Green's Theorem

Theorem Statement

Let \( C \) be a positively oriented, piecewise-smooth, simple closed curve in the plane, and let \( D \) be the region bounded by \( C \).

If \( P(x,y) \) and \( Q(x,y) \) have continuous first partial derivatives on an open region containing \( D \), then:

\[ \oint_{C} \left( P \, dx + Q \, dy \right) = \iint_{D} \left( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \right) \, dA \]

Key Concepts

Orientation: The curve \( C \) must be traversed counterclockwise (positive orientation).

Connection: Relates a line integral around a closed curve to a double integral over the region it encloses.

Physical Interpretation

For a vector field \( \vec{F} = \langle P, Q \rangle \), Green's theorem states that the circulation of \( \vec{F} \) around \( C \) equals the integral of the scalar curl over \( D \).

Special Case: Area Calculation

The area of region \( D \) can be found using:

\[ \text{Area} = \oint_{C} x \, dy = -\oint_{C} y \, dx = \frac{1}{2} \oint_{C} (-y \, dx + x \, dy) \]

This follows from Green's theorem by choosing \( P \) and \( Q \) such that \( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} = 1 \).

Visual Representation

Boundary Curve C

Region D

Vector Field F

Example Calculation

For the vector field \( \vec{F} = \langle -y, x \rangle \) shown:

\[ \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} = 1 - (-1) = 2 \]

By Green's theorem:

\[ \oint_C \vec{F} \cdot d\vec{r} = \iint_{D} 2 \, dA = 2 \times \text{Area}(D) \]

For an interactive exploration of Green's theorem with different curves and vector fields, check out:

Explore Green's Theorem on GeoGebra