Green's Theorem: Statement, Orientation and Worked Examples

Green's theorem relates a line integral around a closed curve in the plane to a double integral over the region it encloses. It says the total circulation around the boundary equals the sum of all the small circulations inside.

Statement of Green's theorem

Let

1. C be a positively oriented (counterclockwise), piecewise smooth, simple closed curve in the plane, and
2. D be the region bounded by C, and
3. P(x, y) and Q(x, y) have continuous partial derivatives on an open region that contains D.

Then

In vector form, with \mathbf{F} = \langle P, Q \rangle, the left side is the circulation \oint_C \mathbf{F}\cdot d\mathbf{r} and the integrand on the right is the scalar curl Q_x - P_y.

What Green's theorem means

Chop D into tiny squares and add up the circulation of \mathbf{F} around each one. Along every interior edge, two neighboring squares traverse it in opposite directions, so those contributions cancel. Only the outer boundary survives. Meanwhile, the circulation around a tiny square is approximately (Q_x - P_y) times its area. Adding up both ways gives Green's theorem.

Why each hypothesis matters

Orientation. Traversing C clockwise reverses the sign of the line integral, while the double integral does not change. Always check the direction.

Continuity everywhere inside. The vortex field

has Q_x - P_y = 0 at every point except the origin, where it is undefined. Around the unit circle x = \cos t, y = \sin t, the line integral is \int_0^{2\pi} (\sin^2 t + \cos^2 t)\,dt = 2\pi, not 0. Green's theorem does not apply because the origin, inside D, is a point where P and Q are not even defined.

Closed curve. For a curve that does not close up, there is no enclosed region, so the theorem says nothing.

Worked examples

Example 1: compute \oint_C xy\,dx + x^2\,dy around the region in the graph.

Step 1. P = xy and Q = x^2, so Q_x - P_y = 2x - x = x.

Step 2. The region is x^2 \le y \le x for 0 \le x \le 1:

Check directly. Along y = x^2 (x = t, 0 \to 1) the integrand is t^3\,dt + t^2(2t\,dt) = 3t^3\,dt, contributing \frac{3}{4}. Along y = x back from (1, 1) to (0, 0) it is 2t^2\,dt with t from 1 to 0, contributing -\frac{2}{3}. The total is \frac{3}{4} - \frac{2}{3} = \frac{1}{12}. Same answer, more work.

Example 2: area of an ellipse. Choose P = -\frac{y}{2} and Q = \frac{x}{2}, so Q_x - P_y = 1 and the double integral is the area:

For the ellipse x = a\cos t, y = b\sin t, 0 \le t \le 2\pi: x\,dy - y\,dx = (ab\cos^2 t + ab\sin^2 t)\,dt = ab\,dt. So A = \frac{1}{2}\cdot 2\pi ab = \pi ab.

Example 3: a messy field, an easy answer. Evaluate \oint_C (x^2 - y)\,dx + (x + y^3)\,dy around the unit circle, counterclockwise. Here Q_x - P_y = 1 - (-1) = 2, so the integral is 2 \cdot \text{area} = 2\pi.

Common mistakes

1. Swapping the order in the integrand. It is Q_x - P_y, not P_y - Q_x or P_x - Q_y.
2. Ignoring orientation. A clockwise curve gives the negative of the double integral.
3. Using it across a singularity. Check that P and Q are smooth on all of D, not just on C.

Related theorems and guides

Green's theorem is the flat case of Stokes' theorem, and its flux form is the two-dimensional divergence theorem. Since Q_x - P_y = 0 for gradient fields by Clairaut's theorem, conservative fields have zero circulation. For the double integrals on the right side, see double integrals.