Double Integrals in Polar Coordinates

Learning Objectives

Recognize the format of a double integral over a polar rectangular region.
Evaluate a double integral in polar coordinates by using an iterated integral.
Recognize the format of a double integral over a general polar region.

Polar Rectangular Regions of Integration

When we defined the double integral for a continuous function in rectangular coordinates—say, $g$ over a region $R$ in the $xy$ -plane—we divided $R$ into subrectangles with sides parallel to the coordinate axes. These sides have either constant $x$ -values and/or constant $y$ -values. In polar coordinates, the shape we work with is a , whose sides have constant $r$ -values and/or constant $\theta$ -values. This means we can describe a polar rectangle as in Figure 1(a), with ${R} = {\left \{{(r,{\theta})}{\mid}{a} \ {\leq} \ {r} \ {\leq} \ {b,{\alpha}} \ {\leq} \ {\theta} \ {\leq} \ {\beta} \right \}}$ .

In this section, we are looking to integrate over polar rectangles. Consider a function ${f}{(r,{\theta})}$ over a polar rectangle $R$ . We divide the interval $[a,b]$ into $m$ subintervals ${[{r_{i-1}},{r_i}]}$ of length ${\Delta}{r} = {(b-a)}{/m}$ and divide the interval ${[{\alpha}, {\beta}]}$ into $n$ subintervals ${[{{\theta}_{j-1}},{{\theta}_{j}}]}$ of width ${\Delta}{\theta} = {({\beta}-{\alpha})}{/n}$ . This means that the circles ${r} = {r_{i}}$ and rays ${\theta} = {{\theta}_{j}}$ for ${1} \ {\leq} \ {i} \ {\leq} \ {m}$ and ${1} \ {\leq} \ {j} \ {\leq} \ {n}$ divide the polar rectangle $R$ into smaller polar subrectangles ${R}_{ij}$ (Figure 1(b)).

This figure consists of three figures labeled a, b, and c. In figure a, a sector of an annulus is shown in the polar coordinate plane with radii a and b and angles alpha and beta from the theta = 0 axis. In figure b, this sector of an annulus is cut up into subsectors in a manner similar to the way in which previous spaces were cut up into subrectangles. In figure c, one of these subsectors is shown with angle Delta theta, distance between inner and outer radii Delta r, and area Delta A = r* sub theta Delta r Delta theta, where the center point is given as (r* sub i j, theta* sub i j).

Figure 1. (a) A polar rectangle $R$ (b) divided into subrectangles $R_{ij}$ (c) Close-up of a subrectangle.

As before, we need to find the area ${\Delta}{A}$ of the polar subrectangle ${R}_{ij}$ and the “polar” volume of the thin box above ${R}_{ij}$ . Recall that, in a circle of radius $r$ , the length $s$ of an arc subtended by a central angle of ${\theta}$ radians is ${s} = {r}{\theta}$ . Notice that the polar rectangle ${R}_{ij}$ looks a lot like a trapezoid with parallel sides ${r}_{i-1}{\Delta}{\theta}$ and ${r}_{i}{\Delta}{\theta}$ and with a width ${\Delta}{r}$ . Hence the area of the polar subrectangle ${R}_{ij}$ is

$\Large{\Delta}{A} = {\dfrac{1}{2}}{\Delta}{r}{({r_{i-1}}{\Delta}{\theta}+{r_1}{\Delta}{\theta})}$ .

Simplifying and letting ${{r}^{*}_{ij}} = {\dfrac{1}{2}}{({r_{i-1}}+{r_i})}$ , we have ${\Delta}{A} = {{r}^{*}_{ij}}{\Delta}{r}{\Delta}{\theta}$ . Therefore, the polar volume of the thin box above ${R}_{ij}$ (Figure 2) is

$\Large{f}{({{r}^{*}_{ij}},{{\theta}^{*}_{ij}})}{\Delta}{A} = {f}{({{r}^{*}_{ij}},{{\theta}^{*}_{ij}})}{{r}^{*}_{ij}}{\Delta}{r}{\Delta}{\theta}$ .

In x y z space, there is a surface f (r, theta). On the x y plane, a series of subsectors of annuli are drawn as in the previous figure with radius between annuli Delta r and angle between subsectors Delta theta. A subsector from the surface f(r, theta) is projected down onto one of these subsectors. This subsector has center point marked (r* sub i j, theta* sub i j).

Figure 2. Finding the volume of the thin box above polar rectangle $\small{R_{ij}}$

Using the same idea for all the subrectangles and summing the volumes of the rectangular boxes, we obtain a double Riemann sum as

$\Large{\displaystyle\sum_{i=1}^{m}\displaystyle\sum_{j=1}^{n}}{f}{({{r}^{*}_{ij}},{{\theta}^{*}_{ij}})}{{r}^{*}_{ij}}{\Delta}{r}{\Delta}{\theta}$ .

As we have seen before, we obtain a better approximation to the polar volume of the solid above the region $R$ when we let $m$ and $n$ become larger. Hence, we define the polar volume as the limit of the double Riemann sum,

$\Large{V} = {\displaystyle\lim_{m,n \rightarrow \infty}} \ {\displaystyle\sum_{i=1}^{m}\displaystyle\sum_{j=1}^{n}}{f}{({{r}^{*}_{ij}},{{\theta}^{*}_{ij}})}{{r}^{*}_{ij}}{\Delta}{r}{\Delta}{\theta}$ .

This becomes the expression for the double integral.

definition

The double integral of the function ${f}{(r,{\theta})}$ over the polar rectangular region $R$ in the ${r}{\theta}$ -plane is defined as

$\large{\underset{R}{\displaystyle\iint}f(r,\theta)dA = \displaystyle\lim_{m,n \rightarrow \infty}\displaystyle\sum_{i=1}^{m}\displaystyle\sum_{j=1}^{n}f(r^*_{ij},\theta^*_{ij})\Delta{A} = \displaystyle\lim_{m,n \rightarrow \infty}\displaystyle\sum_{i=1}^{m}\displaystyle\sum_{j=1}^{n}f(r^{*}_{ij},\theta^{*}_{ij})r^{*}_{ij}\Delta{r}\Delta\theta}$ .

Again, just as in Double Integrals over Rectangular Regions, the double integral over a polar rectangular region can be expressed as an iterated integral in polar coordinates. Hence,

$\Large{\underset{R}{\displaystyle\iint}f(r,\theta)dA = \underset{R}{\displaystyle\iint}f(r,\theta)r \ dr \ d\theta = \displaystyle\int_{\theta=\alpha}^{\theta=\beta}\displaystyle\int_{r=a}^{r=b}f(r,\theta)r \ dr \ d\theta}$ .

Notice that the expression for $dA$ is replaced by ${r} \ {dr} \ {d{\theta}}$ when working in polar coordinates. Another way to look at the polar double integral is to change the double integral in rectangular coordinates by substitution. When the function $f$ is given in terms of $x$ and $y$ , using $x=r\cos\theta,y = r\sin\theta$ and ${dA} = {{r} \ {dr} \ {d{\theta}}}$ changes it to

$\Large{\underset{R}{\displaystyle\iint}f(x,y)dA=\underset{R}{\displaystyle\iint}f(r\cos\theta,r\sin\theta)r \ dr \ d\theta}$ .

Note that all the properties listed in Double Integrals over Rectangular Regions for the double integral in rectangular coordinates hold true for the double integral in polar coordinates as well, so we can use them without hesitation.

Example: sketching a polar rectangular region

Sketch the polar rectangular region ${R} = {\left \{{(r,{\theta})}{\mid}{1} \ {\leq} \ {r} \ {\leq} \ {3,0} \ {\leq} \ {\theta} \ {\leq} \ {\pi} \right \}}$ .

Show Solution

As we can see from Figure 3, $r=1$ and $r=3$ are circles of radius $1$ and $3$ and ${0} \ {\leq} \ {\theta} \ {\leq} \ {\pi}$ covers the entire top half of the plane. Hence the region $R$ looks like a semicircular band.

Half an annulus R is drawn with inner radius 1 and outer radius 3. That is, the inner semicircle is given by x squared + y squared = 1, whereas the outer semicircle is given by x squared + y squared = 9.

Figure 3. The polar region $R$ lies between two semicircles.

Now that we have sketched a polar rectangular region, let us demonstrate how to evaluate a double integral over this region by using polar coordinates.

example: evaluating a double integral over a polar rectangular region

Evaluate the integral $\underset{R}{\displaystyle\iint}{3x}{dA}$ over the region ${R} = {\left \{{(r,{\theta})}{\mid}{1} \ {\leq} \ {r} \ {\leq} \ {2,0} \ {\leq} \ {\theta} \ {\leq} \ {\pi} \right \}}$ .

Show Solution

First we sketch a figure similar to Figure 3 but with outer radius 2. From the figure we can see that we have

$\begin{align} \underset{R}{\displaystyle\iint}3x \ dA &= \displaystyle\int_{\theta=0}^{\theta=\pi}\displaystyle\int_{r=1}^{r=2}3r\cos\theta{r} \ dr \ d\theta &\quad &\text{Use an iterated integral with correct limits of integration.} \\ &=\displaystyle\int_{\theta=0}^{\theta=\pi}\cos\theta\left[r^3\bigg|_{r=1}^{r=2}\right]d\theta &\quad &\text{Integrate first with respect to }r. \\ &=\displaystyle\int_{\theta=0}^{\theta=\pi}7\cos\theta{d}\theta=7\sin\theta\bigg|_{\theta=0}^{\theta=\pi}=0. \end{align}$

try it

Sketch the region ${R} = {\left \{{(r,{\theta})}{\mid}{1} \ {\leq} \ {r} \ {\leq} \ {2,-{\frac{\pi}{2}}} \ {\leq} \ {\theta} \ {\leq} \ {\frac{\pi}{2}} \right \}}$ , and evaluate $\underset{R}{\displaystyle\iint}{x} \ {dA}$ .

Show Solution

$\frac{14}3$

Example: Evaluating a double integral by converting from rectangular coordinates

Evaluate the integral $\underset{R}{\displaystyle\iint}{(1-{x^2}-{y^2})}{dA}$ where $R$ is the unit circle on the $xy$ -plane.

Show Solution

The region $R$ is a unit circle, so we can describe it as ${R} = {\left \{{(r,{\theta})}{\mid}{0} \ {\leq} \ {r} \ {\leq} \ {1,0} \ {\leq} \ {\theta} \ {\leq} \ {2{\pi}} \right \}}$ .

Using the conversion $x=r\cos\theta,y=r\sin\theta$ , and ${dA} = {{r} \ {dr} \ {d{\theta}}}$ , we have

$\hspace{4cm}\begin{align} \underset{R}{\displaystyle\iint}(1-x^2-y^2)dA&=\displaystyle\int_0^{2\pi}\displaystyle\int_0^1(1-r^2)r \ dr \ d\theta =\displaystyle\int_0^{2\pi}\displaystyle\int_0^1(r - r^3)dr \ d\theta \\ &=\displaystyle\int_0^{2\pi}\left[\frac{r^2}2-\frac{r^4}4\right]_0^1{d}\theta=\displaystyle\int_0^{2\pi}\frac14d\theta=\frac{\pi}2. \end{align}$

Example: Evaluating a double integral by converting from rectangular coordinates

Evaluate the integral $\underset{R}{\displaystyle\iint}{(x+y)}{dA}$ where ${R} = {\left \{{(x,y)}{\mid}{1} \ {\leq} \ {{x^2}+{y^2}} \ {\leq} \ {4,x} \ {\leq} \ {0} \right \}}$ .

Show Solution

We can see that $R$ is an annular region that can be converted to polar coordinates and described as ${R} = {\left \{{(r,{\theta})}{\mid}{1} \ {\leq} \ {r} \ {\leq} \ {2,{\frac{\pi}{2}}} \ {\leq} \ {\theta} \ {\leq} \ {\frac{3{\pi}}{2}} \right \}}$ (see the following graph).

Two semicircles are drawn in the second and third quadrants, with equations x squared + y squared = 1 and x squared + y squared = 2.

Figure 4. The annular region of integration $R$ .

Hence, using the conversion $x =r\cos\theta,y =r\sin\theta$ , and ${dA} = {{r} \ {dr} \ {d{\theta}}}$ , we have
$\hspace{5cm}\begin{align}\underset{R}{\displaystyle\iint}(x+y)dA&=\displaystyle\int_{\theta=\pi/2}^{\theta=3\pi/2}\displaystyle\int_{r=1}^{r=2}(r\cos\theta+r\sin\theta)r \ dr \ d\theta \\&=\left(\displaystyle\int_{r=1}^{r=2}r^2dr\right)\left(\displaystyle\int_{\pi/2}^{3\pi/2}(\cos\theta+\sin\theta)d\theta\right) \\&=\left[\frac{r^3}3\right]_1^2[\sin\theta-\cos\theta]\bigg|_{\pi/2}^{3\pi/2} \\&=-\frac{14}3.\end{align}$

try it

Evaluate the integral $\underset{R}{\displaystyle\iint}{(4-{x^2}-{y^2})}{dA}$ where $R$ is the circle of radius 2 on the $xy$ -plane.

Show Solution

$8\pi$

Watch the following video to see the worked solution to the above Try It

You can view the transcript for “CP 5.18” here (opens in new window).

General Polar Regions of Integration

To evaluate the double integral of a continuous function by iterated integrals over general polar regions, we consider two types of regions, analogous to Type I and Type II as discussed for rectangular coordinates in Double Integrals over General Regions. It is more common to write polar equations as ${r} = {f}{(\theta)}$ than ${\theta} = {f}{(r)}$ , so we describe a general polar region as ${R} = {\left \{{(r,{\theta})}{\mid}{\alpha} \ {\leq} \ {\theta} \ {\leq} \ {{\beta},{h_1}}{(\theta)} \ {\leq} \ {r} \ {\leq} \ {{h_2}{(\theta)}} \right \}}$ (see the following figure).

A region D is shown in polar coordinates with edges given by theta = alpha, theta = beta, r = h2(theta), and r = h1(theta).

Figure 5. A general polar region between $\alpha < 0<\beta$ and $h_{1}(\theta) < r<h_{2}(\theta)$ .

theorem: double integrals over a general polar regions

If ${f}{(r,{\theta})}$ is continuous on a general polar region $D$ as described above, then

$\large{\underset{R}{\displaystyle\iint}f(r,\theta)r \ dr \ d\theta=\displaystyle\int_{\theta=\alpha}^{\theta=\beta}\displaystyle\int_{r=h_1(\theta)}^{r=h_2(\theta)}f(r,\theta)r \ dr \ d\theta}$

Example: evaluating a double integral over a general polar region

Evaluate the integral $\underset{D}{\displaystyle\iint}{r^2}{\sin\theta}{r} \ {dr} \ {d{\theta}}$ where $D$ is the region bounded by the polar axis and the upper half of the cardioid ${r} = {1} + {\cos\theta}$ .

Show Solution

We can describe the region $D$ as $\{(r,\theta)\mid0\leq\theta\leq\pi,0\leq{r}\leq1+\cos\theta\}$ as shown in the following figure.

A region D is given as the top half of a cardioid with equation r = 1 + cos theta.

Figure 6. The region $D$ is the top half of a cardioid.

Hence, we have

\begin{array}{ccc} \iint_{D} r^{2} \sin θ r d r d θ & = \int_{θ = 0}^{θ = π} \int_{r = 0}^{r = 1 + \cos θ} (r^{2} \sin θ) r d r d θ \\ = \frac{1}{4} \int_{θ = 0}^{θ = π} [r^{4}]_{r = 0}^{r = 1 + \cos θ} \sin θ d θ \\ = \frac{1}{4} \int_{θ = 0}^{θ = π} (1 + \cos θ)^{4} \sin θ d θ \\ = - \frac{1}{4} [\frac{(1 + \cos θ)^{5}}{5}]_{0}^{π} = \frac{8}{5} . \end{array}

$\begin{array}{ccc} \underset{D}{\displaystyle\iint}r^2\sin\theta{r}dr \ d\theta \hfill &= \hfill \displaystyle\int_{\theta=0}^{\theta=\pi}\displaystyle\int_{r=0}^{r=1+\cos\theta}(r^2\sin\theta)r dr d\theta \\ \hfill &= \hfill \frac{1}{4}\displaystyle\int_{\theta=0}^{\theta=\pi}[r^4]_{r=0}^{r=1+\cos\theta}\sin\theta d\theta \\ \hfill &= \hfill \frac{1}{4}\displaystyle\int_{\theta=0}^{\theta=\pi}(1+\cos\theta)^4\sin\theta d\theta \\ \hfill &= \hfill -\frac{1}{4}[\frac{(1+\cos\theta)^5}{5}]_0^{\pi}=\frac{8}{5}. \end{array}$

Try It

Evaluate the integral $\underset{D}{\displaystyle\iint}r^2\sin^2{2}\theta{r} \ dr \ d\theta$ where ${D} = {\left \{{(r,{\theta})}{\mid}{-{\frac{\pi}{4}}} \ {\leq} \ {\theta} \ {\leq} \ {\frac{\pi}{4}} \ {\leq} \ {r} \ {\leq} \ {{2}{\sqrt{\cos2{\theta}}}} \right \}}$ .

Show Solution

$\pi/4$

Watch the following video to see the worked solution to the above Try It

You can view the transcript for “CP 5.19” here (opens in new window).

Module 5: Multiple Integration