Can I use Gauss theorem in this case

Question

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?

Answer 1

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?

Answer 2

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \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}