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. It is a special two-dimensional case of the more general Stokes' theorem, and is named after British scientist George Green.

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 the integral symbol (\oint_{C}) to indicate that the curve C is closed. For positive orientation, an arrow pointing in the counterclockwise direction may be drawn in this circle.

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 when D is a simple region

If D is a simple region with its boundary consisting of the curves C1, C2, C3, C4, Green's theorem can be demonstrated.
If D is a simple region with its boundary consisting of the curves C1, C2, C3, C4, Green's theorem can be demonstrated.

The following is a proof of the theorem for the simplified area D, a type I region where C2 and C4 are vertical lines. A similar proof exists for when D is a type II region where C1 and C3 are straight lines.

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 in the first case.

Defined the type I region D as pictured on the right by:

D = \{(x,y)|a\le x\le b, g_1(x) \le y \le g_2(x)\}

where g1 and g2 are continuous functions. Compute the double integral in (1):

\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)}


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. Then

\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

The integral over C3 is negated because it goes in the negative direction from b to a, as C is oriented positively (counterclockwise). 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)\, dx + \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 (1). Similar computations give (2).

[edit] See also

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