Green's Theorem

A conservative field has zero circulation around every loop. What about a field that is not conservative — how much does it circulate? Green's theorem answers exactly, and the answer is one of the most beautiful trades in mathematics: a boundary integral equals an interior integral.

Let C be a positively oriented (counterclockwise) simple closed curve bounding a region R, and let \mathbf{F} = (P, Q) have continuous partials on R. Then the circulation of \mathbf{F} around C equals the double integral of its "curl" over R:

\oint_C \mathbf{F}\cdot d\mathbf{r} = \oint_C P\,dx + Q\,dy = \iint_R \left( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \right) dA.

The loop integral on the left lives only on the edge; the area integral on the right lives in the interior. Green's theorem says they are the same number.

Verifying Green's theorem on a square, line by line

Trust grows by checking. Take the field \mathbf{F} = (-y,\; x) — the rotation field, the most circulation-heavy field there is — over the unit square R = [0, 1] \times [0, 1], with boundary C traversed counterclockwise. We compute both sides and watch them match.

The right-hand side (the double integral)

Step 1 — form the integrand. With P = -y and Q = x,

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

Step 2 — integrate the constant over the square.

\iint_R 2 \, dA = 2 \cdot \operatorname{area}(R) = 2 \cdot 1 = 2.

The left-hand side (the loop integral)

Step 3 — walk the four edges counterclockwise. Bottom (0,0)\to(1,0), right (1,0)\to(1,1), top (1,1)\to(0,1), left (0,1)\to(0,0). The integrand is P\,dx + Q\,dy = -y\,dx + x\,dy.

Step 4 — bottom edge. Here y = 0, so P = -y = 0, and dy = 0. Contribution: 0.

Step 5 — right edge. Here x = 1 so dx = 0, and Q = x = 1 with y running 0\to1:

\int_0^1 1 \, dy = 1.

Step 6 — top edge. Here y = 1 so P = -1 and dy = 0, with x running 1\to0:

\int_1^0 (-1)\, dx = -\,(0 - 1) = 1.

Step 7 — left edge. Here x = 0 so dx = 0 and Q = x = 0. Contribution: 0.

Step 8 — sum the edges.

\oint_C \mathbf{F}\cdot d\mathbf{r} = 0 + 1 + 1 + 0 = 2.

Both sides equal 2. Green's theorem checks out: the total counterclockwise push around the edge equals twice the area, exactly as the constant curl 2 predicted.

The area corollary

Run the trade backwards. If we can make the integrand Q_x - P_y equal to 1, the double integral becomes the plain area of R — so Green's theorem turns a boundary walk into an area.

Step 1 — choose a field with unit curl. Take P = -\tfrac{y}{2}, Q = \tfrac{x}{2}. Then

\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} = \tfrac12 - \left(-\tfrac12\right) = 1.

Step 2 — apply Green's theorem.

\iint_R 1\, dA = \oint_C \left(-\tfrac{y}{2}\,dx + \tfrac{x}{2}\,dy\right) = \tfrac12 \oint_C (x\,dy - y\,dx).

Step 3 — read off the area formula.

\operatorname{area}(R) = \frac{1}{2}\oint_C (x\,dy - y\,dx).

You can measure the area of a region by walking around its edge — never stepping inside. This is not a party trick; it is exactly how a planimeter works.

Green's theorem is the first member of a family that dominates advanced calculus: the integral over a region equals an integral over its boundary. The fundamental theorem of calculus already said it in one dimension — \int_a^b f' \, dx = f(b) - f(a), the interior derivative on the left, the two boundary points on the right. Green lifts it to the plane. Lift again and you get the divergence theorem (flux through a closed surface = integral of divergence inside) and Stokes' theorem (circulation around a space curve = flux of curl through any spanning surface). In the language of differential forms all four are a single statement,

\int_{\partial \Omega} \omega = \int_{\Omega} d\omega,

the generalized Stokes' theorem: integrate a form over a boundary, or its derivative over the inside, and get the same number. Green's theorem is the case \dim\Omega = 2.

The area corollary is a real instrument. A planimeter is a little linkage with a tracer arm; you run its pointer once around the boundary of a shape — a lake on a map, a region on an X-ray, an irregular machined part — and a dial reads off the enclosed area, without any formula for the boundary and without computing anything inside. Mechanically it is integrating \tfrac12\oint(x\,dy - y\,dx) as the wheel rolls and slips along the curve. Engineers and cartographers used them for over a century — a brass embodiment of Green's theorem you could hold in your hand.

Circulation over a region

Here is the rotation field \mathbf{F} = (-y, x) with a region R and its counterclockwise boundary C (the orientation arrows show the positive direction). Switch the region: for each, the boundary circulation \oint_C \mathbf{F}\cdot d\mathbf{r} equals \iint_R 2\,dA = 2\,\operatorname{area}(R) — boundary on the edge, curl on the inside, one number.