Electromagnetic Fields and Waves: Notes, Solved Problems and Interactive Lab

2.3.9 Integration in spherical coordinates

The surface and volume elements, Eqs. (2.3.40) and (2.3.41), turn integrals over spheres, which would be awkward in Cartesian coordinates, into products of simple one-dimensional integrals. The procedure is always the same.

Recipe · Evaluating line, surface and volume integrals

Choose the coordinate system in which the boundary of the region is a coordinate surface (a sphere centred at the origin is $r = const$ , a cylinder around the $z$ axis is $ρ = const$ ).
Take the corresponding element $d ℓ$ , $d S$ or $d V$ from Eqs. (2.3.39) to (2.3.41) (or Table 2.3.4), scale factors included, and fix the sign of $d S$ by the required orientation.
Choose the limits so that every point of the region is covered exactly once ( $0 \leq θ \leq π$ , $0 \leq ϕ < 2 π$ for a full sphere).
If the integrand is a vector dotted with $d S$ or $d ℓ$ , evaluate the dot product before integrating. If the integrand is a vector, convert its unit vectors to Cartesian ones.
If the limits are constants and the integrand is a product of functions of one variable each, split the multiple integral into a product of single integrals.

The three examples of this subsection show how this works. Figure 2.3.27 gives a useful picture for the first one.

Slicing a sphere of radius b into rings. The ring between the cones \theta and \theta+d\theta is a narrow band of width b\,d\theta wrapped around a circle of radius b\sin\theta, so its area is 2\pi b\sin\theta\cdot b\,d\theta. This is what the \phi integration of dS=b^2\sin\theta\,d\theta\,d\phi produces. Rings near the poles are short, rings near the equator are long. — **Figure 2.3.27.** Slicing a sphere of radius $b$ into rings. The ring between the cones $θ$ and $θ + d θ$ is a narrow band of width $b d θ$ wrapped around a circle of radius $b sin θ$ , so its area is $2 π b sin θ \cdot b d θ$ . This is what the $ϕ$ integration of $d S = b^{2} sin θ d θ d ϕ$ produces. Rings near the poles are short, rings near the equator are long.

Example 2.3.16Surface area of a sphere Example 2.3.17Volume of a sphere Example 2.3.18Flux through a sphere

Quick check

Without a new calculation: what is the area of the northern hemisphere ( $0 \leq θ \leq π /2$ ) of a sphere of radius $b$ , and what is the area of the polar cap $0 \leq θ \leq π /3$ ?

$2 π b^{2} \int_{0}^{θ_{0}} sin θ d θ = 2 π b^{2} (1 - cos θ_{0})$ . For $θ_{0} = π /2$ this is $2 π b^{2}$ (half the sphere), and for $θ_{0} = π /3$ it is $π b^{2}$ , a quarter of the sphere, although the cap reaches one third of the way from pole to pole.

2.3.9 Integration in spherical coordinates

In this part

Examples · 3