TITLE: Every closed $C^1$ curve in $\mathbb R^3 \setminus \{ 0 \}$ is the boundary of some $C^1$ 2-surface $\Sigma \subset \mathbb R^3 \setminus \{ 0 \}$
QUESTION [5 upvotes]: How can I prove it?
This problem looks similar to Plateau's problem - but it is much more specific. I believe there exists some elementary proof.
(Proving this will help me apply Stokes' theorem to another problem)
thanks in advance.
EDIT: I have a proof that works in some cases (I hope). Assume $\Gamma$ is a such a curve: 

Calculate the center of mass for the curve.
Translate the curve by its center of mass.
Define $\Sigma'=\cup_{0 \leq \lambda \leq 1} \lambda \Gamma$, where $\lambda \Gamma=\{   \lambda p:p \in \Gamma\}$
$\Sigma$ will be $\Sigma'$ translated backwards by the center of mass.

REPLY [2 votes]: Since your goal is to apply Stokes' theorem, you need an orientable surface. The orientability makes the problem tricky, which is why it took until 1930 to sort it out. Such a surface indeed exists for any smooth $\Gamma$; it is called a Seifert surface of $\Gamma$ and can be obtained using Seifert's algorithm. This page has a schematic drawing of a Seifert surface of the trefoil knot; more detailed images can be found with a Google search; e.g., Visualization of Seifert surfaces. 
Your candidate for $\Sigma'$ is not a smooth surface unless $\Gamma$ is a planar curve (look at what happens at the center of mass). It may have self-intersections issues too.