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!

  • 2
    $\begingroup$ $\int\limits_{0}^{2\pi}\int\limits_{0}^{\pi}\int\limits_{0}^{1}\cos^2(\phi) r² \sin \psi drd\phi d\psi=\left(\int\limits_{0}^{1} r² dr\right)\left(\int\limits_{0}^{\pi}\cos^2(\phi)\right)\left(\int\limits_{0}^{2\pi}\sin \psi d\psi\right)$ $\endgroup$ May 21, 2020 at 19:55
  • $\begingroup$ @Digitallissimo No, i mean to integrate over x²+y²+z² = 1... I think $S_2$ has fixed radius r=1. $\endgroup$
    – melmo99
    May 22, 2020 at 10:38
  • $\begingroup$ @AlexeyBurdin Thank you, I think I was able to solve it this way. $\endgroup$
    – melmo99
    May 22, 2020 at 10:40

1 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.


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