What's my mistake? Computing $\int\limits_{S^2}(x^k+y^k+z^k)dS$

Question

I need to compute for a positive integer the following integral: $$I=\int\limits_{S^2}(x^k+y^k+z^k)dS$$ where $S^2$ is the unit sphere in $\mathbb{R}^3$.

I'll decribe my attempt.

First, by symmetry, $$\int\limits_{S^2}(x^k+y^k+z^k)dS=3\int\limits_{S^2}z^kdS$$ (The right integral is the same for a replacement of $z$ with any of the other 2 variables) - I'm quite certain about that.

Now, we can use the map $r: (0, 2\pi)\times (0, \pi)\rightarrow S^2$ defined by shperical coordinates: $$r(\varphi, \theta)=(\cos\varphi \sin\theta, \sin\varphi \sin\theta, \cos\theta)$$ This map covers all the sphere, except for a neasure-zero set w.r.t to $S^2$.
In addition, our lecturer showed us that $$\sqrt{\Gamma \left ( \frac{\partial r}{\partial \varphi}, \frac{\partial r}{\partial \theta} \right )}= \sin^2\theta$$ so I have no doubts about that either.
Therefore, by definition of integration on manifolds, $$I=3\int _{(0, 2\pi)\times (0, \pi)}(\cos\theta)^k(\sin\theta)^2 d\varphi d\theta=3\cdot 2\pi \int \limits_{0}^{\pi}(\cos\theta)^k(\sin\theta)^2 d\theta$$

And that's not right. It can be seen thorugh plugging $k=2$. $I$ should be the surface area of the unit sphere which is $4\pi$, but the obtained result gives something far different.

Where did I go wrong?

Answer 1

I think you've just missed a square root: the area measure on the unit sphere is $\sin{\theta } \, d\theta \, d\phi$.

Note that $$r_{\theta} = (\cos{\phi}\cos{\theta},\sin{\phi}\cos{\theta},-\sin{\theta}),\\ r_{\phi} = (-\sin{\phi}\sin{\theta},\cos{\phi}\sin{\theta},0), $$ so $g_{\theta\theta} = 1$, $g_{\theta\phi}=g_{\phi\theta}=0$, $g_{\phi\phi} = \sin^2{\theta}$, and so $\sqrt{\det{g}} = \sin{\theta}$ since $0<\theta<\pi$.

Answer 2

I would avoid spherical coordinates and just apply Cavalieri's principle. Given some $a\in[-1,1]$ the section $S^2\cap\{z=a\}$ is a circle with length $2\pi\sqrt{1-a^2}$, hence

$$ I(k)=\int_{S^2}x^k+y^k+z^k\,d\mu = 3\int_{-1}^{1}2\pi\sqrt{1-a^2}a^k\,da $$ clearly equals zero if $k$ is odd, while assuming $k=2m$ we have: $$ I(k)=12\pi \int_{0}^{1}\sqrt{1-a^2}a^{2m}\,da = 6\pi\int_{0}^{1}(1-u)^{1/2} u^{m-1/2}\,du = 6\pi\cdot\frac{\Gamma\left(\frac{3}{2}\right)\Gamma\left(m+\frac{1}{2}\right)}{\Gamma(m+2)}$$ by Euler's Beta function.