Green's Theorem

Quick answer

Green's theorem connects the circulation of a field around a closed curve with what happens inside it: ∮ P dx + Q dy equals the double integral of ∂Q/∂x − ∂P/∂y over the enclosed region, when the curve is traversed counterclockwise. The integrand measures circulation per unit area at each point, and the circulation of small squares adds up to the circulation around the boundary. The theorem turns hard line integrals into routine double integrals and gives area formulas.

What you'll learn

State Green's theorem with the correct orientation
Explain why circulation around small squares adds up
Evaluate line integrals as double integrals and back
Compute areas and handle regions with holes

Circulation around the edge, swirl inside

A line integral around a closed curve measures circulation: how much a field runs along the curve. Green’s theorem says the same number can be found from the inside of the curve, by adding up how much the field swirls at each point.

Green’s theorem. Let $C$ be a simple closed curve, traversed counterclockwise, that encloses a region $D$ . If $P$ and $Q$ have continuous partial derivatives on $D$ and $C$ , then

$\displaystyle\oint_C P\,dx + Q\,dy = \iint_D \left(\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}\right) dA$

The counterclockwise direction is called positive orientation: walking along $C$ , the region stays on your left. The expression $\tfrac{\partial Q}{\partial x} - \tfrac{\partial P}{\partial y}$ is the circulation density, or scalar curl, of $\mathbf{F} = (P, Q)$ .

The triangle of Example 1 with its positive orientation

Why the small circulations add up

Start with a tiny rectangle with corners $(x, y)$ and $(x + \Delta x, y + \Delta y)$ , and go around it counterclockwise. The bottom edge contributes about $P(x, y)\,\Delta x$ and the top edge, traversed leftward, about $-P(x, y + \Delta y)\,\Delta x$ . Together they give about $-\tfrac{\partial P}{\partial y}\,\Delta y\,\Delta x$ . The right and left edges give $\tfrac{\partial Q}{\partial x}\,\Delta x\,\Delta y$ in the same way. So the circulation around the tiny rectangle is about

\left(\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}\right)\Delta A

Now cut $D$ into many such rectangles and add their circulations. Every inside edge is shared by two neighbors and traversed once in each direction, so its two contributions cancel. Only the outer edges survive, and they make up the boundary $C$ . The sum of the small circulations is a Riemann sum for the double integral, and what survives is the line integral around $C$ .

Worked examples

Example 3 — Area from the boundary

Show that the area of $D$ is $\tfrac{1}{2}\oint_C x\,dy - y\,dx$ , and use it to find the area of the ellipse $\tfrac{x^2}{a^2} + \tfrac{y^2}{b^2} = 1$ .

With $P = -\tfrac{y}{2}$ and $Q = \tfrac{x}{2}$ , the curl is $\tfrac{1}{2} + \tfrac{1}{2} = 1$ , so Green’s theorem gives $\iint_D 1\,dA$ , the area. For the ellipse, $\mathbf{r}(t) = (a\cos t,\ b\sin t)$ :

\frac{1}{2}\int_0^{2\pi} \big(a\cos t \cdot b\cos t - b\sin t \cdot (-a\sin t)\big)\,dt = \frac{1}{2}\int_0^{2\pi} ab\,dt = \pi ab

Surveyors’ planimeters measure areas on maps this way, by tracing only the boundary.

Example 5 — A ring

Let $D$ be the ring $1 \le x^2 + y^2 \le 4$ and $\mathbf{F} = (-y^3,\ x^3)$ . Evaluate the integral around the boundary of $D$ with positive orientation.

The boundary has two circles. Keeping $D$ on the left means going counterclockwise around the outer circle and clockwise around the inner one. Green’s theorem holds for this region, with both pieces included. The curl is $3x^2 + 3y^2 = 3r^2$ , so in polar coordinates

\iint_D 3r^2\,dA = \int_0^{2\pi} \int_1^2 3r^3\,dr\,d\theta = 2\pi \cdot \frac{3(16 - 1)}{4} = \frac{45\pi}{2}

Separately, the outer circle counterclockwise gives $24\pi$ and the inner circle counterclockwise gives $\tfrac{3\pi}{2}$ ; the difference is $\tfrac{45\pi}{2}$ , as it should be.

A ring and its positively oriented boundary

Common mistakes

Practice problems

Use Green’s theorem to evaluate $\oint_C y\,dx - x\,dy$ around the unit circle, counterclockwise.

Answer

$-2\pi$

Full solution

$\tfrac{\partial Q}{\partial x} - \tfrac{\partial P}{\partial y} = -1 - 1 = -2$ , and $-2$ times the area $\pi$ is $-2\pi$ .
Evaluate $\oint_C xy\,dx + x^2\,dy$ , where $C$ is the boundary of the unit square $[0, 1] \times [0, 1]$ , counterclockwise.

Answer

$\tfrac{1}{2}$

Full solution

The curl is $2x - x = x$ , and $\int_0^1 \int_0^1 x\,dy\,dx = \tfrac{1}{2}$ .
Evaluate $\oint_C (x^2 + y)\,dx + (3x + y^2)\,dy$ around the circle $x^2 + y^2 = 4$ , counterclockwise.

Answer

$8\pi$

Full solution

The curl is $3 - 1 = 2$ , and $2$ times the area $4\pi$ is $8\pi$ .
Evaluate $\oint_C -y^2\,dx + xy\,dy$ around the rectangle $[0, 2] \times [0, 3]$ , counterclockwise.

Answer

$27$

Full solution

The curl is $y - (-2y) = 3y$ , and $\int_0^2 \int_0^3 3y\,dy\,dx = 2 \cdot \tfrac{27}{2} = 27$ .
Use the area formula to find the area of the ellipse $\tfrac{x^2}{9} + \tfrac{y^2}{4} = 1$ .

Answer

$6\pi$

Full solution

Example 3 with $a = 3$ and $b = 2$ gives $\pi \cdot 3 \cdot 2$ .
Show that $\oint_C x\,dy$ and $-\oint_C y\,dx$ also equal the area enclosed by $C$ , and check the first on the circle of radius $2$ .

Answer

Both fields have curl $1$ ; on the circle, $\oint_C x\,dy = 4\pi$ .

Full solution

For $(0, x)$ the curl is $1 - 0$ , and for $(-y, 0)$ it is $0 - (-1)$ . On $\mathbf{r}(t) = (2\cos t, 2\sin t)$ , $\oint x\,dy = \int_0^{2\pi} 4\cos^2 t\,dt = 4\pi$ , the area of the disk.
Explain with Green’s theorem why a field with $\tfrac{\partial Q}{\partial x} = \tfrac{\partial P}{\partial y}$ on the whole plane has zero circulation around every simple closed curve.

Answer

The double integral of a zero integrand is zero.

Full solution

The whole plane has no holes, so every simple closed curve encloses a region where the field is defined and the curl is $0$ . Green’s theorem gives $\iint_D 0\,dA = 0$ . This is the reason the test for conservative fields works.
A student evaluates $\oint_C x^4\,dx + xy\,dy$ around the triangle of Example 1 traversed clockwise and gets $\tfrac{1}{6}$ . Correct the answer.

Answer

$-\tfrac{1}{6}$

Full solution

Green’s theorem gives $\tfrac{1}{6}$ for the counterclockwise direction. Reversing the direction of a line integral changes its sign.

Frequently asked questions

What does Green's theorem say?

If C is a simple closed curve traversed counterclockwise, enclosing the region D, then the line integral of P dx + Q dy around C equals the double integral over D of (∂Q/∂x − ∂P/∂y).

What does ∂Q/∂x − ∂P/∂y measure?

The circulation of the field per unit area at a point: how strongly the field swirls counterclockwise there. It is the scalar curl of the field in the plane.

What is positive orientation?

Traversing the boundary so that the region stays on the left. For a single closed curve that is counterclockwise; for the inner boundary of a region with a hole it is clockwise.

How do you find an area with Green's theorem?

Choose a field with ∂Q/∂x − ∂P/∂y = 1, such as (−y/2, x/2). Then the area of D equals ½∮(x dy − y dx) around its boundary.

How is Green's theorem related to conservative fields?

If ∂Q/∂x = ∂P/∂y throughout a region without holes, Green's theorem makes every loop integral zero, which is why the test for conservative fields works.

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