Gauss-Green Theorem from generalized Stoke's Theorem. I am trying to deduce the next identity (Green-Gauss theorem)  $$\int_\Omega \dfrac{\partial u}{\partial x_i} dx = \int_{\partial \Omega} uv_i dS$$
from the generalized Stoke's theorem for manifolds. Here $u$ is a $C^\infty$ function on an open bounded set $\Omega$ with $C^1$ boundary and $v_i$ denotes the i-th component of the unit normal vector field. 
I've been trying to define $n-1$ forms on $\Omega$ whose derivatives are $\dfrac{\partial u}{\partial x_i} dx$ to apply the theorem but I cannot prove that those forms have the same integral as $uv_i$ on $\partial \Omega$. 
Any ideas? 
Also, I would really appreciate references of the proofs of classical calculus integration results using the generalized Stoke's theorem. 
Thanks in advance.
 A: First of all, you need to understand what $dS$ is.
$dS$ is the form $dV$ contracted by the vector field $\nu$ on $\partial \Omega$.  In other words, to find $dS(v_2,v_3,...,v_{n})$ for tangent vectors $v_2,v_3,...,v_{n}$ to $\partial \Omega$, you just evaluate $dV(\nu, v_2,v_3,...,v_{n})$.  The motivation for this definition is that the area of a parallelogram $P$ is the same as the volume of the parallelepiped with base $P$ and height $1$.
So now note that,
$
 \frac{\partial f}{\partial x_1} dV = d(f dx_2 \wedge dx_3 \wedge ...\wedge dx_n)
$
Letting $ \widehat{dx_1} = dx_2 \wedge dx_3 \wedge ...\wedge dx_n$
Let $p \in \partial \Omega$ and $v_2,v_3,...,v_n \in T_p( \partial \Omega)$ then
$
\begin{align*}
f(p) \widehat{dx_1}(v_2,v_3,...,v_n) &= f(p) dV(e_1,v_2,...,v_n)\\
&=f(p) dV(\textrm{proj}_{\nu(p)}(e_1),v_2,...,v_n)\\
&=f(p) dV(\nu_1(p)\nu(p),v_2,...,v_n)\\
&=f(p)\nu_1(p) dV(\nu(p),v_2,...,v_n)\\
&=f(p)\nu_1(p) dS(v_2,...,v_n)
\end{align*}
$
so as forms on $\partial S$, we have $f\widehat{dx_1} = f\nu_1dS$.  The big step is the second line in the equality above, which is justified by writing $e_1$ as a linear combination of $\nu$ and $v_i$'s.  All of the $v_i$ terms die because of the alternating property of forms.
Now we conclude that 
$\begin{align*}
\int_\Omega \frac{\partial f}{\partial x_1} dV &= \int_\Omega d(f \widehat{dx_i})\\
&=\int_{\partial \Omega} f \widehat{dx_i} \text{by Stokes' theorem}\\ 
&=\int_{\partial \Omega} f \nu_1 dS
\end{align*}
$
If any of this does not make sense, please ask for follow up in the comments.  Notation in differential geometry is horrible, and everyone makes it up for themselves I think.  So I understand if some part of this is hard to parse.  I am also no differential geometer, but I do a lot of calculus in high dimensional spaces (I work in several complex variables).  This kind of thing used to frustrate me a lot.
A: Steven's answer is good, let me reformulate it in more formal flavor. By Stokes, it suffices to prove 
$$l^*(dx^2 \wedge...\wedge dx^n) = dS := l^*(i_v(dx^1\wedge...\wedge dx^n))$$
Where $l:\partial U \to U$ the inclusion map, $i_v$ is the contraction operator with respect to $v$.
Let $i_{\frac{\partial}{\partial x^1}}$ be the contrition operator with respect to $\frac{\partial}{\partial x^1}$, then we have $l^*(dx^2 \wedge...\wedge dx^n) = l^*i_{\frac{\partial}{\partial x^1}}(dx^1,...,dx^n)$.
Then we just need to prove $l^*i_{\frac{\partial}{\partial x^1}}(dx^1,...,dx^n) = \langle \frac{\partial}{\partial x^1}, v \rangle l^*(i_v(dx^1\wedge...\wedge dx^n))$. To see this, decompose $\frac{\partial}{\partial x^1} = \langle \frac{\partial}{\partial x^1}, v \rangle v + \alpha$, where $\alpha$ is tangent to $\partial U$.(essentially we used Lemma 16.30 in Lee's introduction to smooth manifold.), then we have 
$$l^*i_{\frac{\partial}{\partial x^1}}(dx^1,...,dx^n) = \langle \frac{\partial}{\partial x^1}, v \rangle l^*(i_v(dx^1\wedge...\wedge dx^n))+l^*i_\alpha (dx^1\wedge...\wedge dx^n)$$
Now it remains to prove
$l^*i_\alpha (dx^1\wedge...\wedge dx^n) = 0$, which is because choosing $v_2,...,v_n$ tangent to $\partial U$, then $\alpha$ and $v_2,...,v_n$ must be linear dependent thus have zero determinant, so $l^*i_\alpha (dx^1\wedge...\wedge dx^n)(v_2,...,v_n)=(dx^1 \wedge...\wedge dx^n)(\alpha, v_2,...,v_n)=0$
