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Categories: Multivariate calculus | Theorems

Green's theorem

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 states:

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

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

Sometimes the notation

$\oint_{C} L\, dx + M\, dy$

is used to indicate the line integral is calculated using the positive orientation of the closed curve C.

Proof of Green's theorem, general edition

TODO

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

If we show Equations 1 and 2

$EQ.1 = \int_{C} L dx = \int\!\!\!\int_{D} \left(- \frac{\partial L}{\partial y}\right) dA$

and

$EQ.2 = \int_{C} M\, dy = \int\!\!\!\int_{D} \left(\frac{\partial M}{\partial x}\right)\, dA$

are true, we would prove Green's theorem.

If we express D as a region such that:

$D = \{(x,y)|a\le x\le b, g_1(x) \le y \le g_2(x)\}$

where g₁ and g₂ are continuous functions, we can compute the double integral of equation 1:

$EQ.4 = \int\!\!\!\int_{D} \left(\frac{\partial L}{\partial y}\right)\, dA = \int_a^b\!\!\int_{g_1(x)}^{g_2(x)} \left(\frac{\partial L}{\partial y} (x,y)\, dy\, dx \right) = \int_a^b [L(x,g_2(x)) - L(x,g_1(x))]\, dx$

Now we break up C 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:

$\int_{C_1} L(x,y)\, dx = \int_a^b [L(x,g_1(x))]\, dx$

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

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

With C₂ and C₄, x is a constant, meaning:

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

Therefore,

$\int_{C} L\, dx = \int_{C_1} L(x,y)\, dx + \int_{C_2} L(x,y)\, dx + \int_{C_3} L(x,y) + \int_{C_4} L(x,y)\, dx$

$= - \int_a^b [L(x,g_2(x))]\, dx + \int_a^b [L(x,g_1(x))]\, dx$

Combining this with equation 4, we get:

$\int_{C} L(x,y)\, dx = \int\!\!\!\int_{D} \left(- \frac{\partial L}{\partial y}\right)\, dA$

A similar proof can be employed on Eq.2.

Categories: Multivariate calculus | Theorems

Science Fair Project Encyclopedia Contents Page

10-26-2009 08:16:03
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