Generalized Stokes Theorem, applied to 2D/3D Generalized Stokes Theorem says $\int_{\partial S} \omega = \int_S d \omega$ where $\omega$ is a $(k-1)$-form and $S$ is a $k$-dimensional manifold. I understand that, for $n=3$,


*

*If $k=3$ this becomes the Divergence Theorem

*If $k=2$ this becomes the Kelvin-Stokes Theorem, the one with the curl

*If $k=1$ what does this become? My guess is that this looks like the FTC of conservative field, because $\omega$ is a curve and the boundaries are the two points, and $d$ operator on a 0 form is the "grad" operator. Is that right? I have a trouble formalizing this guess.
But for $n=2$, what does this look like, for $k=1$ and $k=2$? My guess is Green's theorem but I have a hard time seeing the connection.
Specifically, I have seen Green's Theorem formulated two ways: A: $\int_C P dx + Q dy = \int \int_S (\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}) dA$
B: $\int_C (P \vec{i} + Q \vec{j}).\vec{n} . ds = \int \int_S(\frac{\partial P}{\partial x} + \frac{\partial Q}{\partial y}) dA $
Both are equivalent, but formulation A looks like a 2-D version of Kelvin-Stokes Theorem (the RHS looks very much like a curl), and formulation B looks like a 2-D version of Divergence theorem (LHS looks like a 2D flux, and RHS looks like a div). So I'm wondering which one is a direct consequence of Generalized Stokes, and the other formulation was just done to "make it convenient"
A: This may be a bit late but if k=1, we have that $\omega$ is a $0$-form scalar function $f$. It follows that $d\omega$ is a $1$-form. Hence, S is a 1-manifold curve C, while $\partial$$S$ is simply the two terminal points of the curve.
Now if you define a 1-manifold curve X($t$) and let $\omega$=$f(x)$ in $R^n$, and simply apply the Generalized Stokes Theorem, you can prove the following result:
$\int_C$ ($\nabla$$f$$\bullet$ $\overrightarrow T$)$ds$=$f(B)-f(A)$, where $\overrightarrow T$ is a (unit) tangent vector to the curve. Also, A and B are the endpoints of the curve.
Hint: Note that $\int_A^B$$f$=$f(B)-f(A)$. The difference in signs is due to the orientation of the endpoints of the curve itself. 
For k=1 in $R^2$, ($\omega$ is a $0$-form), you simply get $\int_C$ ($\nabla$$f$$\bullet$ $\overrightarrow T$)$ds$=$f(B)-f(A)$ again, but $\nabla$$f$ only has 2 components.
For k=1 in $R$, the Generalized Stokes Theorem implies the Fundamental Theorem of Calculus. You can prove this rigorously or simply note that $(\nabla$$f$$\bullet$ $\overrightarrow T$)$ds$=$f ' (x) dx$ in $R$.
In $R^2$, the Generalized Stokes Theorem implies formulation A when working with k=2 ($\omega$ is a $1$-form). Note that formulation A (Green's Theorem) is precisely the Kelvin-Stokes Theorem in $R^2$. Also, note that Formulation B is equivalent to the divergence theorem in $R^2$. 
