We have seen that sometimes double integrals are simplified by doing them in polar coordinates; not surprisingly, triple integrals are sometimes simpler in cylindrical coordinates or spherical coordinates. To set up integrals in polar coordinates, we had to understand the shape and area of a typical small region into which the region of integration was divided. We need to do the same thing here, for three dimensional regions.

The cylindrical coordinate system is the simplest, since it is just the polar coordinate system plus a $z$ coordinate. A typical small unit of volume is the shape shown in figure 15.2.1 "fattened up'' in the $z$ direction, so its volume is $r\Delta r\Delta \theta\Delta z$, or in the limit, $r\,dr\,d\theta\,dz$.

Example 15.6.1 Find the volume under $z=\sqrt{4-r^2}$ above the quarter circle inside $x^2+y^2=4$ in the first quadrant.

We could of course do this with a double integral, but we'll use a triple integral: $$\int_0^{\pi/2}\int_0^2\int_0^{\sqrt{4-r^2}} r\,dz\,dr\,d\theta= \int_0^{\pi/2}\int_0^2 \sqrt{4-r^2}\; r\,dr\,d\theta= {4\pi\over3}.$$ Compare this to example 15.2.1.

Example 15.6.2 An object occupies the space inside both the cylinder $x^2+y^2=1$ and the sphere $x^2+y^2+z^2=4$, and has density $x^2$ at $(x,y,z)$. Find the total mass.

We set this up in cylindrical coordinates, recalling that $x=r\cos\theta$: \eqalign{ \int_0^{2\pi}\int_0^1\int_{-\sqrt{4-r^2}}^{\sqrt{4-r^2}} r^3\cos^2(\theta)\,dz\,dr\,d\theta &=\int_0^{2\pi}\int_0^1 2\sqrt{4-r^2}\;r^3\cos^2(\theta)\,dr\,d\theta\cr &=\int_0^{2\pi} \left({128\over15}-{22\over5}\sqrt3\right)\cos^2(\theta)\,d\theta\cr &=\left({128\over15}-{22\over5}\sqrt3\right)\pi\cr }

Spherical coordinates are somewhat more difficult to understand. The small volume we want will be defined by $\Delta\rho$, $\Delta\phi$, and $\Delta\theta$, as pictured in figure 15.6.1. The small volume is nearly box shaped, with 4 flat sides and two sides formed from bits of concentric spheres. When $\Delta\rho$, $\Delta\phi$, and $\Delta\theta$ are all very small, the volume of this little region will be nearly the volume we get by treating it as a box. One dimension of the box is simply $\Delta\rho$, the change in distance from the origin. The other two dimensions are the lengths of small circular arcs, so they are $r\Delta\alpha$ for some suitable $r$ and $\alpha$, just as in the polar coordinates case.