Section 5.13 Simple Scalar Surface Integrals
¶What is the surface area of a sphere of radius \(a\text{?}\) You surely know the answer: \(4\pi a^2\text{.}\) But do you know why?
How do you chop up a sphere? In spherical coordinates, of course. As we have seen in Section 5.12, an infinitesimal rectangle on the surface of the sphere has sides \(r\,d\theta\) and \(r\,\sin\theta\,d\phi\text{,}\) so the scalar surface element on a sphere is
\begin{equation}
dA = r^2\sin\theta\,d\theta\,d\phi\tag{5.13.1}
\end{equation}
and we know that \(r=a\text{.}\) The surface area is simply the sum of these infinitesimal areas:
\begin{align*}
A = \int dA \amp
= \int_0^{2\pi}\int_0^\pi a^2 \sin\theta\,d\theta\,d\phi\\
\amp = 2\pi a^2 (-\cos\theta)\Big|_0^\pi = 4\pi a^2 \nonumber
\end{align*}
(How did we choose the limits of integration?)
Now suppose the charge density on the surface of a sphere is not constant and is given by
\begin{equation}
\sigma=\frac{15}{8\pi}\sin^2\theta\,\cos^2\theta .\tag{5.13.2}
\end{equation}
How much total charge is there on the surface of the sphere?
The limits of integration are the same, so now we have
\begin{align*}
Q = \int \sigma\, dA
\amp= \int_0^{2\pi}\int_0^\pi \frac{15}{8\pi}
\sin^3\theta\,\cos^2\theta\,d\theta\,d\phi \nonumber\\
\amp= \frac{15}{4} \int_0^\pi \left( \cos^2\theta-\cos^4\theta \right)
\sin\theta\, d\theta\\
\amp= \frac{15}{4} \left(
\frac{\cos^5\theta}{5}-\frac{\cos^3\theta}{3} \right)
\Bigg|_0^\pi = 1 .\nonumber
\end{align*}