# Green's Theorem

## Theorem

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

Let $U = \Int \Gamma$, 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:

$\ds \oint_\Gamma \paren {A \rd x + B \rd y} = \iint_U \paren {\frac {\partial B} {\partial x} - \frac {\partial A} {\partial y} } \rd x \rd y$

where the left hand side is a contour integral.

## 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:

 $\ds \iint_R \frac {\partial B} {\partial x} \rd x \rd y$ $=$ $\ds \int_c^d \int_a^b \frac {\partial B} {\partial x} \rd x \rd y$ $\ds$ $=$ $\ds \int_c^d \paren {\map B {b, y} - \map B {a, y} } \rd y$ $\ds$ $=$ $\ds \int_c^d \map B {b, y} \rd y + \int_d^c \map B {a, y} \rd y$ $\ds$ $=$ $\ds \int_{C_2} B \rd y + \int_{C_4} B \rd y$

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

So:

$\ds \int_{C_1} B \rd y = \int_{C_3} B \rd y = 0$

Hence:

 $\ds \int_{C_2} B \rd y + \int_{C_4} B \rd y$ $=$ $\ds \int_{C_1} B \rd y + \int_{C_2} B \rd y + \int_{C_3} B \rd y + \int_{C_4} B \rd y$ $\ds$ $=$ $\ds \oint_C B \rd y$

A similar argument demonstrates that:

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

and hence:

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

$\blacksquare$

## Source of Name

This entry was named for George Green.