Green's Theorem

The statement

Let D be a region in the plane bounded by a simple, piecewise-smooth, positively oriented curve C = ∂D, and let F = ⟨P(x,y), Q(x,y)⟩ have continuous first partials on an open set containing D. Then

∮_C (P dx + Q dy)  =  ∬_D ( ∂Q/∂x − ∂P/∂y ) dA.

The left side is a line integral — the work, or circulation, of F around the boundary. The right side is a double integral of the (scalar) 2D curl over the interior. Green's theorem is the slogan "boundary captures interior" written down in two dimensions.

Sketch the picture: the region D is some blob in the plane, ∂D is its boundary curve, an arrow on the curve points counterclockwise. Inside D, imagine a fine grid of tiny squares. Each square contributes a small circulation roughly equal to (∂Q/∂x − ∂P/∂y) · (area). When you sum over the grid, every interior edge is traversed twice in opposite directions and cancels, leaving only the outer boundary — that is the line integral. Telescoping over a 2D grid is the heart of every proof of Green's.

Worked example — verifying on the unit disk

Take F = ⟨−y, x⟩, with P = −y and Q = x. Let D be the closed unit disk and C the unit circle x = cos θ, y = sin θ for 0 ≤ θ ≤ 2π.

Line integral. dx = −sin θ dθ, dy = cos θ dθ:

∮_C (P dx + Q dy) = ∫₀^{2π} (−sin θ)(−sin θ) dθ + (cos θ)(cos θ) dθ
                  = ∫₀^{2π} (sin²θ + cos²θ) dθ
                  = ∫₀^{2π} 1 dθ  =  2π.

Double integral. ∂Q/∂x − ∂P/∂y = 1 − (−1) = 2. Integrating the constant 2 over the unit disk gives 2 · π = 2π. The two sides agree — Green's theorem checks out.

Worked example — area by line integral

Pick P = 0, Q = x. Then ∂Q/∂x − ∂P/∂y = 1, so

Area(D) = ∬_D 1 dA = ∮_{∂D} x dy.

For an ellipse parametrised by x = a cos θ, y = b sin θ, dy = b cos θ dθ:

Area = ∮ x dy = ∫₀^{2π} (a cos θ)(b cos θ) dθ = ab ∫₀^{2π} cos²θ dθ = πab.

The classical πab formula for the ellipse falls out for free. The symmetric form ½∮(x dy − y dx) is what mechanical planimeters compute by tracing a closed curve on a map; it is also the routine surveyors use to find the acreage of a polygon from its corner coordinates.

Worked example — conservative test

Take F = ⟨2xy + 3, x² − 4y⟩. Is the line integral ∫_C F·dr path-independent?

∂Q/∂x = 2x,    ∂P/∂y = 2x.    Their difference: 0.

The 2D curl vanishes, so by Green's theorem the loop integral is zero around any simple closed curve, and the field is conservative on the whole plane. We can find its potential — φ(x, y) = x²y + 3x − 2y² satisfies ∇φ = F. The line integral from A to B is therefore φ(B) − φ(A).

Green's vs Stokes' vs the divergence theorem

Green's theorem is the planar instance of two more general theorems. Side by side:

	Green's (circulation form)	Stokes'	Divergence (Gauss)
Setting	Plane region D ⊂ ℝ²	Surface S ⊂ ℝ³	Solid V ⊂ ℝ³
Boundary	Closed curve ∂D	Closed curve ∂S	Closed surface ∂V
Boundary integral	∮_∂D (P dx + Q dy)	∮_∂S F·dr	∬_∂V F·dS
Interior integrand	∂Q/∂x − ∂P/∂y	(∇×F)·n̂	∇·F
Operator on F	2D curl (scalar)	3D curl (vector)	Divergence (scalar)
Dimension lost	1 (1D ← 2D)	1 (1D ← 2D)	1 (2D ← 3D)
Special case	Stokes' on flat surface	Generalises Green's	3D analogue of Green's flux form

Green's flux form, ∮ F·n̂ ds = ∬ (∂P/∂x + ∂Q/∂y) dA, sits between Green's circulation form and the divergence theorem — it is the 2D divergence theorem. All three identities are instances of the general Stokes' theorem ∫_∂Ω ω = ∫_Ω dω from differential geometry, with ω a differential form of the appropriate degree.

Why Green's theorem is true

The cleanest proof handles a region of the form a ≤ x ≤ b, g(x) ≤ y ≤ h(x). Compute

∬_D −∂P/∂y dA = ∫_a^b ∫_{g(x)}^{h(x)} −∂P/∂y dy dx
              = ∫_a^b [P(x, g(x)) − P(x, h(x))] dx
              = ∮_C P dx,

where the last step recognises the integrals on top and bottom as the contributions of the lower and upper boundary curves. The same argument with x and y swapped gives ∬ ∂Q/∂x dA = ∮ Q dy. Add the two halves and you have the theorem. A general region is decomposed into pieces of this form, and the interior cancellations leave only the outermost boundary.

Where Green's theorem appears

• Planimeters. A mechanical instrument for measuring the area of an irregular planar region — a draftsman's tool with a tracer arm and an integrating wheel. The wheel computes ∮ x dy as you trace the boundary, and Green's theorem turns the loop integral into the enclosed area. Patented in 1854; still in use for blueprint takeoff.
• CAD polygon area. The shoelace formula, Area = ½|Σ (x_i y_{i+1} − x_{i+1} y_i)|, is Green's theorem applied to a polygon. Used in every GIS package and in 3D graphics for triangle-orientation tests.
• Circulation in fluid flow. A flow with velocity v has circulation around a closed curve C equal to ∮ v·dr = ∬ (∇×v)·dA over any disk spanning C. Lift on a 2D airfoil obeys Kutta–Joukowski: L = ρ v_∞ Γ, where Γ is the circulation around the wing.
• Cauchy's theorem in complex analysis. ∮ f(z) dz = 0 for f holomorphic on a simply-connected domain — an immediate corollary, since the Cauchy–Riemann equations make the relevant 2D curls vanish.
• Test for conservative fields. Compute ∂Q/∂x − ∂P/∂y. If it is zero on a simply-connected domain, the field has a potential and ∫ F·dr is path-independent.
• Pick's theorem. The classic result that lattice-polygon area equals (interior lattice points) + ½(boundary lattice points) − 1 has a Green-theorem proof using lattice-edge increments.

Holes — multiply connected regions

Suppose D is the annulus between two concentric circles. The boundary now has two pieces: the outer circle (counterclockwise) and the inner circle (clockwise — so D stays on your left). Green's theorem becomes

∮_{C_outer} F·dr  −  ∮_{C_inner} F·dr  =  ∬_D (∂Q/∂x − ∂P/∂y) dA,

where both line integrals on the left are computed counterclockwise. If F is curl-free on the annulus, the two loop integrals are equal. That is exactly how complex analysis proves the residue theorem — a closed loop's integral depends only on which singularities it surrounds, not its shape.

Common mistakes

• Wrong orientation. Green's theorem requires the boundary to be traversed counterclockwise. If you parametrise it clockwise, the line integral flips sign — and your answer disagrees with the double integral by a minus. Always pick the parametrisation so that the region is on the left.
• Applying it across a singularity. If P or Q is undefined or non-smooth somewhere inside D, the theorem does not apply. F = ⟨−y, x⟩/(x²+y²) is the textbook trap: curl looks zero where the formula is defined, but the origin punctures the region and the loop integral is 2π, not 0.
• Confusing curl form with flux form. ∮ P dx + Q dy uses the tangent (it computes circulation). ∮ F·n̂ ds uses the normal (it computes flux). Both are special cases of Green's, with different signs and different right-hand sides (curl vs divergence). Pick one and stay with it.
• Missing the simply-connected hypothesis. ∂Q/∂x = ∂P/∂y on a domain with a hole does not imply F is conservative there. You need both the equation and the topology.
• Sign error in the right-hand side. The integrand is ∂Q/∂x − ∂P/∂y, in that order. Swapping gives the negative of the correct answer. Mnemonic: x derivatives go with x partners (Q couples to ∂/∂x).
• Trying it in 3D. Green's lives strictly in the plane. The 3D analogue is Stokes' theorem, where the integrand becomes a vector curl dotted with a surface normal. Embedding a 2D problem in 3D as ⟨P, Q, 0⟩ on a flat surface recovers Green's, but going the other way is not automatic.