Divergence theorem: why is it $F_n \cdot dS$? I am learning the calculus 3 interpretation of Stokes' theorems while just dipping my toe in the water of differential forms. Since I have only limited knowledge of the real differential geometry involved, I'm having trouble putting things together.
Reading the treatment of Stokes' theorems in Buck's Advanced Calculus, I find the following lemma relating the 2-form to the normal:
$$\int_{\Sigma} f dy dz + g dz dx + hdx dy \\= \int_{D} f \frac{ \partial(y,z) }{ \partial(u,v) }du dv + g \frac{ \partial(z,x) }{ \partial(u,v) }du dv + h\frac{ \partial(x,y) }{ \partial(u,v) }du dv\\ = \int_D \vec{U} \cdot \mathbf{n}(u,v)\, dA$$
where $\vec{U} = (f,g,h)$ and $\sigma: \mathbb{R}^2 \to \mathbb{R}^3$ is a parameterized surface in $\mathbb{R}^3$ given by $\sigma(u,v) = (x(u,v), y(u,v), z(u,v))$.
The referring to $dudv$ as "$dA$" is a little confusing to me. "$dA$" seems to be something that if we integrated it, we should end up with the surface area of $\Sigma$ (I would like to think of it like "parameterizing by arclength".) But if we integrate $dudv$, we end up with the area in the parameter space. 
OK, so perhaps the $dA$ refers to the area of the parameter space. But then here on Wikipedia I find the statement 
$$\iiint _V (\nabla \cdot F) dV = \iint_{S} F \cdot n \,dS.$$
which I read as translating
$$\int_V (f_x + g_y + h_z) dx dy dz \\=\int_V d(f\, dy dz + g \,dz dx + h \,dx dy) + \int_{\partial V}f\, dy dz + g \,dz dx + h \,dx dy.$$
Surely this $dS$ here does not refer to any particular parameterization space...? Here $dS$ is something which, if we integrated it, would yield the surface area of $S$, right?
Question 2: What 2-form would we integrate if we want the surface area of a parameterized surface in $\mathbb{R}^3$?
 A: When you do a line integral, you are not obtaining the arc length. You are integrating a 1-form along the line. That is, this:
$$
\int_L A\cdot dl = \int_L\left( A_x\,dx + A_ydy + A_z\,dz \right)  = \int_L \left( A_x\frac{dx}{dt}+A_y\frac{dy}{dt}+A_z\frac{dz}{dt}\right)\,dt
$$
is something different from this:
$$
\int_L |dl| = \int_L \sqrt{dx^2+dy^2+dz^2}=\int_L \sqrt{\left(\frac{dx}{dt}\right)^2+\left(\frac{dy}{dt}\right)^2+\left(\frac{dz}{dt}\right)^2}\,dt.
$$
Just as well, there is a difference between integrating something (a 2-form) over a surface, and integrating the modulus of the area to obtain its absolute magnitude.
The first is:
$$
\int_S U\cdot n\,dS,
$$
which is exactly what you wrote, the second would be:
$$
\int_S \bigg| \frac{\partial \vec x}{\partial u}\wedge \frac{\partial \vec x}{\partial v}\bigg| \,du\,dv.
$$
The terms $dS$ could refer to both, and it is confusing, but they are different quantities. 
In the first case, the vector $\vec n\,dS$ is the quantity:
$$
\frac{\partial \vec x}{\partial u}\wedge \frac{\partial \vec x}{\partial v}\,du\,dv;
$$
in the second case, the scalar $dS$ is the modulus of the vector above.
The situation is pretty much the same with $dl$ and $|dl|$ in the line case.
