Flux through sphere I wish to find the flux of $\mathbf{F}=(x^2,y^2,z^2)$ through $S: (x-1)^2+(y-3)^2+(z+1)^2$
Here is what I tried:
I "moved" the sphere to $(0,0)$ by changing the variables to:
$u=x-1$ , $v=y-3$ , $w=z+1$
so now we have $F=((u+1)^2,(v+3)^2,(w-1)^2)$ and $S$ is the unit sphere.
So my calculation is (after switching to polar):
$$\int_0^{2\pi}\int_0^1\int_{-\sqrt{1-r^2}}^{\sqrt{1-r^2}}{\rm div}\,\mathbf{F}\,r\,{\rm d}z\,{\rm d}r\,{\rm d}\theta=\int_0^{2\pi}\int_0^1\int_{-\sqrt{1-r^2}}^{\sqrt{1-r^2}}(2r\cos\theta +2r\sin\theta +2z+6)r\,{\rm d}z\,{\rm d}r\,{\rm d}\theta$$       
but I got $0$ instead $8\pi$
What did I do wrong?
 A: One's procedure is excellent but there are some mistakes in your final calculation. Here is my way to compute the triple integral:
$$
\begin{align}
\int_0^{2\pi}\int_0^1\int_{-\sqrt{1-r^2}}^{\sqrt{1-r^2}}\left( 2r\cos\theta +2r\sin\theta +2z+6\right) r\,dz\,dr\,d\theta
&
 \stackrel{[A]}{=}\int_0^1\int_{-\sqrt{1-r^2}}^{\sqrt{1-r^2}}\int_0^{2\pi}\left( 2r^2\cos\theta +2r^2\sin\theta +2rz+6r\right)\,d\theta \,dz\,dr
\\&
\stackrel{[B]}{=}\int_0^1\int_{-\sqrt{1-r^2}}^{\sqrt{1-r^2}}\left( 4 \pi rz+12\pi r\right) \,dz\,dr
\\&
=\int_0^1\left.\left( 2 \pi rz^2+12\pi r z\right)\right| ^{\sqrt{1-r^2}}_{-\sqrt{1-r^2}} \,dr 
\\ &
\stackrel{[C]}{=} \int_0^1\left( 24\pi r \sqrt{1-r^2}\right) \,dr 
\\&
=8\pi
\end{align}
$$
$[A]$ comes from Fubini's theorem.
$[B]$ comes from the linearity of integration.
$[C]$ comes from U-substitution. 
Hopefully one can borrow some of strategies.
A: To pass to spherical coordinates you do as  follows
$$2 (u+v+w+3)\quad u= r \cos (\phi) \sin (\theta),v= r \sin (\phi) \sin (\theta),w= r \cos (\theta)$$
and don't forget the Jacobian $r^2 \sin\theta$
$\int_0^{2 \pi } \left(\int_0^{\pi } \left(\int_0^1 2 r^2 \sin (\theta) (r \sin (\phi) \sin (\theta)+r \cos (\phi) \sin (\theta)+r \cos (\theta)+3) \, dr\right) \, dt\right) \, d\phi=8\pi$
A: thanks to zack I found my miscalculation:
from where i stopped:
$\int_0^{2\pi}\int_0^1\int_{-\sqrt{1-r^2}}^{\sqrt{1-r^2}}divFrdzdrd\theta=\int_0^{2\pi}\int_0^1\int_{-\sqrt{1-r^2}}^{\sqrt{1-r^2}}(2r\cos\theta +2r\sin\theta +2z+6)rdzdrd\theta=
\int_0^1\int_{-\sqrt{1-r^2}}^{\sqrt{1-r^2}}[r(2r\sin\theta-2r\cos\theta+(2z+6)\theta]_0^{2\pi}=2\pi\int_0^1\int_{-\sqrt{1-r^2}}^{\sqrt{1-r^2}}2zr+6r=2\pi\int_0^1[rz^2+6rz]_{-\sqrt{1-r^2}}^{\sqrt{1-r^2}}=2\pi\int_0^112r\sqrt{1-r^2}=[-4(1-r^2)]_0^1=8\pi$
