Gauss's theorem in 2d: how can it be expressed in differential forms?

Question

How do we express the 2d version of Gauss's theorem in the language of differential forms? In 3d, I know it is

$$d \left(Fdydz + Gdzdx + Hdxdy\right) = F_x + G_y + H_z dxdydz$$

so by Stokes' theorem, we have that

$$\iint_{\partial R} Fdydz + Gdzdx + Hdxdy = \int_R F_x + G_y + H_z dxdydz,$$

and the left side can be identified with the integral of $(F,G,H)\cdot \vec{n}dS$. But how does this work in two dimensions? I.e., how do we show

$$\int_{\partial R} (F,G)\cdot \vec{n}dr = \iint_R F_x + G_ydxdy?$$

Answer 1

The correspondence you're using when you transform $$Fdx + Gdy + Hdz \mapsto F dy \wedge dz + G dz \wedge dx + H dx \wedge dy$$ is the Hodge star. In 2D it is simply $$F dx + G dy \mapsto F dy - G dx$$ and thus the differential identity corresponding to the 2D divergence theorem is $$d(F dy - G dx) = F_x dx \wedge dy - G_y dy \wedge dx = (F_x + G_y) dx\wedge dy.$$

Answer 2

Do you mean $$ \oint_{\partial D} -G\, dx + F\, dy = \iint_{D} (F_{x} + G_{y})\, dx\, dy\dots? $$ To see that the integrand on the left is the flux density of $(F, G)$ across $\partial D$, draw a picture of the oriented boundary with its outward normal $N$ and unit tangent field $T$ (a.k.a. $N$ rotated a quarter turn counterclockwise), and the vector fields $V = (F, G)$ and $J(V) = (-G, F)$ (a.k.a. $V$ rotated a quarter turn counterclockwise). Then note that $V \cdot N = J(V) \cdot T$.