  ## Encyclopedia > Green's theorem

Article Content

# 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.

All Wikipedia text is available under the terms of the GNU Free Documentation License

Search Encyclopedia
 Search over one million articles, find something about almost anything!

Featured Article
 Bugatti ... produced, the most famous being the Type 35 Grand Prix car, the huge "Royale[?]", and the Type 55 sports car. Bugatti also designed a successful motorized railcar, the ...  