Evaluating $\iint_V x^4+y^4+z^4$ with the divergence theorem 
Use Gauss divergence theorem to evaluate
$$\iint_S \left(x^{4} + y^{4} + z^{4}\right)$$ over sphere S of radius $a$.

So I wrote this as
$$
\begin{align}
&a\iint_{\partial V} \Big(x^3 \hat{i}+y^3\hat{j}+y^3\hat{k}\Big)\cdot\Big(\frac{x\hat{i}+y\hat{j}+z\hat{k}}{a}\Big)\\
=\ &a\iiint_V \operatorname{div}(x^3,y^3,z^3)\\
=\ &3a\iiint_Va^2=3a^3\frac{4\pi}{3}a^3=4\pi a^6
\end{align}$$
But my answer is not matching. The answer key says it should be $\frac{12 \pi a^6}{5}$
 A: The normal vector $\mathrm{n}$ at at point $(x,y,z)$ on a sphere is given by $\frac{(x,y,z)}{\sqrt{x^2+y^2+z^2}}$, hence it you want to write $x^4+y^4+z^4$ as the dot product $\mathrm{F}\cdot\mathrm{n}$ you have to take $\mathrm{F}$ as $a(x^3+y^3+z^3)$, then $\text{div }\mathrm{F}=3a(x^2+y^2+z^2)$ and the original integral becomes
$$ 3a\iiint_{x^2+y^2+z^2\leq a^2}(x^2+y^2+z^2)\,d\mu=3a\int_{0}^{a}4\pi\rho^2\cdot\rho^2\,d\rho=\frac{12\pi}{5}a^6$$
by integrating along shells.
A: Using spherical coordinates you get :
$$
\iiint_{B(0,a)} x^2 + y^2 + z^2 dx dy dz = \int_{r=0}^a \int_{\phi=0}^{2\pi} \int_{\theta=0}^\pi r^2 \cdot(r^2  \sin(\theta))\ dr d\phi d\theta =
$$
$$
\int_{\phi=0}^{2\pi} d\phi \int_{\theta=0}^\pi \sin(\theta) d\theta \int_{r=0}^a r^4 dr = 2\pi \cdot 2 \cdot a^5/5
$$
where $(r^2 \sin(\theta))$ is the Jacobian and $B(0,a)$ is the sphere centered at $0$ or radius $a$.
EDIT:
This was the mistake of the OP, I did not detailed all the previous steps he did right. Combining the constant in front of his integral and this result gives the correct answer.
