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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, $g$ 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 polar rectangle , whose sides have constant $r$ -values and/or constant $\theta $ -values. This means we can describe a polar rectangle as in [link] (a), with $R=\left\{\left(r,\theta \right)|a\le r\le b,\alpha \le \theta \le \beta \right\}.$
In this section, we are looking to integrate over polar rectangles. Consider a function $f\left(r,\theta \right)$ over a polar rectangle $R.$ We divide the interval $\left[a,b\right]$ into $m$ subintervals $\left[{r}_{i-1},{r}_{i}\right]$ of length $\text{\Delta}r=\left(b-a\right)\text{/}m$ and divide the interval $\left[\alpha ,\beta \right]$ into $n$ subintervals $\left[{\theta}_{i-1},{\theta}_{i}\right]$ of width $\text{\Delta}\theta =\left(\beta -\alpha \right)\text{/}n.$ This means that the circles $r={r}_{i}$ and rays $\theta ={\theta}_{i}$ for $1\le i\le m$ and $1\le j\le n$ divide the polar rectangle $R$ into smaller polar subrectangles ${R}_{ij}$ ( [link] (b)).
As before, we need to find the area $\text{\Delta}A$ 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}\text{\Delta}\theta $ and ${r}_{i}\text{\Delta}\theta $ and with a width $\text{\Delta}r.$ Hence the area of the polar subrectangle ${R}_{ij}$ is
Simplifying and letting ${r}_{ij}^{*}=\frac{1}{2}({r}_{i-1}+{r}_{i}),$ we have $\text{\Delta}A={r}_{ij}^{*}\text{\Delta}r\text{\Delta}\theta .$ Therefore, the polar volume of the thin box above ${R}_{ij}$ ( [link] ) is
Using the same idea for all the subrectangles and summing the volumes of the rectangular boxes, we obtain a double Riemann sum as
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,
This becomes the expression for the double integral.
The double integral of the function $f\left(r,\theta \right)$ over the polar rectangular region $R$ in the $r\theta $ -plane is defined as
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,
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