0
$\begingroup$

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!

$\endgroup$
3
  • 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 '20 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 '20 at 10:38
  • $\begingroup$ @AlexeyBurdin Thank you, I think I was able to solve it this way. $\endgroup$
    – melmo99
    May 22 '20 at 10:40
0
$\begingroup$

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.

$\endgroup$
0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.