Computing Gauss's of a sphere 
The vector field given as $\vec{F}=\frac{\left \langle x,y,z \right \rangle}{\sqrt{x^{2}+y^{2}+z^{2}}}$
The region $D=\left \{ a^{2}\leq x^{2}+y^{2}+z^{2}\leq b^{2} \right \}$

I've some insights into this problem but would like to explore this further.
Evidently, $x^{2}+y^{2}+z^{2}=f\left ( x,y,z \right )$
But would it be easier, say, if we reformulate instead
$\sqrt{a^{2}-x^{2}-y^{2}}\leq z\leq \sqrt{b^{2}-x^{2}-y^{2}}$
so we have
$\sqrt{a^{2}-x^{2}-y^{2}}\leq g\left ( x,y \right )\leq \sqrt{b^{2}-x^{2}-y^{2}}$ as a graph of the surface z?
Spherical coordinates would greatly simplify the crux of the matter here.
 A: We are given the vector $\vec F=\frac{\vec r}{r^2}=\hat r$, where $\hat r$ is the radial unit vector.  The divergence of $\vec F$ for $0<a\le r\le b$ is 
$$\begin{align}\nabla \cdot \vec F&=\frac{1}{r^2}\frac{\partial (r^2F_r)}{\partial r}+\frac{1}{r\sin \theta}\frac{\partial (\sin \theta F_{\theta})}{\partial \theta}+\frac{1}{r\sin \theta}\frac{\partial ( F_{\phi})}{\partial \phi}\\\\
&=\frac2r\end{align}$$
since $F_r=1$ and $F_{\theta}=F_{\phi}=0$.
The volume integral of $\nabla \cdot \vec F$ over the volume between a sphere of radius $b$ and a sphere of radius $a$ is given by 
$$\begin{align}\int_V \nabla \cdot \vec F\,dV&=\int_0^{2\pi}\int_0^\pi\int_a^b\frac2r\,r^2\sin \theta\,dr\,d\theta\,d\phi\\\\&=4\pi(b^2-a^2)\end{align}$$
Finally, the Divergence Theorem asserts that 
$$\begin{align}\int_V \nabla \cdot \vec F\,dV&=\oint_S \hat n \cdot \vec F\,dS\\\\
&=\int_0^{2\pi}\int_0^\pi (\hat r \cdot \hat r) \,b^2\sin \theta \,d\theta\,d\phi+\int_0^{2\pi}\int_0^\pi (-\hat r \cdot \hat r) \,a^2\sin \theta \,d\theta\,d\phi\\\\
&=4\pi(b^2-a^2)\end{align}$$
as expected!
