# Thread: [SOLVED] Question regarding the flux through a surface

1. ## [SOLVED] Question regarding the flux through a surface

Imagine you have the following closed surface : paraboloid $\displaystyle -x^2-y^2+16$ closed by the $\displaystyle xy$ plane ($\displaystyle z=0$).
Now imagine the divergence of the given field is worth $\displaystyle 0$. For example the field $\displaystyle F(x,y,z)=y \vec i + x \vec j + x \vec k$.
Does that mean that there's no flux passing through the surface of the closed paraboloid? Or it means that the flux entering the paraboloid is worth the same amount than the flux leaving the paraboloid via the $\displaystyle z=0$ plane?
Or does it mean that the total flux is $\displaystyle 0$ and I can't know how much flux pass through the open paraboloid and the $\displaystyle z=0$ plane?

This is a very important question to me! Thanks in advance.

2. Not quite.
Since $\displaystyle {\rm div}{\bf F}=0$ there is a vector function $\displaystyle {\bf f}$ such that $\displaystyle {\bf F}={\rm rot}{\bf f}$. So, using Stoke's theorem, the flux on the paraboloid $\displaystyle S$ becomes

$\displaystyle \int\int_S {\bf F}\cdot {\rm d}{\bf S}=\int\int_S {\rm rot}{\bf f}\cdot {\rm d}{\bf S}=\oint_{\partial S}{\bf f}\cdot{\rm d}{\bf r}$

where $\displaystyle \partial S=\{x^2+y^2=16, \ z=0\}$. This means that the flux depends only on the values of $\displaystyle {\bf f}\vert_{\partial S}$.

3. Originally Posted by Rebesques
Not quite.
Since $\displaystyle {\rm div}{\bf F}=0$ there is a vector function $\displaystyle {\bf f}$ such that $\displaystyle {\bf F}={\rm rot}{\bf f}$.
I wasn't aware of this.
So I can't say anything regarding the flux passing by the closed surface just because the divergence of the field is worth 0. I'd have first to calculate the vector function $\displaystyle \bold f$ and then $\displaystyle \oint_{\partial S}{\bf f}\cdot{\rm d}{\bf r}$.
Thanks a lot.
I have the confusion since I read this thread : Integral De Superficie - Foro fmat.cl (in Spanish), where the guy states that $\displaystyle \vec\nabla\cdot\vec G \ = 0$ then he says that it implies that the flux passing through any closed surface is 0.
He went on to say that the flux passing by a cone plus the flux passing by 2 parallel planes closing the cone is worth 0, hence flux cone + flux planes =0.
I don't know how he did, but he reached the good answer.

4. He's right when $\displaystyle \partial S$ is the union of simple and closed curves, because then the curvilinear integrals $\displaystyle \oint$ are zero.

(lol? it has to be a bad boundary for the zero flux to fail. Thanks for that, clarifies things for me also :P)

5. Originally Posted by Rebesques
He's right when $\displaystyle \partial S$ is the union of simple and closed curves, because then the curvilinear integrals $\displaystyle \oint$ are zero.

(lol? it has to be a bad boundary for the zero flux to fail. Thanks for that, clarifies things for me also :P)
Ok, so in my first post, now I believe
Or it means that the flux entering the paraboloid is worth the same amount than the flux leaving the paraboloid via the plane?
is right. Am I right?

6. Hoora! I get it now!
Or it means that the flux entering the paraboloid is worth the same amount than the flux leaving the paraboloid via the plane?
is yes!
From wikipedia :
In light of the physical interpretation, a vector field with constant zero divergence is called incompressible or solenoidal – in this case, no net flow can occur across any closed surface.
The intuition that the sum of all sources minus the sum of all sinks should give the net flow outwards of a region is made precise by the divergence theorem.
.

7. , that was disgraceful, I aced that class back in uni.

(seems like a full ten years ago... though it's only been nine and a half.)