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 $\mathbb {R} ^{2}$ 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

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

Proof

The divergence theorem reads

$\iiint \limits _{V}\nabla \cdot \mathbf {F} \,dV=\iint \limits _{\partial V}\mathbf {F} \cdot d\mathbf {S}$

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

Insert

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

and use

${\begin{aligned}{\boldsymbol {\nabla }}\cdot \mathbf {F} &={\big (}{\boldsymbol {\nabla }}\phi {\big )}\cdot {\big (}{\boldsymbol {\nabla }}\psi {\big )}-{\big (}{\boldsymbol {\nabla }}\psi {\big )}\cdot {\big (}{\boldsymbol {\nabla }}\phi {\big )}+\phi {\boldsymbol {\nabla }}^{2}\psi -\psi {\boldsymbol {\nabla }}^{2}\phi \\&=\phi {\boldsymbol {\nabla }}^{2}\psi -\psi {\boldsymbol {\nabla }}^{2}\phi \end{aligned}}$

so that we obtain the result to be proved,

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