Green's Theorem

Theorem
Let $$\Gamma \ $$ be a positively oriented piecewise smooth simple closed curve in $$\R^2 \ $$.

Let $$U = \operatorname{Int} (\Gamma) \ $$, i.e. the interior of $$\Gamma$$.

If $$A \ $$ and $$B \ $$ are functions of $$\left({x, y}\right) \ $$ defined on an open region containing $$U \ $$ and have continuous partial derivatives in such a set, then:


 * $$\oint_{\Gamma} (A\, dx + B\, dy) = \iint_{U} \left(\frac{\partial B}{\partial x} - \frac{\partial A}{\partial y}\right)\ dxdy \ $$

Proof
It suffices to demonstrate the theorem for rectangular regions in the $$xy \ $$-plane. The Riemann-sum nature of the double integral will then guarantee the proof of the theorem for arbitrary regions.

Let $$R = \left\{{(x,y)|a\leq x\leq b, c \leq x \leq d }\right\} \ $$ be a rectangular region and its boundary $$C \ $$ oriented counterclockwise.

We break the boundary into $$4 \ $$ pieces: $$C_1, \ $$ which runs from $$(a,c) \ $$ to $$(b,c) \ $$, $$C_2, \ $$ which runs from $$(b,c) \ $$ to $$(b,d) \ $$, $$C_3, \ $$ which runs from $$(b,d) \ $$ to $$(a,d) \ $$, and $$C_4, \ $$ which runs from $$(a,d) \ $$ to $$(a,c) \ $$.

Then:


 * $$\iint_R \frac{\partial B}{\partial y} dxdy = \int_c^d \int_a^b \frac{\partial B}{\partial x} dxdy = \int_c^d \left({ B(b,y)-B(a,y) }\right) dy = \int_c^d B(b,y)dy + \int_d^c B(a,y)dy = \int_{C_2} B dy + \int_{C_4} Bdy \ $$.

We note that $$y \ $$ is constant along $$C_1 \ $$ and $$C_3 \ $$, so $$\int_{C_1} Bdy = \int_{C_3} Bdy = 0 \ $$, hence:


 * $$\int_{C_2} B dy + \int_{C_4} Bdy = \int_{C_1} Bdy + \int_{C_2} B dy + \int_{C_3} Bdy + \int_{C_4} Bdy = \oint_C Bdy \ $$

A similar argument demonstrates that:


 * $$\iint_R \frac{\partial A}{\partial y} dxdy = - \oint_C Adx \ $$

and hence:


 * $$\oint_{C} (A\, dx + B\, dy) = \iint_{R} \left(\frac{\partial B}{\partial x} - \frac{\partial A}{\partial y}\right)\ dxdy \ $$