Example 2.3.18: Flux through a sphere

On the sphere $r=3$ the outward surface element is, by Eq. (2.3.40), $dS=a_r{r}^2\sin \theta d\theta d\varphi =a_{r}9\sin \theta d\theta d\varphi$, and the field there is $3{\sin }^2\theta a_r$. Since $a_r\cdot a_r=1$, the vector character of the integrand disappears and only the magnitude of $dS$ is left:

With ${\sin }^3\theta =(1-{\cos }^2\theta )\sin \theta$ and the substitution $u=\cos \theta$, ${\int }_0^{\pi }{\sin }^3\theta d\theta ={\int }_{-1}^1(1-u^2)du=2-\frac{2}{3}=\frac{4}{3}$. Hence

The field $r{\sin }^2\theta a_r$ points everywhere radially outwards but is strongest at the equator and vanishes at the poles. The integral is its net outward flux through the sphere. (Here $a_r$ was harmless: it met $dS$ in a dot product before any integration took place.)