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Consider the following:

For any smooth Jordan curve $\gamma\subset D$, where $D$ is simply connected, we can find a smooth surface $\Sigma \subset D$ such that $\partial\Sigma = \gamma$.

I am curious about the validity of the statement and how could one possibly go about proving it with more rigour rather than intuition. This was used in a proof that if the vector field is irrotational and the domain is simply connected then it is integrable. Just stating it felt very hand-wavy

Thank you for any suggestions or ideas!

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  • $\begingroup$ What is $D$, aside from simply connected? It sounds like it's a region in $\Bbb R^3$. I believe you also want $\Sigma$ to be an orientable surface (with boundary) if you're wanting to apply Stokes's Theorem. At any rate, you might read about the Seifert surface of a knot. $\endgroup$
    – Ted Shifrin
    Commented Apr 20 at 23:10
  • $\begingroup$ @TedShifrin exactly what I meant, I should have said it:) Thank you! $\endgroup$
    – Teodoras Paura
    Commented Apr 22 at 5:27

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