How can I evaluate this integral over a sphere, with surface area element instead of volume element?

Question

$$\int_S (x^2 + y^2)d\sigma,$$

where S is the sphere of radius 1 centered at (0,0,0) and $\sigma$ is surface area.

I would like some hints on how to proceed. This is tricky, since I am not being asked for a volume integral computation, so I can't use spherical coordinates, I think.

Thanks,

Answer 1

Use spherical coordinates with $\rho =1$

In detail:

if you parameterize the sphere by setting

$$\textbf u=\cos \theta \sin \phi \textbf i+\sin \theta \sin \phi \textbf j+\cos \phi \textbf k$$ and then compute the Jacobian (surface element) by taking $\vert \textbf u_{\theta }\times \textbf u_{\phi }\vert $ you get $\sin \phi $.

Also $$x^{2}+y^{2}=1-z^{2}=1-\cos ^{2}\phi =\sin^{2}\phi $$

So your integral is

$$\int_{0}^{\pi }\int_{0}^{2\pi }\sin^{3} \phi d\theta d\phi=2\pi \int_{0}^{\pi }\sin^{3} \phi d\phi =2\pi \left ( \frac{4}{3} \right )=\frac{8}{3}\pi$$

Answer 2

As described in this answer, the area of an annulus on a sphere of radius $r$ between $z=a$ and $z=b$ is $$ 2\pi r(b-a) $$ This is also mentioned in this Wikipedia article.

Thus, on the the surface of a sphere, the integral of a function dependent only on $z$ is $$ 2\pi r\int_{-r}^rf(z)\,\mathrm{d}z $$ In the case here, we have $r=1$ and $f(z)=1-z^2$, so we get $$ 2\pi\int_{-1}^1(1-z^2)\,\mathrm{d}z=\frac{8\pi}3 $$