the 2-D divergence theorem and Green's Theorem I read somewhere that the 2-D Divergence Theorem is the same as the Green's Theorem.
So for Green's theorem
$$\oint_{\partial \Omega} {\textbf{F}} \cdot d {\textbf{S}} = \iint_{\Omega} \text{2d-curl}{\textbf{F}} \, d\Omega$$
and also by Divergence (2-D) Theorem,
$$\oint_{\partial \Omega} {\textbf{F}} \cdot d {\textbf{S}} = \iint_{\Omega} \text{div}{\textbf{F}} \, d\Omega$$.
Since they can evaluate the same flux integral, then
$$\iint_{\Omega} \text{2d-curl}{\textbf{F}} \, d\Omega = \int_{\Omega} \text{div}{\textbf{F}} \, d\Omega.$$
Is there an intuition for why the summing of divergence in a region is equal to the summing of the curl in a region in 2-D?
 A: This is not quite right: they are equivalent, but they don't use the same vector field or the same vector on the boundary. The divergence theorem says
$$ \iint_{\Omega} \operatorname{div}{\mathbf{F}} \, dx \, dy = \oint_{\partial \Omega} \mathbf{F} \cdot \mathbf{n} \, dl, $$
where $\mathbf{n}$ is an outward-pointing normal and $dl$ is the line element. Now, $\mathbf{n} \, dl $ is perpendicular to $d\mathbf{l}$ (being a normal). $d\mathbf{l} = (dx,dy)$, so the outward-pointing normal is $(dy,-dx)$ (rotate it by $\pi/2$ anticlockwise). So if we take $\mathbf{F}=(M,-L)$, we find this becomes
$$ \iint_{\Omega} \left( \frac{\partial M}{\partial x}-\frac{\partial L}{\partial y} \right) dx \, dy = \oint_{\partial\Omega} (-L) \,(-dx) +M \, dy, $$
which is Green's theorem.
What's actually going on here is that in two dimensions, $\operatorname{curl}{\mathbf{F}}$ can be written as the divergence of the field $\mathbf{F}_{\perp} = (F_2,-F_1)$, the rotation of $\mathbf{F}$ through a right angle. So
$$ \oint_{\partial\Omega} \mathbf{F} \cdot d\mathbf{l} \stackrel{\text{Stokes}}{=} \iint_{\Omega} \operatorname{curl}{\mathbf{F}} \, dx \, dy = \iint_{\Omega} \operatorname{div}{\mathbf{F}_{\perp}} \, dx \, dy \stackrel{\text{div thm}}{=} \oint_{\partial\Omega} \mathbf{F}_{\perp} \cdot \mathbf{n} \, dl.  $$
We can now also understand the equality between the line integrals by the equality $\mathbf{F} \cdot d\mathbf{l} = \mathbf{F}_{\perp} \cdot \mathbf{n} \, dl $, since $ \mathbf{n} dl = (d\mathbf{l})_{\perp}$. So what in effect has happened is that both vectors have been rotated by the same amount, and hence the dot product gives the same value: $ \mathbf{F} \cdot d\mathbf{l} = \mathbf{F}_{\perp} \cdot (d\mathbf{l})_{\perp} $.
