Green's Theorem

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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.


[edit] 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}

[edit] 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

[edit] 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}.

[edit] Proof

The divergence theorem reads

\iiint\limits_V \nabla \cdot \mathbf{F} \, d V =
\iint\limits_{\partial V}\mathbf{F} \cdot d\mathbf{S}

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


\mathbf{F} = \phi \boldsymbol{\nabla}\psi - \psi \boldsymbol{\nabla}\phi

and use

\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

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} .
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