Prove that if $F:\mathbb R^3\to \mathbb R^3$ is a vector field so that $\nabla\times F=0$ $\forall x\in \Omega\subset \mathbb R^3$ (where $\Omega$ is an open simply connected set), then $F$ is a conservative field
I know that I need to use Stokes theorem, the version that was given to me by my teacher is:
Let $S\subset U\subset \mathbb R^3$ be an orientable surface (parametrized by $\sigma:A\subset \mathbb R^2 \to \mathbb R^3$, $\sigma\in C^1$), $\gamma:[a,b]\subset \mathbb R \to \mathbb R^2$ a parametrization of the boundary of $A$ (which is a jordan smooth curve) and $F:U\subset \mathbb R^3 \to \mathbb R^3$ be a $C^1$ vector field then $$\int_S \nabla \times F \ dS=\oint_{\partial S} F dr$$ where $r=\sigma \circ \gamma: [a,b]\subset \mathbb R \to \mathbb R^3$ and $\partial S= \sigma(\partial A)$
I also know that $F:U\subset \mathbb R^3 \to \mathbb R^3$ is conservative field iff $$\oint_\Gamma F d\gamma=0$$ for all jordan smooth curve $\Gamma \subset U$ (parametrized by $\gamma$)
For the proof, I need to take an arbitrary $\Gamma\subset \Omega$ jordan smooth curve (parametrized by a function $\gamma$), but in order to use Stokes theorem I need to prove that $\Gamma$ "generates" an orientable surface $S$ whose boundary $\partial S$ is $\Gamma$ and then using the theorem the result follows, but this is exactly my problem I have no idea how to prove that $\Gamma$ generates and orientable surface $S$
Any comments, suggestions, hints would be really appreciated
Consider $\Gamma$ as 2 distinct curves, $\gamma_1$ and $\gamma_2$ between 2 distinct points on $\Gamma$. Parametrize the curves over [0,1]. The set $$ \{a \gamma_1(t) + (1-a) \gamma_2(t) | \ a \in [0,1] \ \& \ t \in [0,1]\} $$ forms a surface with coordinates $(a,t)$ and boundary $\Gamma$. The coordinate system defines an orientation.