# Calculation of an integral over surface

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!

• $\int\limits_{0}^{2\pi}\int\limits_{0}^{\pi}\int\limits_{0}^{1}\cos^2(\phi) r² \sin \psi drd\phi d\psi=\left(\int\limits_{0}^{1} r² dr\right)\left(\int\limits_{0}^{\pi}\cos^2(\phi)\right)\left(\int\limits_{0}^{2\pi}\sin \psi d\psi\right)$ May 21, 2020 at 19:55
• @Digitallissimo No, i mean to integrate over x²+y²+z² = 1... I think $S_2$ has fixed radius r=1. May 22, 2020 at 10:38
• @AlexeyBurdin Thank you, I think I was able to solve it this way. May 22, 2020 at 10:40

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