<s>[INST] 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?
 [/INST] thanks to zack I found my miscalculation:
from where i stopped:
$\int_0^{2\pi}\int_0^1\int_{-\sqrt{1-r^2}}^{\sqrt{1-r^2}}divFrdzdrd\theta=\int_0^{2\pi}\int_0^1\int_{-\sqrt{1-r^2}}^{\sqrt{1-r^2}}(2r\cos\theta +2r\sin\theta +2z+6)rdzdrd\theta=
\int_0^1\int_{-\sqrt{1-r^2}}^{\sqrt{1-r^2}}[r(2r\sin\theta-2r\cos\theta+(2z+6)\theta]_0^{2\pi}=2\pi\int_0^1\int_{-\sqrt{1-r^2}}^{\sqrt{1-r^2}}2zr+6r=2\pi\int_0^1[rz^2+6rz]_{-\sqrt{1-r^2}}^{\sqrt{1-r^2}}=2\pi\int_0^112r\sqrt{1-r^2}=[-4(1-r^2)]_0^1=8\pi$
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