Green's theorem

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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 two-dimensional case of the more general Stokes' theorem.

The theorem statement is the following. 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 = \iint_{D} \left(\frac{\partial M}{\partial x} - \frac{\partial L}{\partial y}\right)\, dA.

Sometimes a small circle is placed on top of the integral symbol:

\oint_{C}

This indicates that the curve C is closed. To indicate positive orientation, an arrow pointing in the counter-clockwise direction is sometimes drawn in the circle over the integral symbol.

In physics, Green's theorem is mostly used to solve two-dimensional flow integrals, stating that the sum of fluid outflows at any point inside a volume is equal to the total outflow summed about an enclosing area.

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

If D is the simple region so that x ∈ [a, b] and g1(x) < y < g2(x) and the boundary of D is divided into the curves C1, C2, C3, C4, we can demonstrate Green's theorem.
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If D is the simple region so that x ∈ [a, b] and g1(x) < y < g2(x) and the boundary of D is divided into the curves C1, C2, C3, C4, we can demonstrate Green's theorem.

We will prove the theorem for the simplified area D where C2 and C4 are vertical lines, however the theorem remains valid for any area D as defined above.

If it can be shown that

\int_{C} L\, dx = \iint_{D} \left(- \frac{\partial L}{\partial y}\right) dA\qquad\mathrm{(1)}

and

\int_{C} M\, dy = \iint_{D} \left(\frac{\partial M}{\partial x}\right)\, dA\qquad\mathrm{(2)}

are true, then Green's theorem is proven.

We define a region D that is simple enough for our purposes. If region D is expressed 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, the double integral in (1) can be computed:

\iint_{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 \Big\{L[x,g_2(x)] - L[x,g_1(x)] \Big\} \, dx\qquad\mathrm{(3)}


Now C can be rewritten 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 \Big\{L[x,g_1(x)]\Big\}\, 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

On C2 and C4, x remains 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\qquad\mathrm{(4)}

Combining (3) with (4), we get:

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

A similar proof can be employed on equation (2).

[edit] See also

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