Green's Theorem

_C (P dx + Q dy) = _D (Q_x - P_y) dA — circulation around the boundary equals total curl inside.

The idea

Theorem (Green's theorem).

Let $D$ be a region of the plane whose boundary $C$ is a simple closed curve traversed counterclockwise, and let $P(x, y)$ and $Q(x, y)$ have continuous partial derivatives on $D$. Then $\oint_C (P\,dx + Q\,dy) = \iint_D \left(\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}\right) dA.$

The two sides are integrals of different dimensions over different sets: on the left, a one-dimensional integral once around the edge of $D$; on the right, a two-dimensional integral over every point of $D$. The theorem says the circuit around the boundary already determines the interior total.

The mechanism is cancellation. Cut $D$ in two with an interior wall, so that each half has its own boundary. Both of those boundaries run along the wall, but in opposite directions, because each half keeps its own region on the left. Adding the two circuits, the wall contributes twice with opposite signs and cancels exactly, leaving the boundary of the original $D$. Every interior cut cancels the same way, so a sum of circuits around many small pieces collapses to a single circuit around the outside. Green's theorem is that collapse taken to the limit, with $Q_x - P_y$ recording what each infinitesimal piece contributes.

The counterclockwise orientation is part of the statement: reversing the direction of travel around $C$ negates the left side and leaves the right side alone.

Ways to work on it

• Walkthrough. Unpack the statement (line vs. double integral, scalar curl, orientation), then apply on a triangle.
• Practice. Apply Green's theorem to a constant-curl field on a rectangle.
• Hardest. Area via line integral: derive A = 12 (x dy - y dx) and use it on an ellipse.

Not sure where to start? Take the ten-question placement test.