Green's Theorem

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Theorem

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

Let $U = \operatorname{Int} \left({\Gamma}\right)$, that is, the interior of $\Gamma$.

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

Then:

$\displaystyle \oint_\Gamma \left({A \, \mathrm d x + B \, \mathrm d y}\right) = \iint_U \left({\frac {\partial B} {\partial x} - \frac {\partial A} {\partial y} }\right) \, \mathrm d x \, \mathrm d y$


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 $R = \left\{ {\left({x, y}\right): a \le x \le b, c \le y \le d}\right\}$ 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 $\left({a, c}\right)$ to $\left({b, c}\right)$
$C_2$, which runs from $\left({b, c}\right)$ to $\left({b, d}\right)$
$C_3$, which runs from $\left({b, d}\right)$ to $\left({a, d}\right)$
$C_4$, which runs from $\left({a, d}\right)$ to $\left({a, c}\right)$.


Then:

\(\displaystyle \iint_R \frac{\partial B}{\partial x} \, \mathrm d x \, \mathrm d y\) \(=\) \(\displaystyle \int_c^d \int_a^b \frac {\partial B} {\partial x} \, \mathrm d x \, \mathrm d y\) $\quad$ $\quad$
\(\displaystyle \) \(=\) \(\displaystyle \int_c^d \left({B \left({b, y}\right) - B \left({a, y}\right) }\right) \, \mathrm d y\) $\quad$ $\quad$
\(\displaystyle \) \(=\) \(\displaystyle \int_c^d B \left({b, y}\right) \, \mathrm d y + \int_d^c B \left({a, y}\right) \, \mathrm d y\) $\quad$ $\quad$
\(\displaystyle \) \(=\) \(\displaystyle \int_{C_2} B \, \mathrm d y + \int_{C_4} B \, \mathrm d y\) $\quad$ $\quad$


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

So:

$\displaystyle \int_{C_1} B \, \mathrm d y = \int_{C_3} B \, \mathrm d y = 0$

Hence:

\(\displaystyle \int_{C_2} B \, \mathrm d y + \int_{C_4} B \, \mathrm d y\) \(=\) \(\displaystyle \int_{C_1} B \, \mathrm d y + \int_{C_2} B \, \mathrm d y + \int_{C_3} B \, \mathrm d y + \int_{C_4} B \, \mathrm d y\) $\quad$ $\quad$
\(\displaystyle \) \(=\) \(\displaystyle \oint_C B \, \mathrm d y\) $\quad$ $\quad$


A similar argument demonstrates that:

$\displaystyle \iint_R \frac{\partial A}{\partial y} \, \mathrm d x \, \mathrm d y = - \oint_C A \, \mathrm d x$

and hence:

$\displaystyle \oint_C \left({A \, \mathrm d x + B \, \mathrm d y}\right) = \iint_R \left({\frac {\partial B} {\partial x} - \frac {\partial A} {\partial y} }\right) \, \mathrm d x \, \mathrm d y$


Source of Name

This entry was named for George Green.