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:

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

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:

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

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

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

A similar argument demonstrates that:

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

and hence:

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

Source of Name

This entry was named for George Green.