# Flux of a vector field across an ellipsoid.

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?

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.
If you do everything right, you should get an answer of $$4 \pi.$$
• I actually tried to do it and I think I got stuck. I chose the unitary sphere $(\theta, \phi) \mapsto (\sin\theta\cos\phi, \sin\theta\sin\phi, \cos\theta)$ so the normal vector is $(\sin^2\theta \cos\phi, \sin^2\theta\sin\phi, \cos\theta\sin\theta)$ and the integral becomes $\int \int sin \theta d\theta d\phi$, is it correct? May 13, 2020 at 8:07
• @Alberto Yes, that's the right integral. Actually even regardless of the radius of the sphere, the final integral will be the one you describe, because the radius of the sphere in the denominator of your vector field will cancel out thr radius from the area element $dA$ of the surface integral. May 13, 2020 at 8:12
• Perfect, done it and the result is $4\pi$ as wished. Thanks a lot, I'll post different questions about this so I can deepen my knowledge about it. Answer accepted. Thanks again. May 13, 2020 at 8:14