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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?

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  • $\begingroup$ Consider $F = a \hat{x} + b \hat{y} + c \hat{z}$ with $a$, $b$ and $c$ constants not depending on location. $\endgroup$
    – AHusain
    Jun 28 '21 at 9:26
  • $\begingroup$ @AHusain Sorry, I do not understand your hint: could you explain better, please? $\endgroup$ Jun 28 '21 at 9:26
  • $\begingroup$ The vector field $\vec F$ does not have to be zero or constant. $\endgroup$
    – Math Lover
    Jun 28 '21 at 9:37
  • $\begingroup$ @MathLover Could you explain better, please? So the statement is false? $\endgroup$ Jun 28 '21 at 9:37
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    $\begingroup$ Then $\vec F$ should be zero $\endgroup$
    – Math Lover
    Jun 28 '21 at 9:50
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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.

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  • $\begingroup$ So if we choose the surface $S$ to be a circle centered at $P$ and contined in the intersection of $V$ and a plane whose normal is $\nu(p)$ then $$\int_S F\cdot\hat n\ge\int_S \frac{1}2=\frac{1}2\cdot\text{Area}(S)\neq0 $$ that is impossible, right? $\endgroup$ Jun 28 '21 at 10:05
  • $\begingroup$ I think you mean "disk" rather than "circle". $\endgroup$ Jun 28 '21 at 10:08
  • $\begingroup$ @JohnHughes Oh yeah! My bad, sorry. $\endgroup$ Jun 28 '21 at 10:09
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    $\begingroup$ @AntonioMariaDiMauro I assume you meant "a disk...whose normal is $\nu(p)$..." Then yes, the integral is non-zero and your estimation of the integral is correct. Just a small nitpick about "that is impossible". The way I wrote my answer, it was phrased as a contrapositive proof, not a proof by contradiction. So, saying $\int_SF\cdot n\,dS\neq 0$ already completes the proof. If you want to write it up as a proof by contradiction, that's also fine, but of course your phrasing would have to be slightly different. $\endgroup$
    – peek-a-boo
    Jun 28 '21 at 10:09
  • $\begingroup$ @peek-a-boo Okay, thanks too much for your assistance!!! See you soon. $\endgroup$ Jun 28 '21 at 10:11
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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.

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  • $\begingroup$ Okay, and is correct my commment is the @peek-a-boo's answer? $\endgroup$ Jun 28 '21 at 10:08

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