Green's Theorem

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

Let $U = \map {\operatorname {Int} } \Gamma$, that is, the interior of $\Gamma$.

Let $A$ and $B$ be functions of $\tuple {x, y}$ defined on an open region containing $U$ and have continuous partial derivatives in such a set.

Then:
 * $\displaystyle \oint_\Gamma \paren {A \rd x + B \rd y} = \iint_U \paren {\frac {\partial B} {\partial x} - \frac {\partial A} {\partial y} } \rd x \rd y$

Proof
It suffices to demonstrate the theorem for rectangular regions in the $x y$-plane.

The Riemann-sum nature of the double integral will then guarantee the proof of the theorem for arbitrary regions, because a Riemann-sum is technically a summation of the areas of arbitrarily small rectangles.

As the proof is for a rectangle, the proof will work for arbitrary regions, which can be approximated by collections of ever smaller rectangles.

Let $R = \set {\tuple {x, y}: a \le x \le b, c \le y \le d}$ be a rectangular region.

Let the boundary $C$ of $R$ be oriented counterclockwise.

We break the boundary into $4$ pieces:
 * $C_1$, which runs from $\tuple {a, c}$ to $\tuple {b, c}$
 * $C_2$, which runs from $\tuple {b, c}$ to $\tuple {b, d}$
 * $C_3$, which runs from $\tuple {b, d}$ to $\tuple {a, d}$
 * $C_4$, which runs from $\tuple {a, d}$ to $\tuple {a, c}$

Then:

We note that $y$ is constant along $C_1$ and $C_3$.

So:
 * $\displaystyle \int_{C_1} B \rd y = \int_{C_3} B \rd y = 0$

Hence:

A similar argument demonstrates that:


 * $\displaystyle \iint_R \frac {\partial A} {\partial y} \rd x \rd y = -\oint_C A \d x$

and hence:


 * $\displaystyle \oint_C \paren {A \rd x + B \rd y} = \iint_R \paren {\frac {\partial B} {\partial x} - \frac {\partial A} {\partial y} } \rd x \rd y$

Also see

 * Gauss-Ostrogradsky Theorem