5.3 Double integrals in polar coordinates  (Page 2/7)

Notice that the expression for $dA$ is replaced by $rdrd\theta$ 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 $x=r\text{cos}\theta ,y=r\text{sin}\theta ,\text{and}dA=rdrd\theta$ changes it to

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.

Sketching a polar rectangular region

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

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

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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.

Evaluating a double integral by converting from rectangular coordinates

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

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

Hence, using the conversion $x=r\text{cos}\theta ,y=r\text{sin}\theta ,$ and $dA=rdrd\theta ,$ we have

$\begin{array}{cc}\hfill {\displaystyle \underset{R}{\iint }\left(x+y\right)dA}& ={\displaystyle \underset{\theta =\pi \text{/}2}{\overset{\theta =3\pi \text{/}2}{\int }}{\displaystyle \underset{r=1}{\overset{r=2}{\int }}\left(r\text{cos}\theta +r\text{sin}\theta \right)}}rdrd\theta \hfill \\ & =\left({\displaystyle \underset{r=1}{\overset{r=2}{\int }}{r}^2}dr\right)\left({\displaystyle \underset{\pi \text{/}2}{\overset{3\pi \text{/}2}{\int }}\left(\text{cos}\theta +\text{sin}\theta \right)d\theta }\right)\hfill \\ & ={\left[\frac{r^3}{3}\right]}_1^2{\left[\text{sin}\theta -\text{cos}\theta \right]|}_{\pi \text{/}2}^{3\pi \text{/}2}\hfill \\ & =-\frac{14}{3}.\hfill \end{array}$

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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 $r=f\left(\theta \right)$ than $\theta =f\left(r\right),$ so we describe a general polar region as $R=\left\{\left(r,\theta \right)|\alpha \le \theta \le \beta ,h_1\left(\theta \right)\le r\le h_2\left(\theta \right)\right\}$ (see the following figure).