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9.4. Spherical and Cylindrical Coordinates

In cylindrical coordinates (R, φ, z), z measures distance along the axis, R measures distance from the axis, and φ is an angle that wraps around the axis.

cylindrical-coordinates
Figure F. Cylindrical coordinates.

The differential of volume in cylindrical coordinates can be written as dv = R dR dz dφ. This follows from adding a third dimension, along the z axis, to the rectangle in Figure D.

Example 100

◊ Show that the expression for dv has the right units.

◊ Angles are unitless, since the definition of radian measure involves a distance divided by a distance. Therefore the only factors in the expression that have units are R, dR, and dz. If these three factors are measured, say, in meters, then their product has units of cubic meters, which is correct for a volume.

Example 101

◊ Find the volume of a cone whose height is h and whose base has radius b.

◊ Let's plan on putting the z integral on the outside of the sandwich. That means we need to express the radius rmax of the cone in terms of z. This comes out nice and simple if we imagine the cone upside down, with its tip at the origin. Then since we have rmax(z=0) = 0, and rmax(h) = b, evidently rmax = zb/h.

\begin{align} v &= \int dv \\ &= \int_{z=0}^h \int_{r=0}^{zb/h} \int_{\phi=0}^{2\pi} R d\phi dR dz \\ &= 2\pi \int_{z=0}^h \int_{r=0}^{zb/h} R dR dz \\ &= 2\pi \int_{z=0}^h (zb/h)^2/2 dz \\ &= \pi (b/h)^2 \int_{z=0}^h z^2 dz \\ &= \frac{\pi b^2 h}{3} \end{align}

As a check, we note that the answer has units of volume. This is the classical result, known by the ancient Egyptians, that a cone has one third the volume of its enclosing cylinder.

In spherical coordinates (r, θ, φ), the coordinate r measures the distance from the origin, and θ and φ are analogous to latitude and longitude, except that θ is measured down from the pole rather than from the equator.

spherical-coordinates
Figure G. Spherical coordinates.

The differential of volume in spherical coordinates is dv = r2sinθ drdθdφ.

Example 102

◊ Find the volume of a sphere.

\begin{align} v &= \int dv \\ &= \int_{\theta=0}^\pi \int_{r=0}^{r=b} \int_{\phi=0}^{2\pi} r^2\sin\theta d\phi drd\theta \\ &= 2\pi \int_{\theta=0}^\pi \int_{r=0}^{r=b} r^2\sin\theta drd\theta \\ &= 2\pi\cdot\frac{b^3}{3} \int_{\theta=0}^\pi \sin\theta d\theta \\ &= \frac{4\pi b^3}{3} \end{align}