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?
Best Answer
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.