{y}^{2}=1$ in the first quadrant.
In Exercises 11–14., an iterated integral in rectangular coordinates is given. Rewrite the integral using polar coordinates and evaluate the new double integral.
${\int}_{0}^{5}}{\displaystyle {\int}_{-\sqrt{25-{x}^{2}}}^{\sqrt{25-{x}^{2}}}}\sqrt{{x}^{2}+{y}^{2}}\mathrm{d}y\mathrm{d}x$
${\int}_{-4}^{4}}{\displaystyle {\int}_{-\sqrt{16-{y}^{2}}}^{0}}\left(2y-x\right)\mathrm{d}x\mathrm{d}y$
${\int}_{0}^{2}}{\displaystyle {\int}_{y}^{\sqrt{8-{y}^{2}}}}\left(x+y\right)\mathrm{d}x\mathrm{d}y$
In Exercises 15–16., special double integrals are presented that are especially well suited for evaluation in polar coordinates.
The surface of a right circular cone with height $h$ and base radius $a$ can be described by the equation $f(x,y)=h-h\sqrt{{\displaystyle \frac{{x}^{2}}{{a}^{2}}}+{\displaystyle \frac{{y}^{2}}{{a}^{2}}}}$, where the tip of the cone lies at $(0,0,h)$ and the circular base lies in the $x$-$y$ plane, centered at the origin.
Confirm that the volume of a right circular cone with height $h$ and base radius $a$ is $V={\displaystyle \frac{1}{3}}\pi {a}^{2}h$ by evaluating ${\iint}_{R}}f(x,y)\mathrm{d}A$ in polar coordinates.
Use a double integral and polar coordinates to find the area of the region which lies inside the circle $r=\frac{3}{2}$ and to the right of the line $4r\mathrm{cos}\theta =3$.
Use a double integral and polar coordinates to find the area of the region that lies inside the circle $r=3\mathrm{cos}\theta $ and outside the circle $r=\mathrm{cos}\theta $. Check your answer using geometry.
Evaluate ${\iint}_{R}}{\displaystyle \frac{1}{x}}\mathrm{d}A$ where $R$ is the region that lies inside the circle ${x}^{2}+{y}^{2}=1$ and to the right of the parabola $2x+{y}^{2}=1$.
Evaluate ${\iint}_{R}}{\left(1+{x}^{2}+{y}^{2}\right)}^{-3/2}\mathrm{d}A$ where $R$ is that part of the disk ${x}^{2}+{y}^{2}\le 1$ in the first quadrant.
Let $R$ be the annular region ${a}^{2}\le {x}^{2}+{y}^{2}\le 1$. Find the average distance of points in $R$ to the origin. What is the limit of the average distance as $a\to 0$? As $a\to 1$?