What is the surface area of a sphere of radius $a$? You surely know the answer: $4\pi a^2$. But do you know why?

How do you chop up a sphere? In spherical coordinates, of course. As we have seen in § {Scalar Surface Elements}, an infinitesimal rectangle on the surface of the sphere has sides $r\,d\theta$ and $r\,\sin\theta\,d\phi$, so the scalar surface element on a sphere is \begin{equation} dA = r^2\sin\theta\,d\theta\,d\phi \end{equation} and we know that $r=a$. The surface area is simply the sum of these infinitesimal areas: \begin{eqnarray} A = \int dA &=& \int_0^{2\pi}\int_0^\pi a^2 \sin\theta\,d\theta\,d\phi \\ &=& 2\pi a^2 (-\cos\theta)\Big|_0^\pi = 4\pi a^2 \nonumber \end{eqnarray} (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 \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{eqnarray} Q = \int \sigma\, dA &=& \int_0^{2\pi}\int_0^\pi \frac{15}{8\pi} \sin^3\theta\,\cos^2\theta\,d\theta\,d\phi \nonumber\\ &=& \frac{15}{4} \int_0^\pi \left( \cos^2\theta-\cos^4\theta \right) \sin\theta\, d\theta \\ &=& \frac{15}{4} \left( \frac{\cos^5\theta}{5}-\frac{\cos^3\theta}{3} \right) \Bigg|_0^\pi = 1 \nonumber\\ \end{eqnarray}