How to evaluate $\int_S(x^4+y^4+z^4) \, dS$ over surface of the unit sphere. 
Question. Let $S$ denote the unit sphere in $\mathbb{R}^3$. Evaluate: $$\int_S (x^4+y^4+z^4) \, dS$$

My Solution. First I parametrize $S$ by $$r(u,v)=(\cos v \cos u, \cos v \sin u, \sin v)$$ $0\le u \le 2 \pi;~-\frac{\pi}{2}\le v \le \frac{\pi}{2}$ 
Let $~f(x,y,z)=x^4+y^4+z^4$. Here $~|\frac{\partial r}{\partial u} \times \frac{\partial r}{\partial v}|=|\cos v|$
Then $\displaystyle \int_S(x^4+y^4+z^4)\,dS = \int_{-\pi/2}^{\pi/2} \int_0^{2\pi} f[r(u,v)] \left|\frac{\partial r}{\partial u} \times \frac{\partial r}{\partial v}\right|~du~dv$
Thus I try to calculate this integral directly using the definition of surface integral.
But I had so much calculations in this way. Does this particular problem can be solved using any theorems e.g-Gauss' Divergence (By writing $f$ as $F\cdot n$ for some vector field $F$?
Thank you.
 A: I shall denote by $\text{d}\Sigma$ the surface area element.  In the polar form $$(x,y,z)=\big(r\cos(\phi)\sin(\theta),r\sin(\phi)\cos(\theta),r\cos(\theta)\big)\,,$$  I would only compute the integral $$\int_{\partial B_1(\boldsymbol{0})}\,z^4\,\text{d}\Sigma=\int_0^{2\pi}\,\int_0^\pi\,\cos^4(\theta)\,\sin(\theta)\,\text{d}\theta\,\text{d}\phi=2\pi\,\int_{-1}^{+1}\,t^4\,\text{d}t=\frac{4\pi}{5}\,,$$ where $t:=\cos(\theta)$, and then multiply that by $3$ to get the final answer.  Due to symmetry, this is justified.
A: Let $~f(x,y,z)=x^4+y^4+z^4$
The unit outward drawn normal to $S$ is $n=(x,y,z)$. Taking $F=(x^3,y^3,z^3)$ we have $f=F.n$. Hence the given integral:
$\int_{S}f(x,y,z)dS=\int_{S}F.n~dS=\int_{V}div F~dV =3\int_{0}^{1}\int_{-\pi/2}^{-\pi/2}\int_{0}^{2\pi}r^2|\cos v|~dr~du~dv=\frac{12\pi}{5}.$
A: Direct calculation (unit sphere).  $x^4+y^4+z^4=(x^2+y^2+z^2)^2-2(x^2y^2+x^2z^2+y^2z^2)=(1-2h)$    
where $h=cos^4vcos^2usin^2u+cos^2vsin^2v=h_1+h_2$.  The surface integral  is $S=\int_0^{2\pi}\int_{\frac{-\pi}{2}}^{\frac{\pi}{2}}(1-2h)cosvdvdu$.  These can be calculated directly.
$\int_0^{2\pi}\int_{\frac{-\pi}{2}}^{\frac{\pi}{2}}cosvdvdu=4\pi$.  Let 
$I_1=\int_0^{2\pi}\int_{\frac{-\pi}{2}}^{\frac{\pi}{2}}cos^5vdvcos^2usin^2udu$.  Let $I_2=\int_0^{2\pi}\int_{\frac{-\pi}{2}}^{\frac{\pi}{2}}sin^2vcos^3vdv$.  At this point note that $I_1=\frac{I_2}{2}$.  $I_2$ is easier to calculate, so we don't need to calculate  $I_1$.  $I_2=\int_0^{2\pi}\int_{\frac{-\pi}{2}}^{\frac{\pi}{2}}(sin^2v-sin^4v)cosvdv=\frac{8\pi}{15}$.  Therefore $S=\frac{12\pi}{5}$.
