Polar Rectangular Regions of Integration

Double integrals are sometimes much easier to evaluate if we change rectangular coordinates to polar coordinates. However, before we describe how to make this change, we need to establish the concept of a double integral in a polar rectangular region.

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 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 or constant [latex]\theta[/latex]-values. We can describe a polar rectangle as shown in Figure 1(a):

[latex]R = {(r,\theta)\mid a \leq r \leq b, \alpha \leq \theta \leq \beta }[/latex].

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 = \frac{b-a}{m}[/latex]. Similarly, we divide the interval [latex][\alpha, \beta][/latex] into [latex]n[/latex] subintervals [latex][\theta_{j-1},\theta_{j}][/latex] of width [latex]\Delta\theta = \frac{\beta-\alpha}{n}[/latex]. 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)).

As before, we need to find the area [latex]\Delta A[/latex] of the polar subrectangle [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]. The polar rectangle [latex]R_{ij}[/latex] looks like a trapezoid with parallel sides [latex]r_{i-1}\Delta\theta[/latex] and [latex]r_{i}\Delta\theta[/latex], and width [latex]\Delta r[/latex]. Hence the area of the polar subrectangle [latex]R_{ij}[/latex] is:

[latex]\Delta A = \frac{1}{2} \Delta r \left( r_{i-1} \Delta \theta + r_i \Delta \theta \right).[/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]f{({{r}^{*}_{ij}},{{\theta}^{*}_{ij}})}{\Delta}{A} = {f}{({{r}^{*}_{ij}},{{\theta}^{*}_{ij}})}{{r}^{*}_{ij}}{\Delta}{r}{\Delta}{\theta}[/latex].

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

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

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]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.

Just as with rectangular regions, the double integral over a polar rectangular region can be expressed as an iterated integral in polar coordinates:

[latex]\underset{R}{\displaystyle\iint}f(r,\theta)dA = \underset{R}{\displaystyle\iint}f(r,\theta)rdrd\theta = \displaystyle\int_{\theta=\alpha}^{\theta=\beta}\displaystyle\int_{r=a}^{r=b}f(r,\theta)rdrd\theta[/latex]

Notice that the expression for [latex]dA[/latex] is replaced by [latex]rdrd\theta[/latex] when working in polar coordinates. The extra [latex]r[/latex] factor is crucial and comes from the geometry of polar rectangles.

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[/latex], [latex]y = r\sin\theta[/latex], and [latex]dA = rdrd\theta[/latex] changes it to

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

Note that all the properties we listed earlier for the double integral in rectangular coordinates hold true for the double integral in polar coordinates as well.

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.

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.

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).

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]