I'm now starting to compute my first flux integrals and I'd like to ask you if my procedure is correct and if everything I've done is correctly motivated.
I'm asked to compute the flux of $F=r^{-3}(x,y,z)$ where $r=\sqrt{x^2+y^2+z^2}$ across the ellipsoid centered in $O(0,0,0)$ and of semiaxis $1,2,5$.
Our surface can be described as: $$\sigma \colon(\theta, \phi) \longmapsto (\sin\theta \cos\phi, 2\sin\theta \sin\phi, 5\cos\theta)$$ and we can compute the normal vector: $$n=\frac{\partial \sigma}{\partial \theta}\land \frac{\partial\sigma}{\partial\phi}=i(10\sin^2\theta\cos\phi)+j(5\sin^2\theta\sin\phi)+k(\cos\theta\sin\theta(1+\sin^2\phi))$$ but doing so we get a difficult integral.
So I tried to verify what $divF$ is and it's easy to see that $divF=0 \quad \forall(x,y,z) \in \mathbb{R}^3$.
The question is: can I conclude now by applying the divergence theorem and say that the integral is $0$?
If yes, can you explain me better how to formally explain this thing? If not, can you explain me how to prosecute in the exercise?
Thanks in advance.
No, the vector field is not defined at the origin, so it's divergence is not defined at the origin (some might say its infinite); i.e you only have $\text{div}(F)=0$ on $\Bbb{R}^3\setminus \{0\} $. So, the divergence theorem is not applicable directly to the full ellipsoid which contains the origin. Instead, consider a small closed ball centered around the origin such that it lies entirely inside the ellipsoid.
Now, apply the divergence theorem to the region between the ellipsoid and ball to deduce that the surface integral over the ellipse is equal to surface integral over sphere (be very very careful with which way the unit outward normal vector points). This surface integral should be much easier to calculate directly. Basically the idea is to use the divergence theorem to switch surfaces.
If you do everything right, you should get an answer of $4 \pi. $