6.4 Green's Theorem

Big idea

The previous section traded a line integral for a subtraction, but only for conservative fields. This section handles the general case in the plane, and the trade is different: a line integral around a closed curve becomes a double integral over the region the curve encloses. Nothing has to be conservative for this to work. What you gain is a choice, since either side can be the easier computation, and what you pay is that the curve must be closed.

The statement continues the pattern started by the fundamental theorem for line integrals. There, an integral over a curve was determined by data on the curve’s boundary, its two endpoints. Here, an integral over a plane region is determined by data on the region’s boundary, a closed curve. The dimension has gone up by one on both sides, and so has the derivative involved: a gradient there, a combination of first partials here.

There are two versions, and they are the same theorem read through two different dot products. Integrating the tangential component of a field around the boundary measures circulation, and it equals the double integral of $Q_x - P_y$. Integrating the normal component measures flux across the boundary, and it equals the double integral of $P_x + Q_y$. The two integrands are the two derivative combinations that the next section names curl and divergence, so this theorem is also the first evidence that those two combinations are the right ones to care about.

A sign convention is unavoidable. Because reversing a closed curve negates its line integral but does nothing to the double integral, the theorem is only true for one of the two orientations. The convention, called positive orientation, is to walk the boundary so that the region stays on your left, which for a simple region without holes means counterclockwise.

Decoder

For a positively oriented simple closed curve bounding a region in the plane, the circulation of a field around the curve equals the double integral over the region of the $x$-partial of the second component minus the $y$-partial of the first.

Each hypothesis is doing something. “Simple” rules out self-crossings, so the curve really has an inside. “Closed” is what makes a boundary out of the curve; an open arc bounds nothing and the statement would be meaningless. “Positively oriented” fixes the sign. And the field must have continuous partial derivatives on an open set containing the region, including the inside, which is exactly the hypothesis the punctured-plane field of the previous section violates at the origin.

Notice how the conservative case falls out. If $Q_x = P_y$ everywhere on a region with no holes, the double integral is zero, so the circulation around every closed curve inside is zero, and the field is conservative there. That is a proof of the converse half of the cross-partials test, and it also shows why the simply connected hypothesis is needed: the theorem requires the entire inside of the curve to be part of the domain.

The most common error is orientation applied to a region with a hole. For an annulus, positive orientation means counterclockwise on the outer curve and clockwise on the inner one, since in both cases the shaded material must lie on your left as you walk.

Definitions and results

Positive orientation. A simple closed curve is positively oriented with respect to the region it bounds when the region lies to the left of the direction of travel. For a region with no holes this is counterclockwise.

Green’s theorem, circulation form. Let $D$ be a region in the plane whose boundary $C$ is a positively oriented, piecewise smooth, simple closed curve, and let $P$ and $Q$ have continuous partial derivatives on an open set containing $D$. Then $$ \oint_C P\,dx + Q\,dy = \iint_D \left( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \right) dA $$

Green’s theorem, flux form. With $\mathbf{n}$ the outward unit normal along $C$ and the same hypotheses, $$ \oint_C \mathbf{F}\cdot \mathbf{n}\,ds = \iint_D \left( \frac{\partial P}{\partial x} + \frac{\partial Q}{\partial y} \right) dA $$ This is the circulation form applied to the field $\langle -Q, P\rangle$, which is $\mathbf{F}$ rotated a quarter turn, so the two versions carry identical information.

Outward normal along a parametrized curve. If $\mathbf{r}(t) = \langle x(t), y(t)\rangle$ is positively oriented, then $\mathbf{n}\,ds = \langle y'(t), -x'(t)\rangle\,dt$, so the flux integral can also be written $\oint_C P\,dy - Q\,dx$.

Area from the boundary. Taking $\mathbf{F} = \langle -y, x\rangle$ makes the integrand $Q_x - P_y$ equal to $2$, so the double integral is twice the area. Hence $$ A = \frac{1}{2}\oint_C x\,dy - y\,dx $$ The choices $\mathbf{F} = \langle 0, x\rangle$ and $\mathbf{F} = \langle -y, 0\rangle$ give the one-term versions $A = \oint_C x\,dy$ and $A = -\oint_C y\,dx$.

Regions with holes. If $D$ lies between an outer curve and one or more inner curves, the theorem still holds with the boundary taken as all of them, each oriented to keep $D$ on the left. Cutting the region along channels that join the curves and cancelling the two passes along each cut is the standard argument.

Consequence for conservative fields. If $Q_x = P_y$ throughout a simply connected open region, every closed curve in it has zero circulation, so the field is conservative there.

Consequence for hard integrands. Any term in $P$ that depends only on $x$, or in $Q$ that depends only on $y$, contributes nothing to $Q_x - P_y$ and can be dropped before the double integral is set up.

Worked examples

Both sides on a square

Let $\mathbf{F} = \langle xy, \; x^2\rangle$ and let $C$ be the boundary of the unit square $0 \le x \le 1$, $0 \le y \le 1$, counterclockwise.

Double integral first. Here $Q_x = 2x$ and $P_y = x$, so the integrand is $2x - x = x$ and $$ \iint_D x\,dA = \int_0^1\!\!\int_0^1 x\,dx\,dy = \frac{1}{2} $$

Now the four edges. Along the bottom, $y = 0$ and $dy = 0$, so $P\,dx = xy\,dx = 0$ and the edge contributes $0$. Along the right edge, $x = 1$ and $dx = 0$, so the integrand is $Q\,dy = 1\,dy$ and the contribution is $1$. Along the top, $y = 1$ and $dy = 0$, so the integrand is $P\,dx = x\,dx$ with $x$ running from $1$ to $0$, contributing $-\tfrac12$. Along the left edge $x = 0$ and $dx = 0$, so $Q\,dy = 0$ and the contribution is $0$. The total is $0 + 1 - \tfrac12 + 0 = \tfrac12$, matching the double integral.

Turning an impossible line integral into an easy one

Evaluate $$ \oint_C \left(e^{x} - y^3\right)dx + \left(\cos y + x^3\right)dy $$ around the unit circle, counterclockwise.

Parametrizing would produce $e^{\cos t}$ and $\cos(\sin t)$ inside an integral, neither of which has an elementary antiderivative. Green’s theorem never sees them: $Q_x = 3x^2$ and $P_y = -3y^2$, so the integrand of the double integral is $3x^2 + 3y^2 = 3r^2$. In polar coordinates, $$ \iint_D 3r^2\,dA = \int_0^{2\pi}\!\!\int_0^1 3r^2 \cdot r\,dr\,d\theta = 2\pi \cdot \frac{3}{4} = \frac{3\pi}{2} $$ The two unpleasant terms vanished because $e^x$ does not depend on $y$ and $\cos y$ does not depend on $x$.

Area of an ellipse from its boundary

Use $A = \frac12 \oint_C x\,dy - y\,dx$ on the ellipse $\mathbf{r}(t) = \langle a\cos t, \; b\sin t\rangle$, $0 \le t \le 2\pi$, with $a, b > 0$.

Here $dx = -a\sin t\,dt$ and $dy = b\cos t\,dt$, so $$ x\,dy - y\,dx = \left(ab\cos^2 t + ab\sin^2 t\right)dt = ab\,dt $$ and $A = \frac12 \int_0^{2\pi} ab\,dt = \pi a b$. Setting $a = b = r$ returns $\pi r^2$, the right answer for a circle, which is a good check that the constant in the formula is $\frac12$ and not something else.

The flux form on a disk

Take $\mathbf{F} = \langle x, y\rangle$ and the unit circle again. Then $P_x + Q_y = 1 + 1 = 2$, so the flux form predicts $\iint_D 2\,dA = 2\pi$.

Compute the flux directly to confirm. On the unit circle the outward unit normal at $\langle \cos t, \sin t\rangle$ is $\langle \cos t, \sin t\rangle$ itself, and the field there is the same vector, so $\mathbf{F}\cdot\mathbf{n} = \cos^2 t + \sin^2 t = 1$. Then $\oint_C 1\,ds$ is the circumference, $2\pi$. The two agree, and the interpretation is that this field is spreading outward at a steady rate of $2$ units of area per unit area, everywhere at once.

Practice

These problems ask you to convert in both directions: replace a closed line integral by a double integral when the integrand is awkward, and replace a double integral by a boundary integral when the region is easier to walk than to fill. Watch the orientation before you write down a sign, and use the area formula where a region is given by a parametrized boundary.

Quiz

Five items on the circulation and flux forms, orientation conventions, and computing area from a boundary integral.