Suppose we have a surface given in cylindrical coordinates as $z=f(r,\theta)$ and we wish to find the integral over some region. We could attempt to translate into rectangular coordinates and do the integration there, but it is often easier to stay in cylindrical coordinates.

How might we approximate the volume under such a surface in a way that uses cylindrical coordinates directly? The basic idea is the same as before: we divide the region into many small regions, multiply the area of each small region by the height of the surface somewhere in that little region, and add them up. What changes is the shape of the small regions; in order to have a nice representation in terms of $r$ and $\theta$, we use small pieces of ring-shaped areas, as shown in figure 15.2.1. Each small region is roughly rectangular, except that two sides are segments of a circle and the other two sides are not quite parallel. Near a point $(r,\theta)$, the length of either circular arc is about $r\Delta\theta$ and the length of each straight side is simply $\Delta r$. When $\Delta r$ and $\Delta \theta$ are very small, the region is nearly a rectangle with area $r\Delta r\Delta\theta$, and the volume under the surface is approximately $$\sum\sum f(r_i,\theta_j)r_i\Delta r\Delta\theta.$$ In the limit, this turns into a double integral $$\int_{\theta_0}^{\theta_1}\int_{r_0}^{r_1} f(r,\theta)r\,dr\,d\theta.$$

Figure 15.2.1. A cylindrical coordinates "grid''.

Example 15.2.1 Find the volume under $z=\sqrt{4-r^2}$ above the quarter circle bounded by the two axes and the circle $x^2+y^2=4$ in the first quadrant.

In terms of $r$ and $\theta$, this region is described by the restrictions $0\le r\le 2$ and $0\le\theta\le\pi/2$, so we have \eqalign{ \int_{0}^{\pi/2}\int_{0}^{2} \sqrt{4-r^2}\;r\,dr\,d\theta &=\int_{0}^{\pi/2}\left. -{1\over3}(4-r^2)^{3/2}\right|_0^2\,d\theta\cr &=\int_{0}^{\pi/2} {8\over3}\,d\theta\cr &={4\pi\over3}.\cr } The surface is a portion of the sphere of radius 2 centered at the origin, in fact exactly one-eighth of the sphere. We know the formula for volume of a sphere is $(4/3)\pi r^3$, so the volume we have computed is $(1/8)(4/3)\pi 2^3=(4/3)\pi$, in agreement with our answer. (From another point of view, what we've done is prove that the volume of a sphere of radius 2 is $(32/3)$. If you replace 2 by $a$ and do the integral again, it is not any more difficult, and you will prove that the volume of a sphere of radius $a$ is $(4/3)\pi a^3$.) $\square$

This example is much like a simple one in rectangular coordinates: the region of interest may be described exactly by a constant range for each of the variables. As with rectangular coordinates, we can adapt the method to deal with more complicated regions.

Example 15.2.2 Find the volume under $z=\sqrt{4-r^2}$ above the region enclosed by the curve $r=2\cos\theta$, $-\pi/2\le\theta\le\pi/2$; see figure 15.2.2. The region is described in polar coordinates by the inequalities $-\pi/2\le\theta\le\pi/2$ and $0\le r\le2\cos\theta$, so the double integral is $$\int_{-\pi/2}^{\pi/2}\int_{0}^{2\cos\theta} \sqrt{4-r^2}\;r\,dr\,d\theta =2\int_{0}^{\pi/2}\int_{0}^{2\cos\theta} \sqrt{4-r^2}\;r\,dr\,d\theta.$$ We can rewrite the integral as shown because of the symmetry of the volume; this avoids a complication during the evaluation. Proceeding: \eqalign{ 2\int_{0}^{\pi/2}\int_{0}^{2\cos\theta} \sqrt{4-r^2}\;r\,dr\,d\theta &=2\int_{0}^{\pi/2}-{1\over3}\left.(4-r^2)^{3/2}\right|_0^{2\cos\theta}\,d\theta\cr &=2\int_{0}^{\pi/2}-{8\over3}\sin^3\theta+{8\over3}\,d\theta\cr &=\left.2\left(-{8\over3}{\cos^3\theta\over3}-\cos\theta+{8\over3}\theta\right)\right|_0^{\pi/2}\cr &={8\over3}\pi-{32\over9}.\cr } $\square$