Proof of Gauss theorem (divergence theorem) in $\mathbb R^2$ I am trying to solve an exercise in where it is asked to show the divergence theorem, or also known as Gauss theorem, in $\mathbb R^2$ using Green's theorem.
I suppose that the divergence theorem in $\mathbb R^2$ is that for a $C^1$ vector field $F:\mathbb R^2 \to \mathbb R^2$ defined on a $3$ type region $D$ such that $\partial D$ is a closed curve, we have $$\iint_D div(F)dA=\int_{\partial D} Fds$$
I've tried to prove this theorem applying Green's theorem but I coudn't, I would appreciate if someone could provide a solution using Green's theorem (or at least the steps I should follow to prove the equality).
 A: You're missing the normal in your calculation. Green's theorem says that $$\iint_D (Q_x-P_y)dxdy=\int_{\partial D}Pdx+Qdy$$
Consider the form $\omega=-Qdx+Pdy$. Then $d\omega=(P_x+Q_y)dx\wedge dy$, i.e. if $F=(P,Q)$ then $dw=({\rm div\;}F)dx\wedge dy$. Thus the above says that $$\iint_D {\rm div }\; F dxdy=\int_{\partial D}(-Qdx+Pdy)$$
Now suppose $\partial D$ is parametrized by $\gamma:[0,1]\to\Bbb R^2$. Then the derivative is $\gamma'(t)=(\gamma_1(t),\gamma_2(t))$ and the normal is ${\bf n}(t)=(\gamma_2(t),-\gamma_1(t))/\lVert \gamma'(t)\rVert$. Now $$\begin{align}\int_{\partial D}(-Qdx+Pdy)&=\int_0^1 Q(\gamma(t))(-\gamma_1'(t))+P(\gamma(t))\gamma_2'(t)dt\\&=\int_0^1 F(\gamma(t))\cdot {\bf n}(t)\lVert \gamma'(t)\rVert dt\\&=\int_{\partial D}(F\cdot{\bf n})ds\end{align}$$
which is what you wanted. You can do this in three dimensions too, by using the form $\star \omega=Pdy\wedge dz+Qdz\wedge dx+Rdx\wedge dy$, i.e. applying the Hodge star to $\omega =Pdx+Qdy+Rdz$. Then $d(\star \omega)={\rm div}\,F \; dx\wedge dy\wedge dz$ where $F=(P,Q,R)$, and if $\rho:[0,1]\times [0,1] \to\Bbb R^3$ $(s,t)\mapsto \rho(s,t)$ is a parametrization of the boundary of your body, using the normal $\rho_s \times \rho_t$, tortuous calculations will show that $$\iint_{\partial B} \star \omega=\iint_{\partial B} F\cdot {\bf n} dS$$ where the latter is a surface integral.  
A: Recall Green's theorem
$$\int_{\partial D} F ds=\int_{\partial D}F_1 dx+F_2 dy=\iint_D (\partial F_2/\partial x-\partial F_1/\partial y)dxdy$$
From this, setting $G_1=-F_2$ and $G_2=F_1$, we get 
$$\iint_D (\partial F_1/\partial x+\partial F_2/\partial y)dxdy=\iint_D (\partial G_2/\partial x-\partial G_1/\partial y)dxdy=$$$$\int_{\partial D} G_1 dx+G_2dy=\int_{\partial D} -F_2 dx+F_1 dy$$
So $$\iint_D\text{div} Fdxdy=\int_{\partial D} F\cdot\mathbf{n} ds$$
where $\mathbf{n}$ is an outward unit normal vector to $\partial D$.
Notice that this is not the theorem you stated!
