# Show that $\int_{S^2} (xy)^2\, dS = \frac{4\pi}{15}$

I have been constructing some examples related to spherical harmonics. I wanted to know the $L^2$ norm of the function $f(x,y,z) \in S^2 = \{ x^2 + y^2 + z^2 = 1\}$. Wikipedia suggests this normalization:

$$\int_{S^2} (xy)^2\, dS = \frac{4\pi}{15}$$

Are there any derivations that do not involve spherical coordinates? I am hoping for a more algebraic derivation. Maybe it will use spherical coordinates implicitly.

\begin{eqnarray*} x &=& \cos \theta \cos \phi \\ y &=& \sin \theta \cos \phi \\ z &=& \sin \phi \end{eqnarray*}

For smaller degree terms there is a trick. I would do the integral of $1$:

$$\lvert\lvert\,1 \,\rvert\rvert^2 = \frac{1}{4\pi}\int_{S^2} 1 \, dS = 1$$

The averages of homogenous linear terms (or cubic terms) is zero. Quadratics have a trick:

$$\lvert\lvert\,x \,\rvert\rvert^2 = \lvert\lvert\,y \,\rvert\rvert^2 = \lvert\lvert\,z \,\rvert\rvert^2 = \frac{1}{4\pi}\int_{S^2} x^2 \, dS = \frac{1}{3} \frac{1}{4\pi}\int_{S^2} (x^2 + y^2 + z^2) \, dS = \frac{1}{3}$$

I don't think there's such a trick for quadratic polynomials of 4th degree

Through spherical coordinates or not, $$I(k)= \iint_{S^2}(xy)^{2k}\,d\mu$$ boils down to a value of Euler's Beta function. We have $$\iint_{x^2+y^2=R^2}(xy)^{2k}\,d\mu=R^{4k+1}\int_{0}^{2\pi}\left(\cos\theta\sin\theta\right)^{2k}\,d\theta =2\pi R^{4k+1}\cdot\frac{1}{4^{2k}}\binom{2k}{k}$$ hence $I(k)$ can be computed from Cavalieri's principle. Or
$$\begin{eqnarray*} I(k)&=&\int_{0}^{2\pi}\int_{0}^{\pi}\sin(\theta)^{4k+1}\cos(\varphi)^{2k}\sin(\varphi)^{2k}\,d\theta\,d\varphi\\&=&2\sqrt{\pi}\frac{\Gamma\left(k+\tfrac{1}{2}\right)^2}{\Gamma\left(2k+\tfrac{3}{2}\right)}=\color{red}{\frac{4\pi\binom{2k}{k}}{(4k+1)\binom{4k}{2k}}}.\end{eqnarray*}$$
There is $x^2 y^2 + x^2 z^2 = x^2 (1 - x^2)$ and $dS = 2\pi dx$.
• The second one is false as written. Once you integrate around the $x$ axis circle there will be a $2\pi$. – cactus314 Dec 13 '17 at 18:19
• The second one refers to the fact that the area of $S^2$ between the planes $x=x_1$ and $x=x_2$ is equal to $2\pi|x_2 -x_1|$. The only remaining integral is along the x-axis. – random Dec 13 '17 at 18:44
• That's fine. You exploited symmetry to obtain a function of $x$ alone. Then $\int \int dS = \int 2\pi dx$. Your math is cprrect, your writing is just a bit off. Really elegant use of symmetry though. – cactus314 Dec 13 '17 at 18:51
• This could be the washer method from Calculus? This article on the napkin ring problem could be relevant. In any case I have $\int \int dS = \int 2\pi \sqrt{1-x^2} dx$ – cactus314 Dec 13 '17 at 19:10