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# Green, George

Green, George, 1793-1841, English mathematician and physicist. He was largely self-taught until, in 1833, he entered Caius College, Cambridge. In addition to making a number of contributions to the calculus, Green was especially interested in the equilibrium of fluids and was the first to introduce the potential in its application to the theories of the magnetic and electric fields. He also studied light and sound. His papers were edited, with a memoir, by N. M. Ferrers (1871).
In physics and mathematics, Green's theorem gives the relationship between a line integral around a simple closed curve C and a double integral over the plane region D bounded by C. It is the two-dimensional special case of the more general Stokes' theorem, and is named after British scientist George Green.

Let C be a positively oriented, piecewise smooth, simple closed curve in the plane R2, and let D be the region bounded by C. If L and M are functions of (x, y) defined on an open region containing D and have continuous partial derivatives there, then

$int_\left\{C\right\} \left(L, dx + M, dy\right) = iint_\left\{D\right\} left\left(frac\left\{partial M\right\}\left\{partial x\right\} - frac\left\{partial L\right\}\left\{partial y\right\}right\right), dA.$

Sometimes a small circle is placed on the integral symbol $left\left(oint_\left\{C\right\}right\right)$ to indicate that the curve C is closed. For positive orientation, an arrow pointing in the counterclockwise direction may be drawn in this circle.

In physics, Green's theorem is mostly used to solve two-dimensional flow integrals, stating that the sum of fluid outflows at any point inside a volume is equal to the total outflow summed about an enclosing area.

## Proof when D is a simple region

The following is a proof of the theorem for the simplified area D, a type I region where C2 and C4 are vertical lines. A similar proof exists for when D is a type II region where C1 and C3 are straight lines.

If it can be shown that

$int_\left\{C\right\} L, dx = iint_\left\{D\right\} left\left(- frac\left\{partial L\right\}\left\{partial y\right\}right\right), dAqquadmathrm\left\{\left(1\right)\right\}$

and

$int_\left\{C\right\} M, dy = iint_\left\{D\right\} left\left(frac\left\{partial M\right\}\left\{partial x\right\}right\right), dAqquadmathrm\left\{\left(2\right)\right\}$

are true, then Green's theorem is proven in the first case.

Define the type I region D as pictured on the right by:

$D = \left\{\left(x,y\right)|ale xle b, g_1\left(x\right) le y le g_2\left(x\right)\right\}$

where g1 and g2 are continuous functions on [a, b]. Compute the double integral in (1):

$iint_\left\{D\right\} left\left(frac\left\{partial L\right\}\left\{partial y\right\}right\right), dA$ $=int_a^b!!int_\left\{g_1\left(x\right)\right\}^\left\{g_2\left(x\right)\right\} left\left[frac\left\{partial L\right\}\left\{partial y\right\} \left(x,y\right), dy, dx right\right]$
$= int_a^b Big\left\{L\left(x,g_2\left(x\right)\right) - L\left(x,g_1\left(x\right)\right) Big\right\} , dxqquadmathrm\left\{\left(3\right)\right\}$

Now compute the line integral in (1). C can be rewritten as the union of four curves: C1, C2, C3, C4.

With C1, use the parametric equations: x = x, y = g1(x), axb. Then

$int_\left\{C_1\right\} L\left(x,y\right), dx = int_a^b Big\left\{L\left(x,g_1\left(x\right)\right)Big\right\}, dx$

With C3, use the parametric equations: x = x, y = g2(x), axb. Then

$int_\left\{C_3\right\} L\left(x,y\right), dx = -int_\left\{-C_3\right\} L\left(x,y\right), dx = - int_a^b \left[L\left(x,g_2\left(x\right)\right)\right], dx$

The integral over C3 is negated because it goes in the negative direction from b to a, as C is oriented positively (counterclockwise). On C2 and C4, x remains constant, meaning

$int_\left\{C_4\right\} L\left(x,y\right), dx = int_\left\{C_2\right\} L\left(x,y\right), dx = 0$

Therefore,

 $int_\left\{C\right\} L, dx$ $= int_\left\{C_1\right\} L\left(x,y\right), dx + int_\left\{C_2\right\} L\left(x,y\right), dx + int_\left\{C_3\right\} L\left(x,y\right), dx + int_\left\{C_4\right\} L\left(x,y\right), dx$ $= -int_a^b \left[L\left(x,g_2\left(x\right)\right)\right], dx + int_a^b \left[L\left(x,g_1\left(x\right)\right)\right], dxqquadmathrm\left\{\left(4\right)\right\}$

Combining (3) with (4), we get (1). Similar computations give (2).

## Relationship to the divergence theorem

Green's theorem is equivalent to the following two-dimensional analogue of the divergence theorem:
$iint_Dleft\left(nablacdotmathbf\left\{F\right\}right\right)dA=int_C mathbf\left\{F\right\} cdot mathbf\left\{hat n\right\} , ds,$
where $mathbf\left\{hat n\right\}$ is the outward-pointing unit normal vector on the boundary.

To see this, consider the unit normal in the right side of the equation. Since $dmathbf\left\{r\right\} = langle dx, dyrangle$ is a vector pointing tangential along a curve, and the curve C is the positively-oriented (i.e. counterclockwise) curve along the boundary, an outward normal would be a vector which points 90° to the right, which would be $langle dy, -dxrangle$. The length of this vector is $sqrt\left\{dx^2 + dy^2\right\} = ds$. So $mathbf\left\{hat n\right\},ds = langle dy, -dxrangle$.

Now let the components of $mathbf\left\{F\right\} = langle P, Qrangle$. Then the right hand side becomes

$int_C mathbf\left\{F\right\} cdot mathbf\left\{hat n\right\} , ds = int_C P dy - Q dx$
which by Green's theorem becomes
$int_C -Q dx + P dy = iint_\left\{D\right\} left\left(frac\left\{partial P\right\}\left\{partial x\right\} + frac\left\{partial Q\right\}\left\{partial y\right\}right\right), dA = iint_Dleft\left(nablacdotmathbf\left\{F\right\}right\right)dA.$