# 21. Multiple Integrals in Curvilinear Coordinates

## a. Integrating in Polar Coordinates

## 3. Integral over a Non-Rectangular Polar Region

On the previous page, we learned how to integrate over a polar rectangle. Polar coordinates can also be used to compute integrals over some regions which are not polar rectangles, namely those of Type \(r\) or Type \(\theta\).

A region is of Type \(r\) if it consists of all points satisfying \(a \le r \le b\) and \(g(r) \le \theta \le h(r)\). A Type \(r\) integral is an integral over a region of Type \(r\).

A region is of Type \(\theta\) if it consists of all points satisfying \(\alpha \le \theta \le \beta\) and \(g(\theta) \le r \le h(\theta)\). A Type \(\theta\) integral is an integral over a region of Type \(\theta\).

Samples of these regions are shown below:

We most often work with regions of type \(\theta\) because the boundaries \(r=g(\theta)\) and \(r=h(\theta)\) are standard polar curves. In Calculus 2, you may have learned that the area of a region of Type \(\theta\) is \(\displaystyle \dfrac{1}{2}\int_\alpha^\beta (h(\theta)^2-g(\theta)^2)\,d\theta\). If not, we will derive it on the next page from the 2D formula.

To compute double integrals over these types of regions we use the following theorems which say that the double integrals can be written as iterated integrals, with the area differential \[ dA=r\,dr\,d\theta \]

Memorize this!

If \(R\) is a Type \(r\) region with \(a \le r \le b\) and
\(g(r) \le \theta \le h(r)\), then
\[
\iint\limits_R f(r,\theta)\,dA
=\int_a^b\int_{g(r)}^{h(r)} f(r,\theta)\,r\,d\theta\,dr
\]
This is just the iterated integral computed by holding \(r\) fixed
while doing the \(\theta\) integral and then integrating the resulting
function over \(r\).

If \(R\) is a Type \(\theta\) region with \(\alpha \le \theta \le \beta\) and
\(g(\theta) \le r \le h(\theta)\), then
\[
\iint\limits_R f(r,\theta)\,dA
=\int_\alpha^\beta\int_{g(\theta)}^{h(\theta)} f(r,\theta)\,r\,dr\,d\theta
\]
This is just the iterated integral computed by holding \(\theta\) fixed
while doing the \(r\) integral and then integrating the resulting
function over \(\theta\).

Notice that, in both cases, *the inner integral has limits that
depend on the variable of the outer integral, but not vice versa.* So you
cannot, in general, reverse the order of integration. However, it is
frequently useful to convert from rectangular coordinates to polar
coordinates as done on
another page.

Find the volume under the surface \(z=y\) above the region in the \(xy\)-plane inside the cardioid \(r=2-2\cos\theta\) with \(y \ge 0\).

This is a Type \(\theta\) region since \(0 \le \theta \le \pi\) and \(0 \le r \le 2-2\cos\theta\). (Look at a radial line and see where it crosses the curve.) In the integral for the volume, we express \(x\), \(y\) and \(dA\) in polar coordinates: \[ x=r\cos\theta \qquad y=r\sin\theta \qquad \text{and} \qquad dA=r\,dr\,d\theta \] So the volume is: \[\begin{aligned} V&=\iint\limits_R y\,dA =\int_0^\pi\int_0^{2-2\cos\theta} (r\sin\theta)\,r\,dr\,d\theta \\ &=\int_0^\pi\sin\theta \left[\dfrac{r^3}{3}\right]_{r=0}^{2-2\cos\theta}\,d\theta =\dfrac{1}{3}\int_0^\pi (2-2\cos\theta)^3\sin\theta\,d\theta \end{aligned}\] To do the integral, we make the substitution \(u=2-2\cos\theta\). Then \(du=2\sin\theta\,d\theta\) and \(\dfrac{1}{2}du=\sin\theta\,d\theta\). So: \[ V=\dfrac{1}{6}\int_0^4 u^3\,du =\dfrac{1}{6}\left[\dfrac{u^4}{4}\right]_0^4 =\dfrac{1}{6}4^3 =\dfrac{32}{3} \]

Find the area of the region bounded by the curves \(\theta=\dfrac{1}{2} (\pi-r)\) and \(\theta=r\).

Remember the area of a region, \(R\), is the double integral:
\[
\text{Area}=\iint\limits_R 1\,dA
\]
Is this integral Type \(r\) or Type \(\theta\)?

Where do the curves \(\theta=\dfrac{1}{2}(\pi-r)\) and
\(\theta=r\) intersect?

\(\displaystyle \iint\limits_R 1\,dA =\int_0^{\pi/3}\int_r^{(\pi-r)/2} r\,d\theta\,dr =\dfrac{1}{108}\pi^3 \)

It would be very difficult to do this integral as a Type \(\theta\) integral. It is much easier to do it as a Type \(r\) integral because the \(\theta\) bounds are already given as functions of \(r\). To find the upper bound for \(r\), we compute the point of intersection by setting the curves equal: \[ \dfrac{1}{2}\pi-\,\dfrac{1}{2}r=r \quad \Rightarrow \quad \dfrac{\pi}{2}=\dfrac{3}{2}r \quad \Rightarrow \quad r=\dfrac{\pi}{3} \] Thus, the \(r\)-bounds are: \(0 \le r \le \dfrac{\pi}{3}\). The graph confirms that the point of intersection is just past \(r=1\). Travelling counterclockwise (lower values to higher values) in the graph, the \(\theta\)-bounds are \(r \le \theta \le \dfrac{1}{2}(\pi-r)\). So we can compute the area as \[\begin{aligned} \text{Area} &=\iint\limits_R 1\,dA =\int_0^{\pi/3}\int_r^{(\pi-r)/2} r\,d\theta\,dr \\ &=\int_0^{\pi/3} \left(\dfrac{\pi-r}{2}-r\right)r\,dr =\int_0^{\pi/3} \left(\dfrac{\pi}{2}r-\,\dfrac{3}{2}r^2\right)\,dr \\ &=\left[\dfrac{\pi r^2}{4}-\,\dfrac{r^3}{2}\right]_0^{\pi/3} =\dfrac{\pi}{4}\left(\dfrac{\pi}{3}\right)^2 -\,\dfrac{1}{2}\left(\dfrac{\pi}{3}\right)^3 \\ &=\pi^3\left(\dfrac{1}{36}-\,\dfrac{1}{54}\right) =\dfrac{1}{108}\pi^3 \end{aligned}\]

On the next page we will use polar integrals in applications.

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