Green's theorem

First prove it for a rectangle $R = [a,b]\times[c,d]$ with $D = R$. By Fubini's theorem, $\iint_R \frac{\partial Q}{\partial x}\,dA = \int_c^d \left(Q(b,y) - Q(a,y)\right) dy$ and $\iint_R \frac{\partial P}{\partial y}\,dA = \int_a^b \left(P(x,d) - P(x,c)\right) dx$.

These are exactly the line integrals of $Q\,dy$ along the right and left edges of $R$ and of $P\,dx$ along the top and bottom edges, so $\iint_R \left(\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}\right) dA = \oint_{\partial R} (P\,dx + Q\,dy)$, matching the theorem for a rectangle.

For a general region $D$, approximate it by a fine grid of small rectangles covering $D$. Apply the rectangle case to each small piece and sum over all pieces. Every interior edge shared by two adjacent pieces is traversed once in each direction by the two pieces' boundaries, so the two line integral contributions along that shared edge cancel exactly.

What remains after cancellation is precisely the line integral around the outer boundary $\partial D$, while the sum of the double integrals over the small rectangles converges to $\iint_D \left(\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}\right) dA$ as the grid is refined, proving the identity for $D$.