Can I use Gauss theorem in this case There is something unclear about the following example. Namely, they used symmetry in order to calculate the surface integral by using Gauss. I tried not to use symmetry, and just put that the field is F= x^3*i, and I still got the same solution. Is it okay if I do that, or must y and z be present in F? 

 A: Inasmuch as the Divergence Theorem states 
$$\oint_{x^2+y^2+z^2= a^2}\vec F(x,y,z)\cdot \hat n\,dS=\iiint_{x^2+y^2+z^2\le a^2} \nabla \cdot \vec F(x,y,z)\,dx\,dy\,dz$$
for suitably smooth vector fields, then clearly if $\vec F(x,y,z)=\hat x x^3$, and since $\hat n=\frac{\hat xx+\hat yy+\hat zz}{\sqrt{x^2+y^2+z^2}}=\frac1a (\hat xx+\hat yy+\hat zz)$
$$\begin{align}
\oint_{x^2+y^2+z^2= a^2}\vec F(x,y,z)\cdot \hat n\,dS&=\frac1a \oint_{x^2+y^2+z^2= a^2}x^4\,dS\\\\&=\iiint_{x^2+y^2+z^2\le a^2}  3x^2\,dx\,dy\,dz
\end{align}$$
And you can finish?
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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\begin{align}
\int x^{4}\,\dd S & = \int z^{4}\,\hat{r}\cdot\hat{r}\,\,\dd S =
\int\pars{{z^{4} \over r}\,\vec{r}}\cdot\
\overbrace{\dd\vec{S}\,\,\,}^{\ds{\hat{r}\ \dd S}}\ =\
\int\nabla\cdot\pars{{z^{4} \over r}\,\vec{r}}\dd^{3}\vec{r}
\\[5mm] & =
\int\braces{\bracks{{4z^{3}\,\hat{z} \over r} +
z^{4}\pars{-\,{\vec{r} \over r^{3}}}}\cdot\vec{r} +
{z^{4} \over r}\,3}\dd^{3}\vec{r} =
\int\pars{{4z^{4} \over r} - {z^{4} \over r} + {3z^{4} \over r}}\dd^{3}\vec{r}
\\[5mm] & =
\int_{0}^{a}{6r^{4} \over r}
\underbrace{\pars{\int_{\Omega_{\vec{r}}}\cos^{4}\pars{\theta}
\,\dd\Omega_{\vec{r}}}}_{\ds{\int_{0}^{2\pi}\int_{0}^{\pi}\cos^{4}\pars{\theta}\sin\pars{\theta}\,\dd\theta\,\dd\phi = {4\pi \over 5}}}\ r^{2}\,\dd r
\\[5mm] & =
{4\pi \over 5}\int_{0}^{a}6r^{5}\,\dd r = \bbx{4\pi a^{6} \over 5}
\end{align}
