Green's theorem

From Exampleproblems

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 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.

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

Image:Green's-theorem-simple-region.png

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.

If it can be shown that if

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

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,

$\int_{C} L\, dx$	$= \int_{C_1} L(x,y)\, dx + \int_{C_2} L(x,y)\, dx + \int_{C_3} L(x,y) + \int_{C_4} L(x,y)\, dx$
	$= -\int_a^b [L(x,g_2(x))]\, dx + \int_a^b [L(x,g_1(x))]\, dx\qquad\mathrm{(4)}$