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 \(\theta\).

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 \(r\).
Derivation

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 \(\theta\).
Derivation

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.

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: \[ V=\dfrac{1}{6}\int_0^4 u^3\,du =\dfrac{1}{6}\left[\dfrac{u^4}{4}\right]_0^4 =\dfrac{1}{6}4^3 =\dfrac{32}{3} \]

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