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
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 g
_{1} and g
_{2} 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: C_{1}, C_{2}, C_{3}, C_{4}.
- (Pic could be added here to see how C could be broken up and help explain following proof)
With C_{1}, use the parametric equations[?], x = x, y = g_{1}(x), a ≤ x ≤ b. Therefore:
- <math>\int_{C_1} P(x,y) dx = \int_a^b [P(x,g_1(x))] dx</math>
With -C_{3}, use the parametric equations[?], x = x, y = g_{2}(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 C_{2} and C_{4}, 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.
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