Green's theorem

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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. Green's theorem was named after British scientist George Green and is a special case of the more general Stokes' theorem.

The theorem statement is the following. Let C be a positively oriented, piecewise smooth, simple closed curve in the plane and let D be the region bounded by C. If L and M have continuous partial derivatives on an open region containing D, then

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

Sometimes a small circle is placed on top of the integral symbol:

<math>\oint_{C}</math>

This indicates that the curve C is closed. To indicate positive orientation, an arrow pointing in the counter-clockwise direction is sometimes drawn in the circle over the integral symbol.

Proof of Green's theorem when D is a simple region

Image:Green's-theorem-simple-region.png If it can be shown that

<math>\int_{C} L\, dx = \iint_{D} \left(- \frac{\partial L}{\partial y}\right) dA\qquad\mathrm{(1)}</math>

and

<math>\int_{C} M\, dy = \iint_{D} \left(\frac{\partial M}{\partial x}\right)\, dA\qquad\mathrm{(2)}</math>

are true, then Green's theorem is proven.

We define a region D that is simple enough for our purposes. If region D is expressed such that:

<math>D = \{(x,y)|a\le x\le b, g_1(x) \le y \le g_2(x)\}</math>

where g₁ and g₂ are continuous functions, the double integral in (1) can be computed:

<math> \iint_{D} \left(\frac{\partial L}{\partial y}\right)\, dA</math>	<math>=\int_a^b\!\!\int_{g_1(x)}^{g_2(x)} \left[\frac{\partial L}{\partial y} (x,y)\, dy\, dx \right] </math>
	<math> = \int_a^b \Big\{L[x,g_2(x)] - L[x,g_1(x)] \Big\} \, dx\qquad\mathrm{(3)}</math>

Now 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), a ≤ x ≤ b. Therefore:

<math>\int_{C_1} L(x,y)\, dx = \int_a^b \Big\{L[x,g_1(x)]\Big\}\, dx</math>

With −C₃, use the parametric equations, x = x, y = g₂(x), a ≤ x ≤ b. Then:

<math>\int_{C_3} L(x,y)\, dx = -\int_{-C_3} L(x,y)\, dx = - \int_a^b [L(x,g_2(x))]\, dx</math>

On C₂ and C₄, x remains constant, meaning

<math> \int_{C_4} L(x,y)\, dx = \int_{C_2} L(x,y)\, dx = 0</math>

Therefore,

<math> \int_{C} L\, dx </math>	<math> = \int_{C_1} L(x,y)\, dx + \int_{C_2} L(x,y)\, dx + \int_{C_3} L(x,y) + \int_{C_4} L(x,y)\, dx </math>
	<math> = -\int_a^b [L(x,g_2(x))]\, dx + \int_a^b [L(x,g_1(x))]\, dx\qquad\mathrm{(4)}</math>

Combining (3) with (4), we get:

<math>\int_{C} L(x,y)\, dx = \iint_{D} \left(- \frac{\partial L}{\partial y}\right)\, dA</math>

A similar proof can be employed on (2).

See also

• Planimeter
• Method of image charges - A method used in electro statics that takes strong advantage of the uniqueness theorem (derived from Green's theorem)de:Satz von Green

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