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.
What a learner can do afterwards
- 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
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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.
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.
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.
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.
Additional readings and questions written for Marble. Linked videos: Khan Academy, Green's theorem example 1 and Green's theorem example 2. Flashcards use standard vector-calculus facts.
More to read
Choosing the region before calculating
A boundary integral and a double integral need different descriptions of the same region. Before using Green's theorem, identify the enclosed set D and every curve that belongs to its boundary. For one simple closed boundary, traverse it once counterclockwise.
The field also needs attention. Its components P and Q must have continuous first partial derivatives on an open neighborhood of D and its boundary. Knowing the field only along the curve is insufficient. An undefined interior point prevents direct application to the filled region.
Next compute Q_x - P_y, where the subscripts indicate partial differentiation. Green's circulation formula replaces ∮(P dx + Q dy) with the double integral of that expression over D. A constant result can be multiplied by area. A varying result must be integrated using limits that describe D.
The same relation can recover area. For positive orientation, ∮ x dy equals area because P = 0 and Q = x give derivative difference 1. The expression ∮(x dy - y dx) equals twice the area. If the boundary was traversed in the reverse direction, negate the given integral before using either area formula.
What cancels when regions are joined
Draw a cut that divides a region into two adjacent pieces. Give each piece positive boundary orientation. Along the cut, the two traversals have opposite directions. For the same field, their line integrals are negatives of one another. Adding the two boundary integrals removes the cut's contribution and leaves the complete boundary of the joined region.
This explains why holes require extra boundary terms rather than a blanket rejection of Green's theorem. For a region with finitely many holes, positive orientation keeps the region on the left along every component. The outer boundary runs counterclockwise; the boundaries around holes run clockwise. The field condition must hold near the region and all its boundary components. It need not hold inside an excluded hole.
For an annulus, an inner-circle integral calculated counterclockwise has the reverse of the required direction. Subtract it from the counterclockwise outer-circle integral to obtain the positive complete-boundary integral.
Cancellation also limits what one answer tells you. A zero integral around one loop can result from positive and negative interior contributions. It does not establish that the field is conservative. Conservativeness requires path independence throughout the domain, or equivalently zero circulation around every closed path there.
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