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# 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 based on Stokes' theorem. The theorem states:

Green's Theorem
Let C be a positively oriented, piecewise-smooth, simple closed curve in the plan and let D be the region bounded by C. If P and Q have continuous partial deriviatives on an open region containing D, then

$\int_{C} P dx + Q dy = \int \int_{D} \left(\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}\right) dA$

Sometimes the notation

$\oint_{C} P dx + Q dy$
is used to indicate the line integral is calculuated using the positive orientation of the closed curve C.

• TODO

If we show Equations 1 and 2

$EQ.1 = \int_{C} P dx = \int \int_{D} \left(- \frac{\partial P}{\partial y}\right) dA$
and
$EQ.2 = \int_{C} Q dy = \int \int_{D} \left(\frac{\partial Q}{\partial x}\right) dA$

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:

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

Now we break up C as the union of four curves: C1, C2, C3, C4.

• (Pic could be added here to see how C could be broken up and help explain following proof)

With C1, use the parametric equations[?], x = x, y = g1(x), a ≤ x ≤ b. Therefore:

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

With -C3, use the parametric equations[?], x = x, y = g2(x), a ≤ x ≤ b. Therefore:

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

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

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

Therefore,

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

Combining this with equation 4, we get:

[itex]\int_{C} P(x,y) dx = \int \int_{D} \left(- \frac{\partial P}{\partial y}\right) dA</matH>

A similar proof can be employed on Eq.2.

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