Outward Flux of a Divergenceless Vector Field on an Ellipsoid

tl;dr: How do you evaluate $\iint_S \mathbf{F} \cdot d\mathbf{S}$ where $\mathbf{F}(x,y,z) = \frac{1}{(x^2+y^2+z^2)^{3/2}}\langle x,y,z\rangle$ and $\mathbf{S}$ is the outward oriented surface given by $9x^2+4y^2+16z^2=144$?

Long story: My multivariate calculus teacher recently gave our class the following problem:

Compute the outward flux $\iint_S \mathbf{F} \cdot d\mathbf{S}$ where $$\mathbf{F}(x,y,z)=(y + \frac{x}{(x^2+y^2+z^2)^{3/2}})\mathbf{i} + (x + \frac{y}{(x^2+y^2+z^2)^{3/2}})\mathbf{j} + (z + \frac{z}{(x^2+y^2+z^2)^{3/2}})\mathbf{k}$$ and $S$ is the surface of the ellipsoid given by $9x^2+4y^2+16z^2=144$.

The solution he gave us ran along the following lines:

Let $\mathbf{F} = \mathbf{F_1} + \mathbf{F_2}$ where $$\mathbf{F_1} = \langle y,x,z\rangle;\; \mathbf{F_2} = \frac{1}{(x^2+y^2+z^2)^{3/2}}\langle x,y,z\rangle$$

which gives us $\iint_S \mathbf{F} \cdot d\mathbf{S} = \iint_S \mathbf{F_1} \cdot d\mathbf{S} + \iint_S \mathbf{F_2} \cdot d\mathbf{S}$.

By applying the divergence theorem and spherical parametrization, we can find that $\iint_S \mathbf{F_1} \cdot d\mathbf{S} = 96\pi$. This made sense to me; I had no problem with understanding the transformation and evaluation of the integrals.

After that, though, I got lost. According to my notes, he discussed how you can transpose $\iint_S \mathbf{F_2} \cdot d\mathbf{S}$ onto a unit sphere because $\mathbf{F_2}$ is divergenceless, and as a consequence it is equal to the surface integral of the unit sphere, yielding $\iint_S \mathbf{F_2} \cdot d\mathbf{S} = 4\pi$.

This raised two questions: first, of course, was what the heck did my teacher just do? Second was: what's wrong with $\iint_S \mathbf{F_2} \cdot d\mathbf{S} = \iiint_E\nabla\cdot\mathbf{F_2}\,dV = \iiint_E{0}\,dV = 0$ (where $E$ is the region enclosed by surface $S$)? A friend said that it had to do with the fact that $\mathbf{F_2}$ is undefined at $(0,0,0)$, but I'm still confused.

It would be great if someone could explain to me how to evaluate the outward flux of $\mathbf{F_2}$ on $S$.

Thanks.

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Thanks to user5730 for the explanation! Just as a clarification for anyone who sees this in the future: $\iint \mathbf{F} \cdot d\mathbf{T}$ where $\mathbf{F}=\frac{\langle x,y,z\rangle}{(x^2+y^2+z^2)^{3/2}}$ and $\mathbf{T}$ is the surface of the unit sphere reduces to $$\int_0^{2\pi}\int_0^\pi \left(\frac{r^2}{r^4}\right)\,r^2\sin\phi\,d\phi d\theta$$ because $$\mathbf{F} \cdot d\mathbf{T} = \frac{\langle x,y,z\rangle}{(x^2+y^2+z^2)^{3/2}} \cdot |\langle x,y,z\rangle|= \frac{\langle x,y,z\rangle}{(x^2+y^2+z^2)^{3/2}} \cdot \frac{\langle x,y,z\rangle}{\sqrt{x^2+y^2+z^2}} = \frac{r^2}{r^4}$$ –  draks ... Jan 21 '13 at 10:10
the comment above was orignially posted by user59093... –  draks ... Jan 21 '13 at 10:11

Your friend is right. You can't apply the divergence theorem directly because your $F$ is not defined at the origin.

But you can apply the divergence theorem to calculate a modified flux

$$\int_S F\cdot dS - \int_T F\cdot dT$$

where $T$ is the unit sphere. Notice that the surfaces $S$ and $T$ enclose an "egg-white-shaped" volume $V$ that does not contain $F$'s singularity at the origin. Therefore

$$\int_S F\cdot dS - \int_T F\cdot dT = \int_V \nabla\cdot F dV = 0,$$ and the flux over $T$ is much easier to compute than over $S$: $F\cdot dT = 1$ and so the flux is simply the surface area of the unit sphere, $4\pi$.

Notice that there is nothing special about the unit sphere in the above argument: we could have used a sphere of any radius $r$ (or any other surface that encloses the origin, but spheres are easy to calculate with), and it is worth checking that the answer does not depend on $r$. Using spherical coordinates, $$\int_T F\cdot dT = \int_0^{2\pi}\int_0^\pi \left(\frac{r}{r^3}\right)\,r^2\sin\theta\,d\theta d\phi = -2\pi \cos\phi\Big\vert_{0}^{\pi}=4\pi.$$

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