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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).

The Columbia Electronic Encyclopedia Copyright © 2004.

Licensed from Columbia University Press

Licensed from Columbia University Press

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.## Proof when D is a simple region

Let C be a positively oriented, piecewise smooth, simple closed curve in the plane R^{2}, 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

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

Sometimes a small circle is placed on the integral symbol $left(oint\_\{C\}right)$ 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.

The following is a proof of the theorem for the simplified area D, a type I region where C_{2} and C_{4} are vertical lines. A similar proof exists for when D is a type II region where C_{1} and C_{3} are straight lines.

If it can be shown that

- $int\_\{C\}\; L,\; dx\; =\; iint\_\{D\}\; left(-\; frac\{partial\; L\}\{partial\; y\}right),\; dAqquadmathrm\{(1)\}$

and

- $int\_\{C\}\; M,\; dy\; =\; iint\_\{D\}\; left(frac\{partial\; M\}\{partial\; x\}right),\; dAqquadmathrm\{(2)\}$

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

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

- $D\; =\; \{(x,y)|ale\; xle\; b,\; g\_1(x)\; le\; y\; le\; g\_2(x)\}$

where g_{1} and g_{2} are continuous functions on [a, b]. Compute the double integral in (1):

$iint\_\{D\}\; left(frac\{partial\; L\}\{partial\; y\}right),\; dA$ $=int\_a^b!!int\_\{g\_1(x)\}^\{g\_2(x)\}\; left[frac\{partial\; L\}\{partial\; y\}\; (x,y),\; dy,\; dx\; right]$ $=\; int\_a^b\; Big\{L(x,g\_2(x))\; -\; L(x,g\_1(x))\; Big\}\; ,\; dxqquadmathrm\{(3)\}$ Now compute the line integral in (1). C can be rewritten as the union of four curves: C

_{1}, C_{2}, C_{3}, C_{4}.With C

_{1}, use the parametric equations: x = x, y = g_{1}(x), a ≤ x ≤ b. Then- $int\_\{C\_1\}\; L(x,y),\; dx\; =\; int\_a^b\; Big\{L(x,g\_1(x))Big\},\; dx$

With C

_{3}, use the parametric equations: x = x, y = g_{2}(x), a ≤ x ≤ b. Then- $int\_\{C\_3\}\; L(x,y),\; dx\; =\; -int\_\{-C\_3\}\; L(x,y),\; dx\; =\; -\; int\_a^b\; [L(x,g\_2(x))],\; dx$

The integral over C

_{3}is negated because it goes in the negative direction from b to a, as C is oriented positively (counterclockwise). On C_{2}and C_{4}, x remains constant, meaning- $int\_\{C\_4\}\; L(x,y),\; dx\; =\; int\_\{C\_2\}\; L(x,y),\; dx\; =\; 0$

Therefore,

$int\_\{C\}\; L,\; dx$ $=\; int\_\{C\_1\}\; L(x,y),\; dx\; +\; int\_\{C\_2\}\; L(x,y),\; dx\; +\; int\_\{C\_3\}\; L(x,y),\; dx\; +\; int\_\{C\_4\}\; L(x,y),\; dx$ $=\; -int\_a^b\; [L(x,g\_2(x))],\; dx\; +\; int\_a^b\; [L(x,g\_1(x))],\; dxqquadmathrm\{(4)\}$ 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:- $iint\_Dleft(nablacdotmathbf\{F\}right)dA=int\_C\; mathbf\{F\}\; cdot\; mathbf\{hat\; n\}\; ,\; ds,$

To see this, consider the unit normal in the right side of the equation. Since $dmathbf\{r\}\; =\; langle\; dx,\; dyrangle$ 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 $sqrt\{dx^2\; +\; dy^2\}\; =\; ds$. So $mathbf\{hat\; n\},ds\; =\; langle\; dy,\; -dxrangle$.

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

- $int\_C\; mathbf\{F\}\; cdot\; mathbf\{hat\; n\}\; ,\; ds\; =\; int\_C\; P\; dy\; -\; Q\; dx$

- $int\_C\; -Q\; dx\; +\; P\; dy\; =\; iint\_\{D\}\; left(frac\{partial\; P\}\{partial\; x\}\; +\; frac\{partial\; Q\}\{partial\; y\}right),\; dA\; =\; iint\_Dleft(nablacdotmathbf\{F\}right)dA.$

## See also

- Stokes' theorem
- Divergence theorem
- Planimeter
- Method of image charges - A method used in electrostatics that takes strong advantage of the uniqueness theorem (derived from Green's theorem)
- Green's identities

## External links

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Last updated on Sunday October 05, 2008 at 12:41:57 PDT (GMT -0700)

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