Calculation of an integral over surface

Question

I am trying to solve the following integral: $\int_{S_2} z^2 dS(x)$, where

$S_2=\{(x,y,z)\in \mathbb{R^3}$:$x^2+y^2+z^2=1$} and the Volume element $dS(x)=r^2sin\psi \quad dr d\phi d\psi $.

I parametrized $S_2$: $(0, \psi)$ $\in$ $(0, \pi)\times(0,2\pi)$ $\mapsto$ $(\sin \phi \cos\psi, \sin\phi \sin \psi, \cos \phi)$.

My idea: $\int_{S_2} z^2 dS(x)$ = $\int_{0}^{2\pi}$$\int_{0}^{\pi}$$\int_{0}^{1}$$\cos^2(\phi) r^2 sin \psi drd\phi\psi$.

I don't know what to do next... Would be great if someone could help me out. Thanks!

Answer 1

You could use the measure surface :

If $f \in L^{1}(\mathbb{R}^{n},\mathbb{C})$,$x \in \mathbb{R}^{n}-\left\lbrace 0 \right\rbrace, v = \frac{x}{|x|}$, where $r = |x|$, then

$$\int_{\mathbb{R}^{n}}f(x)d_{\lambda_{n}}(x) = \int_{r=0}^{r=+\infty}(\int_{r\mathbb{S}^{n-1}}f(x)d_{\lambda_{r\mathbb{S}^{n-1}}}(x))dr = \int_{r=0}^{r=+\infty}r^{n-1}(\int_{\mathbb{S}^{n-1}}f(rv)d_{\lambda_{\mathbb{S}^{n-1}}}(v))dr$$

If the function is radial, the integral becomes straightfoward.