Polar Rectangular Regions of Integration

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

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

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

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

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

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

[latex]\Large{f}{({{r}^{*}_{ij}},{{\theta}^{*}_{ij}})}{\Delta}{A} = {f}{({{r}^{*}_{ij}},{{\theta}^{*}_{ij}})}{{r}^{*}_{ij}}{\Delta}{r}{\Delta}{\theta}[/latex].

Figure 2. Finding the volume of the thin box above polar rectangle [latex]\small{R_{ij}}[/latex]

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

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

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

[latex]\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}[/latex].

This becomes the expression for the double integral.

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,

[latex]\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}[/latex].

Notice that the expression for [latex]dA[/latex] is replaced by [latex]{r} \ {dr} \ {d{\theta}}[/latex] 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 [latex]f[/latex] is given in terms of [latex]x[/latex] and [latex]y[/latex], using [latex]x=r\cos\theta,y = r\sin\theta[/latex] and [latex]{dA} = {{r} \ {dr} \ {d{\theta}}}[/latex] changes it to

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

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 [latex]{R} = {\left \{{(r,{\theta})}{\mid}{1} \ {\leq} \ {r} \ {\leq} \ {3,0} \ {\leq} \ {\theta} \ {\leq} \ {\pi} \right \}}[/latex].

Show Solution

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

Figure 3. The polar region [latex]R[/latex] 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 by converting from rectangular coordinates

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

 

Show Solution

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

Figure 4. The annular region of integration [latex]R[/latex].

Hence, using the conversion[latex]x =r\cos\theta,y =r\sin\theta[/latex], and [latex]{dA} = {{r} \ {dr} \ {d{\theta}}}[/latex], we have
[latex]\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}[/latex]

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

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 [latex]{r} = {f}{(\theta)}[/latex] than [latex]{\theta} = {f}{(r)}[/latex], so we describe a general polar region as [latex]{R} = {\left \{{(r,{\theta})}{\mid}{\alpha} \ {\leq} \ {\theta} \ {\leq} \ {{\beta},{h_1}}{(\theta)} \ {\leq} \ {r} \ {\leq} \ {{h_2}{(\theta)}} \right \}}[/latex] (see the following figure).

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

Example: evaluating a double integral over a general polar region

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

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

Figure 6. The region [latex]D[/latex] is the top half of a cardioid.

Hence, we have

[latex]\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}[/latex]

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