Problem. Consider the General Stokes Theorem and $M$ a submanifold of $\mathbb{R}^{n}$, with boundary orientable.
(a) Prove the Green Theorem $$\int_{M}(g_{x} - f_{y})dxdy = \int_{\partial M} fdx + gdy$$ from Stokes Theorem.
(b) Prove the Gauss Theorem $$\int_{M}\mathrm{div}FdV = \int_{\partial M}\langle F, n \rangle dA$$ from Stokes Theorem.
(c) Use Gauss Theorem to prove the Green Theorem.
My attempt
(a) Take the $1$-form $\omega = fdx + gdy$. Then
$\begin{eqnarray*} \int_{\partial M}fdx + gdy &=& \int_{\partial M}\omega\\ &=& \int_{M}d\omega\\ &=& \int_{M}d(fdx + gdy)\\ &=& \int_{M}d(f)\wedge dx + d(g)\wedge dy\\ &=& \int_{M}(f_{x}dx + f_{y}dy)\wedge dx + (g_{x}dx + d_{y}dy)\wedge dy\\ &=& \int_{M}f_{y}dy\wedge dx + g_{x}dx\wedge dy\\ &=& \int_{M}g_{x}dx\wedge dy - f_{y}dx\wedge dy\\ &=& \int_{M}(g_{x}-f_{y})dxdy. \end{eqnarray*}$
(b) Consider $n = 3$ (the general case works with the same idea). Take $\omega = F_{1}dy\wedge dz + F_{2}dz\wedge dx + F_{3}dx\wedge dy$.
$\begin{eqnarray*} \int_{\partial M}\langle F,n \rangle dA &=& \int_{\partial M}F_{1}n_{1} + F_{2}n_{2} + F_{3}n_{3}\\ &=& \int_{\partial M}F_{1}(n_{1}dA) + F_{2}(n_{2}dA) + F_{3}(n_{3}dA)\\ &=& \int_{\partial M}F_{1}dy\wedge dz + F_{2}dz\wedge dx + F_{3}dx\wedge dy\\ &=& \int_{\partial M}\omega\\ &=& \int_{M}d\omega\\ &=& \int_{M}d(F_{1}dy\wedge dz + F_{2}dz\wedge dx + F_{3}dx\wedge dy)\\ &=& \int_{M}d(F_{1})\wedge dy\wedge dz + d(F_{2})\wedge dz\wedge dx + d(F_{3})\wedge dx\wedge dy\\ &=& \int_{M}F_{1}dx\wedge dy\wedge dz + F_{2}dx\wedge dy\wedge dz + F_{3}dx\wedge dy\wedge dz\\ &=& \int_{M}\mathrm{div}FdV. \end{eqnarray*}$
(c) I tried to use Gauss $\to$ Stokes (in $\mathbb{R}^{3}$) $\to$ Green, but I couldnt prove (probably dont work). Can someone help me?
The items (a) and (b) I wrote a sketch of the prove (without the details) because I believe that this idea is correct, but correct me if I'm wrong.