I got the following problem: Let $F$ be a continously differentiable vector field with $div(F)=0$. We define an integral curve as a curve that is tangent to the vector field F.

Now let us take a curve C that is never tangent to F. When we look at the integral curves of F through C we get a surface, call it S.

Prove that the flux of F through every section of S is equal.

Any help? I tried to find a general parametrization to the surface F and compute the flux "by hand". I got that the flux is zero through every section but that seems odd, so I am probably mistaken. Plus, I haven't used that fact that the field is divergence free.

Please help :)


I think the OP is right, regardless of the value of $\nabla \cdot F$.

Why? Recall that the flux through a surface $S$ is

$\displaystyle \int_S F \cdot \vec n \; dS, \tag 1$

where $dS$ is the area element of $S$, and $\vec n$ is a unit normal field; but here we have

$F \cdot \vec n = 0 \tag 2$

on $S$, since $F$ is tangent to the surface. Thus the integral (1) must vanish, independently of $\nabla \cdot F$.

Note that we required no special parametrization to see this.


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