Green's Theorem

Green's Theorem is a vector identity that is equivalent to the curl theorem in two dimensions. It relates the line integral around a simple closed curve $$\partial\Omega$$ with the double integral over the plane region $$\Omega\,$$.

The theorem is named after the British mathematician George Green. It can be applied to various fields in physics, among others flow integrals.

Mathematical Statement in two dimensions
Let $$\Omega\,$$ be a region in $$\R^2$$ with a positively oriented, piecewise smooth, simple closed boundary $$\partial\Omega$$. $$f(x,y)$$ and $$g(x,y)$$ are functions defined on a open region containing $$\Omega\,$$ and have continuous partial derivatives in that region. Then Green's Theorem states that

\oint\limits_{\partial\Omega}(fdx+gdy)=\iint\limits_\Omega \left(\frac{\partial g}{\partial x}-\frac{\partial f}{\partial y}\right)dxdy $$

The theorem is equivalent to the curl theorem in the plane and can be written in a more compact form as

\oint\limits_{\partial \Omega}\mathbf{F}\cdot d\mathbf{S}=\iint\limits_\Omega (\nabla\times\mathbf{F})d\mathbf{A} $$

Application: Area Calculation
Green's theorem is very useful when it comes to calculating the area of a region. If we take $$f(x,y)=y$$ and $$g(x,y)=x$$, the area of the region $$\Omega\,$$, with boundary $$\partial\Omega$$ can be calculated by

A=\frac{1}{2}\oint\limits_{\partial \Omega} xdy-ydx $$ This formula gives a relationship between the area of a region and the line integral around its boundary.

If the curve is parametrized as $$\left(x(t),y(t)\right)$$, the area formula becomes

A=\frac{1}{2}\oint\limits_{\partial \Omega}(xy'-x'y)dt $$

Statement in three dimensions
Different ways of formulating Green's theorem in three dimensions may be found. One of the more useful formulations is

\iiint\limits_V \Big( \phi \boldsymbol{\nabla}^2\psi - \psi \boldsymbol{\nabla}^2\phi\Big)\, d V = \iint\limits_{\partial V} \big(\phi \boldsymbol{\nabla}\psi\big) \cdot d\mathbf{S} - \iint\limits_{\partial V} \big(\psi \boldsymbol{\nabla}\phi\big) \cdot d\mathbf{S}. $$

Proof
The divergence theorem reads where $$d\mathbf{S}$$ is defined by $$d\mathbf{S}=\mathbf{n} \, dS$$ and $$\mathbf{n}$$ is the outward-pointing unit normal vector field.

Insert

\mathbf{F} = \phi \boldsymbol{\nabla}\psi - \psi \boldsymbol{\nabla}\phi $$ and use

\begin{align} \boldsymbol{\nabla}\cdot \mathbf{F} &= \big(\boldsymbol{\nabla}\phi\big)\cdot \big(\boldsymbol{\nabla}\psi\big) -\big(\boldsymbol{\nabla}\psi\big)\cdot \big( \boldsymbol{\nabla}\phi\big) + \phi \boldsymbol{\nabla}^2\psi - \psi \boldsymbol{\nabla}^2\phi \\ &= \phi \boldsymbol{\nabla}^2\psi - \psi \boldsymbol{\nabla}^2\phi \end{align} $$ so that we obtain the result to be proved,

\iiint\limits_V \phi \boldsymbol{\nabla}^2\psi - \psi \boldsymbol{\nabla}^2\phi\, d V = \iint\limits_{\partial V}\phi \boldsymbol{\nabla}\psi \cdot d\mathbf{S} - \iint\limits_{\partial V}\psi \boldsymbol{\nabla}\phi \cdot d\mathbf{S}. $$