I am trying to find the flux through a sphere of radius $R$ centered at the origin, due to the field given by $\overline{E}$, where $a,b,c$ are non-zero constants and $\vec{i}$, $\vec{j}$, $\vec{k}$ denote the unit vectors for the $x$, $y$, $z$ axes.
$$\overline{E}=\frac{ax^2\vec{i}+by^2\vec{j}+cz^2\vec{k}}{ax^3+by^3+cz^3}$$
Flux is by definition $\phi=\oint\overline{E}\cdot\mathrm d\overline{A}$. Using Gauss' law in its differential form, $\epsilon_o\overline{\nabla}\cdot\overline{E}=\rho$ and also $\epsilon_0 \Phi_{\text{net}}=q_{\text{enc}}$. This calculation gives $\rho(x,y,z)$ as $$\boxed{\rho(x,y,z)=\frac{2(ax+by+cz)}{ax^3+by^3+cz^3}-\frac{3(a^2x^4+b^2y^4+c^2z^4)}{(ax^3+by^3+cz^3)^2}}$$
Now the flux would be simply the integral of $\rho\mathrm dV$ over $\Gamma :x^2+y^2+z^2=R^2$. So the following expression is to be evaluated over $\Gamma$.
$$\boxed{\iiint_{\Gamma}\left(\frac{2(ax+by+cz)}{ax^3+by^3+cz^3}-\frac{3(a^2x^4+b^2y^4+c^2z^4)}{(ax^3+by^3+cz^3)^2}\right)\mathrm dx\mathrm dy\mathrm dz}$$
I have only studied single-variable calculus. Any hints are appreciated. Thanks
Best Answer
This is actually pretty easy without using Gauss.
Note that the outward pointing unit Normal is given by $$ \hat n=\frac 1R(x\vec i + y \vec j + z \vec k) $$ so $$ \hat n \cdot \vec E = \frac 1R$$ Integrating over the surface should give you a flux of $$\phi = \frac {4 \pi R^2}R = 4 \pi R$$