Example 2.3.16: Surface area of a sphere

Place the centre of the sphere at the origin, so that the sphere is the coordinate surface $r=b$. On it only $\theta$ and $\varphi$ vary, and by Eq. (2.3.40) a small patch has area $dS=h_{\theta }{h}_{\varphi }d\theta d\varphi =b^2\sin \theta d\theta d\varphi$. To cover the sphere exactly once, $\theta$ runs from $0$ to $\pi$ and $\varphi$ from $0$ to $2\pi$. The limits are constants and the integrand is a product of a function of $\theta$ and a function of $\varphi$, so the double integral factorises:

Note the role of the factor $\sin \theta$: the strips of constant $d\theta$ near the poles are short circles, those near the equator are long (Figure 2.3.27), and ${\int }_0^{\pi }\sin \theta d\theta =2$ accounts for this. Had we forgotten the scale factor $h_{\varphi }=b\sin \theta$ we would have obtained the wrong value $2{\pi }^2{b}^2$.