Not quite.
Since there is a vector function such that . So, using Stoke's theorem, the flux on the paraboloid becomes
where . This means that the flux depends only on the values of .
Imagine you have the following closed surface : paraboloid closed by the plane ( ).
Now imagine the divergence of the given field is worth . For example the field .
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 plane?
Or does it mean that the total flux is and I can't know how much flux pass through the open paraboloid and the plane?
This is a very important question to me! Thanks in advance.
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 and then .
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 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.
Hoora! I get it now!
The answer tois 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.