Surface integral of $x^4+y^4+z^4$ over the sphere $x^2+y^2+z^2=a^2$

Question

After doing regular methodology have reached upto integral shown in figure , but when i eliminate z from it it becomes very complicated to solve .Is there any other way to solve this .Thanks

Answer 1

It seems the following.

Use spherical coordinates $$x=a\sin\varphi\cos\theta,$$ $$y=a\sin\varphi\sin\theta,$$ $$z=a\cos\varphi$$ ($0\le\varphi\le\pi$, $0\le\theta\le 2\pi$). Then $$dS=a^2\sin\varphi d\varphi d\theta$$ and

$$x^4+y^4+z^4=$$ $$a^4(\sin^4\varphi\cos^4\theta+\sin^4\varphi\sin^4\theta+\cos^4\varphi)=$$ $$a^4(\sin^4\varphi( (\cos^2\theta+ \sin^2\theta)^2-2\cos^2\theta\sin^2\theta)+\cos^4\varphi)=$$ $$a^4(\sin^4\varphi\left( 1-\frac{\sin^2 2\theta}2\right)+\cos^4\varphi)=$$ $$a^4\left(1-\frac{\sin^4\varphi \sin^2 2\theta}2\right).$$

So the initial integral is equal to $$a^6\int_0^{2\pi} \int_0^{\pi}\left(1-\frac{\sin^4\varphi \sin^2 2\theta}2\right) \sin\varphi d\varphi d\theta=$$ $$a^6\int_0^{2\pi} \int_0^{\pi}\sin\varphi d\varphi d\theta-a^6\int_0^{2\pi} \int_0^{\pi}\frac{\sin^4\varphi \sin^2 2\theta}2 \sin\varphi d\varphi d\theta=$$ $$2\pi a^6 \int_0^{\pi}\sin\varphi d\varphi-\frac{a^6}2\int_0^{2\pi} \sin^2 2\theta d\theta \int_0^{\pi}\sin^5\varphi d\varphi.$$