Green’s Theorem

Green’s theorem turns a line integral around a closed curve C into a double integral over the region D inside it: ∮_C P dx + Q dy = ∬_D (∂Q/∂x − ∂P/∂y) dA. The curve must be traced counterclockwise, with D on the left. It often replaces several line integrals with one easy double integral, and it gives a formula for area.

Before this lesson: Line Integrals, Conservative Vector Fields and Double Integrals over General Regions

What Green’s theorem says

Think of F = ⟨P, Q⟩ as the velocity of a fluid. The line integral ∮_C F · dr measures the circulation: how much the fluid flows along the boundary. The quantity Q_x − P_y measures how much the fluid spins at each point inside.

Green’s theorem says the total spin inside D equals the circulation around its edge. Inside D, the spins of neighboring tiny squares cancel along their shared edges, so only the outer boundary is left.

Green’s theorem formula

\oint_C P\,dx + Q\,dy = \iint_D \Big(\dfrac{\partial Q}{\partial x} - \dfrac{\partial P}{\partial y}\Big)\,dA

The conditions: C is a simple closed curve (it doesn’t cross itself), it is piecewise smooth, it is traced counterclockwise (positive orientation), and P and Q have continuous partial derivatives on an open region containing D.

Area with Green’s theorem

\text{Area of } D = \oint_C x\,dy = -\oint_C y\,dx = \dfrac{1}{2}\oint_C x\,dy - y\,dx

1. Check that C is closed and traced counterclockwise.
2. Identify P (the coefficient of dx) and Q (the coefficient of dy).
3. Compute Q_x − P_y.
4. Set up the double integral over D, using polar coordinates if D is round, and evaluate.

Green’s theorem examples

Example 1: A rectangle

Evaluate \oint_C (x^2 - y)\,dx + (x + y^2)\,dy, where C is the boundary of the rectangle [0, 3] × [0, 2], counterclockwise.

P = x^2 - y, \quad Q = x + y^2

Q_x = 1, \quad P_y = -1

Q_x - P_y = 1 - (-1) = 2

\oint_C P\,dx + Q\,dy = \iint_D 2\,dA

= 2 \cdot (3 \cdot 2) = 12

Answer: 12. Doing this directly would take four separate line integrals, one per side.

Example 2: A triangle

Evaluate \oint_C xy\,dx + x^2\,dy, where C is the triangle with vertices (0, 0), (1, 0) and (1, 2), counterclockwise.

Q_x - P_y = 2x - x = x

0 \le x \le 1, \quad 0 \le y \le 2x

\iint_D x\,dA = \int_0^1 \int_0^{2x} x\,dy\,dx

\int_0^{2x} x\,dy = 2x^2

\int_0^1 2x^2\,dx = \dfrac{2}{3}

Answer: 2/3. As a check, the three line integrals along the sides give 0, 2 and −4/3, which also add to 2/3.

Example 3: A circle

Evaluate \oint_C -y^3\,dx + x^3\,dy, where C is the circle x² + y² = 4, counterclockwise.

Q_x - P_y = 3x^2 - (-3y^2) = 3x^2 + 3y^2

= 3r^2

\iint_D 3r^2\,dA = \int_0^{2\pi} \int_0^2 3r^2 \cdot r\,dr\,d\theta

\int_0^2 3r^3\,dr = \Big[\dfrac{3r^4}{4}\Big]_0^2 = 12

2\pi \cdot 12 = 24\pi

Answer: 24π ≈ 75.40.

Example 4: Area of an ellipse

Use Green’s theorem to find the area inside the ellipse x²/16 + y²/9 = 1.

x = 4\cos t, \quad y = 3\sin t, \quad 0 \le t \le 2\pi

dx = -4\sin t\,dt, \quad dy = 3\cos t\,dt

x\,dy - y\,dx = 12\cos^2 t\,dt + 12\sin^2 t\,dt

= 12\,dt

A = \dfrac{1}{2}\int_0^{2\pi} 12\,dt = 12\pi

Answer: 12π ≈ 37.70, which matches πab = π(4)(3).

Example 5: Work done on a robot

A garden robot drives once counterclockwise around the edge of a square plot with corners (0, 0), (4, 0), (4, 4) and (0, 4), in meters. Along the way, a force F = ⟨x − y², x² + y⟩ newtons acts on it. Find the work done by F.

Q_x - P_y = 2x - (-2y) = 2x + 2y

W = \int_0^4 \int_0^4 (2x + 2y)\,dy\,dx

\int_0^4 (2x + 2y)\,dy = 8x + 16

\int_0^4 (8x + 16)\,dx = 64 + 64 = 128

Answer: the force does 128 J of work. The field is not conservative (Q_x − P_y is not 0), so a closed trip can do nonzero work.

Common mistakes

Related methods

Conservative fields

If Q_x = P_y everywhere on D, Green’s theorem gives ∮_C F · dr = 0. That matches the test for conservative vector fields.

Regions with holes

For a ring-shaped region, C has two parts: the outer curve counterclockwise and the inner curve clockwise, so D is always on the left. Green’s theorem then uses both parts.

Curl and flux forms

The integrand Q_x − P_y is the k-component of the curl of F, the topic of the next lesson. Green’s theorem also has a flux form that uses the divergence P_x + Q_y.

Practice problems

1. Evaluate \oint_C y\,dx - x\,dy around the unit circle, counterclockwise.
2. Evaluate \oint_C 3y\,dx + 5x\,dy around the square [0, 2] × [0, 2], counterclockwise.
3. Evaluate \oint_C x^2y\,dx + xy^2\,dy around the rectangle [0, 2] × [0, 1], counterclockwise.
4. Use Green’s theorem to find the area of a disk of radius 5.
5. Use Area = ∮_C x dy to find the area of the triangle with vertices (0, 0), (4, 0) and (0, 3).
6. Evaluate \oint_C (e^x + y)\,dx + (3x + \cos y)\,dy around the circle x² + y² = 4, counterclockwise.
7. Evaluate \oint_C -y\,dx + x\,dy around the square [0, 1] × [0, 1], traced clockwise.
8. Show that \oint_C 2xy\,dx + x^2\,dy = 0 for every simple closed curve C.
9. Evaluate \oint_C -x^2y\,dx + xy^2\,dy around the circle x² + y² = 9, counterclockwise.
10. A storm’s wind field is F = ⟨−2y, 3x⟩ km/h, with x and y in km. Find the circulation ∮_C F · dr around the circle of radius 10 km centered at the origin, counterclockwise.

Q_x - P_y = -1 - 1 = -2

\iint_D (-2)\,dA = -2\pi

Q_x - P_y = 5 - 3 = 2

2 \cdot 4 = 8

Q_x - P_y = y^2 - x^2

\int_0^2 \int_0^1 y^2\,dy\,dx = 2 \cdot \dfrac{1}{3} = \dfrac{2}{3}

\int_0^2 \int_0^1 x^2\,dy\,dx = \dfrac{8}{3} \cdot 1 = \dfrac{8}{3}

\dfrac{2}{3} - \dfrac{8}{3} = -2

A = \dfrac{1}{2}\int_0^{2\pi} (25\cos^2 t + 25\sin^2 t)\,dt

= \dfrac{1}{2} \cdot 25 \cdot 2\pi = 25\pi

\text{Bottom: } dy = 0

\text{Slanted side: } \int_0^1 (4 - 4t)(3\,dt) = 12 - 6 = 6

\text{Left side: } x = 0

Q_x - P_y = 3 - 1 = 2

2 \cdot \pi(2)^2 = 8\pi

Q_x - P_y = 1 - (-1) = 2

-\iint_D 2\,dA = -2

Q_x - P_y = 2x - 2x = 0

\iint_D 0\,dA = 0

Q_x - P_y = y^2 - (-x^2) = x^2 + y^2 = r^2

\int_0^{2\pi} \int_0^3 r^2 \cdot r\,dr\,d\theta = 2\pi \cdot \dfrac{81}{4}

= \dfrac{81\pi}{2}

Q_x - P_y = 3 - (-2) = 5

5 \cdot \pi(10)^2 = 500\pi \approx 1570.8

Green’s theorem FAQ

When can you use Green’s theorem?

Use it for a line integral around a simple closed curve in the plane, traced counterclockwise, when P and Q have continuous partial derivatives on the region inside.

What does positive orientation mean?

It means counterclockwise for the outer boundary: as you walk along C, the region D stays on your left. Inner boundaries of a region with holes go clockwise.

How do you find area with Green’s theorem?

Use Area = ½ ∮_C x dy − y dx, or ∮_C x dy alone. Parametrize the boundary and evaluate the line integral.

What is the difference between Green’s theorem and Stokes’ theorem?

Green’s theorem is for flat regions in the xy-plane. Stokes’ theorem extends the same idea to curved surfaces in space.

Summary

• Green’s theorem: ∮_C P dx + Q dy = ∬_D (Q_x − P_y) dA.
• C must be simple, closed and counterclockwise, with D on the left.
• It turns hard line integrals into easier double integrals, and the reverse.
• Area = ½ ∮ x dy − y dx.
• If Q_x = P_y on D, the integral around C is 0.

Previous lesson: Conservative Vector Fields

Next lesson: Curl and Divergence

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