Green's Theorem

Green's Theorem is a result in real analysis. It is a special case of Stokes' Theorem.

Statement

Let $D$ be a bounded subset of $\mathbb{R}^2$ with positively oriented boundary $C$ , and let $L$ and $M$ be functions with continuous partial derivatives mapping an open set containing $D$ into $\mathbb{R}$ . Then $\int_C \bigl[ L(x,y)dx + M(x,y)dy \bigr] = \iint_D \left( \frac{\partial M} {\partial x} - \frac{\partial L}{\partial y} \right) dA.$

Proof

It suffices to show that the theorem holds when $D$ is a square, since $D$ can always be approximated arbitrarily well with a finite collection of squares.

Then let $D$ be a square with vertices $(a,c)$ , $(b,c)$ , $(b,d)$ , $(a,d)$ , with $a\le b$ and $c \le d$ . Then $\begin{align*} \int_C \bigl[ L(x,y)dx + M(x,y)dy \bigr] ={}& \int_{a}^b L(x,c) dx + \int_{c}^d M(b,y)dy \\ &+ \int_b^a L(x,d) dx + \int_{d}^c M(a,y) dy \\ ={}& \int_c^d \bigl[ M(b,y) - M(a,y) \bigr] dy + \int_a^b \bigl[ L(x,d) - L(x,c) \bigr] dx . \end{align*}$ Now, by the Fundamental Theorem of Calculus, $L(x,d) - L(x,c) = \int_c^d \frac{\partial L}{\partial y}(x) dy$ and $M(b,y) - M(a,y) = \int_a^b \frac{\partial M}{\partial x}(y) dx .$ Hence $\begin{align*} \int_c^d \bigl[ M(b,y) - M(a,y) \bigr] dx - \int_a^b \bigl[ L(x,d) - L(x,c) \bigr] dy &= \int_c^d \int_a^b \frac{\partial M}{\partial x} dx dy - \int_a^b \int_c^d \frac{\partial L}{\partial y} dy dx \\ &= \iint_D \left( \frac{\partial M}{\partial x} - \frac{\partial L}{\partial y} \right) dA , \end{align*}$ as desired. $\blacksquare$

Page

Toolbox

Search

Green's Theorem

Statement

Proof

See also