--- title: "Green's Theorem in the Plane" description: "Relate circulation around a positively oriented plane boundary to a double integral over its region. Check the boundary and field conditions, then compute circulation or area." canonical: https://lightmysky.com/learn/mathematics/greens-theorem-in-the-plane-mt_UDgivmPVEJ source: https://lightmysky.com/learn/mathematics/greens-theorem-in-the-plane-mt_UDgivmPVEJ.md retrieved: 2026-11-12 --- > **Agent view.** This is the Markdown twin of the page, for tools and assistants. > When to use this site, and the call that answers each job: https://lightmysky.com/agent-instructions.md > API description (OpenAPI 3.1): https://lightmysky.com/openapi.json · Authentication: https://lightmysky.com/auth.md > Pricing: https://lightmysky.com/pricing.md · Catalog: https://lightmysky.com/llms.txt · Full catalog: https://lightmysky.com/llms-full.txt > Every machine-readable file on this domain: https://lightmysky.com/.well-known/ai-catalog.json > Ask for Markdown with `Accept: text/markdown`, a `.md` address, or `?mode=agent`. # Green's Theorem in the Plane Relate circulation around a positively oriented plane boundary to a double integral over its region. Check the boundary and field conditions, then compute circulation or area. Subject: Mathematics · Area: Calculus & Analysis · Ages 20 to 21 Page: https://lightmysky.com/learn/mathematics/greens-theorem-in-the-plane-mt_UDgivmPVEJ ## Ready when they can - State the theorem with the orientation convention for the boundary - Convert a circulation integral into a double integral and evaluate it - Use the theorem to compute an area from a boundary integral - Use shared-edge cancellation to combine positively oriented boundary circulations of adjacent regions - Explain why zero circulation around one closed curve does not establish conservativeness on the whole domain ## Lesson: Trading a rim walk for the inside Let F = (P, Q). Let C be a simple closed, piecewise smooth plane curve traversed once counterclockwise, enclosing D. P and Q must have continuous first partial derivatives on an open neighborhood of D and C. Then ∮C(P dx + Q dy) = ∬D(Q_x - P_y) dA. Here Q_x = ∂Q/∂x and P_y = ∂P/∂y. The line integral is circulation: it accumulates the tangential component of the field along C. **Example.** If Q_x - P_y is a constant k, its double integral is k times the region's area. For F = (-3y, 2x), Q_x = 2 and P_y = -3, so k = 5. The rectangle 0 <= x <= 2, 0 <= y <= 1 has area 2. Its counterclockwise boundary circulation is 5 × 2 = 10. When the derivative difference varies, keep it inside the double integral. Positive orientation keeps the region on your left. For a single simple boundary, it is counterclockwise. Reversing a curve negates its integral. Green's theorem also allows finitely many holes, with piecewise smooth boundaries and the same derivative condition near the region and its complete boundary. Traverse the outer boundary counterclockwise and each hole boundary clockwise. If all were calculated counterclockwise, subtract the inner integrals from the outer integral. For a positively oriented complete boundary, choose P = -y/2, Q = x/2. Then Q_x - P_y = 1, so area = (1/2)∮(x dy - y dx). Choosing P = 0, Q = x also gives area = ∮ x dy. If Q_x - P_y = k is a nonzero constant, area = circulation/k. These formulas use positive orientation; negate the integral first if the whole boundary is reversed. Split a region into two adjacent pieces and positively orient each piece's boundary. Their shared edge is traversed in opposite directions. Its two integrals are I and -I, so they cancel. Adding the two piece-boundary circulations leaves the complete boundary integral of their union. If the union has no hole, this remaining boundary is its exterior boundary. **Example.** For a <= x <= b and g(x) <= y <= h(x), integrate Q_x - P_y in y from g(x) to h(x), then in x from a to b. For F = (xy, 2x²), the derivative difference is 4x - x = 3x. On 0 <= x <= 1, 0 <= y <= 2x, the inner integral is 6x². The outer integral is [2x³] from 0 to 1 = 2. This is the positive boundary circulation. Finite field values along the boundary are not enough. The derivative condition covers the whole region and its boundary. If the field is undefined or blows up at an included interior point, you cannot apply Green's theorem directly to the filled region. Excluding a small region around that point creates an inner boundary whose integral must also be included. From the previous topic, a conservative field has zero circulation around every closed path. With continuous first partial derivatives, Q_x = P_y, so Green's integrand is zero. One zero circulation does not prove conservativeness: positive and negative contributions can cancel. For F = (0, x²) on -1 <= x <= 1, 0 <= y <= 1, the integrand 2x integrates to zero, although this field is not conservative on the plane. **Recap.** Check the field conditions and complete boundary. With positive orientation, ∮(P dx + Q dy) = ∬(Q_x - P_y) dA. A constant integrand multiplies area; reversing the boundary changes the sign. *Lesson and assessment revisions written for Marble. Calculus references were consulted for mathematical facts and teaching structure.* ## Practice 14 questions on this page, each with its working shown. ## Needs first - [Conservative Fields and Path Independence](https://lightmysky.com/learn/mathematics/conservative-fields-and-path-independence-mt_5djKpI5F33) ## Optional connections - [Double Integrals over General Regions](https://lightmysky.com/learn/mathematics/double-integrals-over-general-regions-mt_UKS9_AhAWw) ## Opens up - [Cauchy's Theorem and Deforming a Contour](https://lightmysky.com/learn/mathematics/cauchys-theorem-and-deforming-a-contour-mt_aacYFVN-HH) - [Divergence and Curl](https://lightmysky.com/learn/mathematics/divergence-and-curl-mt_JbjW8WKOPE)