If the flux of a vector field $F$ is zero for ANY surface $S$ then is $F$ zero?

Question

If $\vec F$ is a vector field defined in a open (contractible) set $U$ of $\Bbb R^3$ such that $$ \int_S\vec F\cdot\hat ndS=0 $$ for ANY surface $S$ then in particular the last equality holds for the boudray of a ball $B$ of radius $\delta$ centered at $x$ so that by the divergence theorem it must be $$ \int_B\text{div}\vec F=0 $$ and thus by the main value integral theorem it follows that $$ \text{div}\vec F=0 $$ so that there exist a vector field $\vec\Phi$ whose curl is $\vec F$ but unfortunately I do not see if this implies that $\vec F$ is zero or equivalentely that $\vec\Phi$ is constant. So could some one help me, please?

Answer 1

If it's for any surface $S$, then $F$ has to be $0$, and the things you claim are equivalences aren't actually equivalent. Firstly, you've lost some information because you went from ALL surfaces to only those which are the boundary of something else (so that you can apply the divergence theorem). Next, from $\text{div}(F)=0$ it does not follow it has a vector potential defined on all of $U$ (Poincare's lemma only gives the local existence of such a vector potential), nor does it follow that $F=0$. The answer to your title question is yes

Here's an almost complete outline for proving $F=0$. Suppose contrapositively that there is a point $p\in U$ where $F(p)\neq 0$. Define the unit vector $\nu(p):=\frac{F(p)}{\|F(p)\|}$. The mapping $x\mapsto G(x):=\langle F(x),\nu(p) \rangle$ is continuous on $U$, and $G(p)=1>0$. So, by continuity of $G$, there is an open neighborhood $V$ of $p$, contained in $U$, such that $G>\frac{1}{2}$ on $V$. Therefore if we choose the surface $S$ to be _______________ then $\int_SF\cdot n\, dS$ is ___________. Fill in the blanks yourself.

Answer 2

If the integral is zero for any surface $S$ (e.g., an arbitrarily small disk or rectangle in any orientation), then the claim is true, and @peek-a-boo shows how you might demonstrate that.

If what you meant was "for any closed surface $S$" (i.e., surface-without-boundary, like a sphere or a torus), then the claim is false, for an everywhere constant field like $F(x, y, z) = (1, 0, 0)$ has the property that its integral over any closed surface is zero.