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

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

Sometimes the notation

<math>\oint_{C} P dx + Q dy</math>

is used to indicate the line integral is calculuated 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

<math>EQ.1 = \int_{C} P dx = \int \int_{D} \left(- \frac{\partial P}{\partial y}\right) dA</math>

and

<math>EQ.2 = \int_{C} Q dy = \int \int_{D} \left(\frac{\partial Q}{\partial x}\right) dA</math>

are true, we would prove Green's Theorem.

If we express D as a region 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, we can compute the double integral of equation 1:

<math> 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</math>

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:

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

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

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

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

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

Therefore,

<math>\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 </math>

<math> = - \int_a^b [P(x,g_2(x))] dx + \int_a^b [P(x,g_1(x))] dx</math>

Combining this with equation 4, we get:

<math>\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.