📐 Concept diagram

07-06 — Green's Theorem

Phase: 7 — Calculus IV: Vector Calculus Subject: 07-06 Prerequisites: 07-05 — Line Integrals, double integrals (Phase 6) Next subject: 07-07 — Curl and Divergence

Learning Objectives

By the end of this subject, you will be able to:

1. State Green's theorem in both circulation-curl and flux-divergence forms
2. Apply Green's theorem to convert line integrals over closed curves to double integrals, and vice versa
3. Verify that Green's theorem holds for a given vector field and region
4. Use Green's theorem to compute areas of planar regions as line integrals
5. Recognize when Green's theorem applies (positively oriented, piecewise-smooth closed curve, simply-connected region, C¹ field)

Core Content

1. Statement of Green's Theorem

⚠️ CRITICAL FOUNDATION: Green's Theorem ∮_C P dx + Q dy = ∬_D (∂Q/∂x - ∂P/∂y) dA converts a line integral around a closed curve into a double integral over the enclosed region. It is the 2D precursor to Stokes' and Divergence theorems.

Let C be a positively oriented, piecewise-smooth, simple closed curve in the plane, and let D be the region bounded by C. Let F(x, y) = ⟨P(x, y), Q(x, y)⟩ be a vector field with continuous partial derivatives on an open region containing D. Then:

Circulation-Curl Form:

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

Flux-Divergence Form (using outward normal n rather than tangent):

$∮_C F · n ds = ∬_D (∂P/∂x + ∂Q/∂y) dA = ∬_D div F dA
$

The first form is most common in calculus texts. The two forms are related by rotating the vector field 90° (replace ⟨P, Q⟩ by ⟨−Q, P⟩ or ⟨Q, −P⟩).

Orientation: "Positively oriented" means C is traversed counterclockwise, keeping the interior D on the left.

2. Circulation-Curl Form — Intuition and Examples

Green's theorem says: the circulation of F around the boundary C equals the integral of the curl (2D curl, ∂Q/∂x − ∂P/∂y) over the interior D.

Intuition: Circulation around the boundary is the sum of all the "microscopic circulation" (vorticity) within the region.

Example 1 — Verifying Green's theorem: Verify the theorem for F(x, y) = ⟨x², xy⟩ and C the boundary of the square [0, 1] × [0, 1].

Line integral (counterclockwise):
C₁ (bottom): y=0, x: 0→1. dy=0. ∫₀¹ x² dx = 1/3.
C₂ (right): x=1, y: 0→1. dx=0. ∫₀¹ 1·y dy = [y²/2]₀¹ = 1/2.
C₃ (top): y=1, x: 1→0. dy=0. ∫₁⁰ x² dx = −1/3.
C₄ (left): x=0, y: 1→0. dx=0. ∫₁⁰ 0·y dy = 0.

∮ F·dr = 1/3 + 1/2 − 1/3 + 0 = 1/2.

Green's theorem:
∂Q/∂x − ∂P/∂y = ∂(xy)/∂x − ∂(x²)/∂y = y − 0 = y.

∬_D y dA = ∫₀¹ ∫₀¹ y dy dx = ∫₀¹ [y²/2]₀¹ dx = ∫₀¹ 1/2 dx = 1/2. ✓

Example 2 — Using Green's to simplify: Evaluate ∮_C (3y + e^{√x}) dx + (5x + sin(y²)) dy where C is the circle x² + y² = 4.

Direct parametrization would be horrific (try integrating e^{√(2 cos t)}).

Use Green's theorem:
P = 3y + e^{√x}, Q = 5x + sin(y²).
∂Q/∂x = 5, ∂P/∂y = 3.
∂Q/∂x − ∂P/∂y = 2.

∮_C P dx + Q dy = ∬_D 2 dA = 2 · Area(D) = 2 · π(2)² = 8π.

The scary parts (e^{√x} and sin(y²)) disappeared! This is the power of Green's theorem.

3. Computing Area with Green's Theorem

Choose P and Q such that ∂Q/∂x − ∂P/∂y = 1. Then:

$Area(D) = ∬_D 1 dA = ∮_C P dx + Q dy
$

Three common choices:

1. P = 0, Q = x: Area = ∮_C x dy
2. P = −y, Q = 0: Area = ∮_C −y dx = −∮_C y dx
3. P = −y/2, Q = x/2: Area = (1/2) ∮_C (x dy − y dx) (most symmetric)

Example 3: Use Green's theorem to find the area of the ellipse x²/a² + y²/b² = 1.

$Parametrize the ellipse: x = a cos t, y = b sin t, t ∈ [0, 2π].

Use form 3: Area = (1/2) ∮_C (x dy − y dx).

x dy − y dx = (a cos t)(b cos t dt) − (b sin t)(−a sin t dt)
            = ab cos²t dt + ab sin²t dt = ab(cos²t + sin²t) dt = ab dt.

Area = (1/2) ∫₀^{2π} ab dt = (1/2) · ab · 2π = πab.
$

This is the familiar formula! Derived cleanly via Green's theorem.

Example 4 — Area of a polygon: Use form 1 (Area = ∮_C x dy) to find the area of the triangle with vertices (0,0), (2,0), (1,3).

$Break into three segments, oriented counterclockwise:
C₁: (0,0)→(2,0). y=0 → dy=0. ∫_C₁ x dy = 0.
C₂: (2,0)→(1,3). Parametrize: r(t) = ⟨2−t, 3t⟩, t∈[0,1]. y=3t, dy=3dt. x=2−t.
    ∫₀¹ (2−t)·3 dt = ∫₀¹ (6−3t) dt = [6t − 3t²/2]₀¹ = 6 − 3/2 = 9/2.
C₃: (1,3)→(0,0). r(t) = ⟨1−t, 3(1−t)⟩, t∈[0,1]. x=1−t, y=3−3t, dy=−3dt.
    ∫₀¹ (1−t)(−3) dt = −3[t − t²/2]₀¹ = −3(1 − 1/2) = −3/2.

Total Area = 0 + 9/2 − 3/2 = 6/2 = 3.

Check: base=2, height=3. Area = (1/2)(2)(3) = 3. ✓
$

4. When Green's Theorem Applies (and When It Doesn't)

Requirements: 1. C is a simple closed curve (no self-intersections) 2. C is piecewise smooth 3. C is positively oriented (counterclockwise) 4. P and Q have continuous partial derivatives on an open region containing D 5. D is a simply-connected region (no holes)

What if the curve has holes? For a region with holes (like an annulus), Green's theorem generalizes:

$∮_{C_outer} − ∮_{C_inner} = ∬_D (∂Q/∂x − ∂P/∂y) dA
$

Each boundary is oriented so D is on the left (outer CCW, inner CW).

What if ∂Q/∂x − ∂P/∂y = 0 but the line integral isn't zero? This can happen if the region has a "hole" where the field isn't defined. Example:

F(x, y) = ⟨−y/(x²+y²), x/(x²+y²)⟩

∂Q/∂x = (y²−x²)/(x²+y²)², ∂P/∂y = (y²−x²)/(x²+y²)².
∂Q/∂x − ∂P/∂y = 0 (away from origin).

But ∮_unit circle F·dr = 2π ≠ 0.
Green's theorem can't be applied because F is not defined at the origin,
which is inside the circle. The region is not simply connected with respect to F.

5. Flux-Divergence Form

If F represents a fluid velocity field, the flux form of Green's theorem measures the net flow out of D:

$Flux = ∮_C F · n ds = ∬_D div F dA
$

Where n is the outward unit normal to C, and div F = ∂P/∂x + ∂Q/∂y (the divergence).

If C is positively oriented (CCW), the outward normal is n = ⟨y'/|r'|, −x'/|r'|⟩ (right-hand rule).

The relationship between the two forms: replace ⟨P, Q⟩ by ⟨−Q, P⟩ in the circulation form, or by ⟨Q, −P⟩.

6. Proof Sketch of Green's Theorem

Green's theorem can be proved by decomposing the region into Type I and Type II subregions.

For a region D that is both Type I and Type II:

$D: a ≤ x ≤ b, g₁(x) ≤ y ≤ g₂(x)  (Type I)
D: c ≤ y ≤ d, h₁(y) ≤ x ≤ h₂(y)  (Type II)
$

For Type I: compute ∮ Q dy along the boundary, reduce to ∬ ∂Q/∂x dA. For Type II: compute ∮ P dx along the boundary, reduce to −∬ ∂P/∂y dA.

Summing gives Green's theorem. For general regions, partition into simple subregions. Interior boundaries cancel.

Common Misconceptions

1. "Green's theorem is just for making line integrals easier." It also works in reverse — to compute a difficult double integral by converting it to a line integral.

2. "Green's theorem always works when ∂Q/∂x − ∂P/∂y = 0." No — if the field has a singularity inside the region, the theorem doesn't apply, and the line integral may be nonzero (as with ⟨−y, x⟩/(x²+y²)).

3. "The orientation doesn't matter." Reversing C flips the sign of the line integral. The positive (CCW) orientation is used in the standard statement.

4. "Green's theorem gives area only when ∂Q/∂x − ∂P/∂y = 1." Yes, that's one use, but the theorem is much more general. Any P, Q with continuous partials work.

Key Terms

• Flux-Divergence Form

Worked Examples

Example 1: Green's Theorem Simplifying a Hard Integral

Problem: Evaluate ∮_C (y + e^{cos x}) dx + (2x + ln(1 + y²)) dy where C is the circle x² + y² = 9.

Solution:

$P = y + e^{cos x}, Q = 2x + ln(1 + y²).
The non-polynomial parts are impossible to antiderivate.

Green's theorem:
∂Q/∂x = 2, ∂P/∂y = 1.
∂Q/∂x − ∂P/∂y = 2 − 1 = 1.

∮ = ∬_D 1 dA = Area(D) = π(3)² = 9π.
$

Example 2: Area of an Astroid

Problem: Find the area enclosed by the astroid x^{2/3} + y^{2/3} = a^{2/3}.

Solution:

$Parametrize: x = a cos³t, y = a sin³t, t ∈ [0, 2π].

Use Area = (1/2)∮ (x dy − y dx).

x dy = a cos³t · 3a sin²t cos t dt = 3a² cos⁴t sin²t dt
y dx = a sin³t · (−3a cos²t sin t) dt = −3a² sin⁴t cos²t dt

x dy − y dx = 3a² cos²t sin²t(cos²t + sin²t) dt = 3a² cos²t sin²t dt

∫₀^{2π} cos²t sin²t dt = (1/4)∫₀^{2π} sin²(2t) dt = (1/4)∫₀^{2π} (1−cos(4t))/2 dt
= (1/8)[t − sin(4t)/4]₀^{2π} = (1/8)(2π) = π/4.

Area = (1/2) · 3a² · (π/4) = 3πa²/8.
$

Example 3: Line Integral from Double Integral

Problem: Use Green's theorem to evaluate ∬_D e^{x²} dA over the triangle with vertices (0,0), (1,0), (0,1). (Hint: choose P and Q so that ∂Q/∂x − ∂P/∂y = e^{x²}.)

Solution:

$We need P, Q such that ∂Q/∂x − ∂P/∂y = e^{x²}.

Since e^{x²} doesn't depend on y, let P = 0 and ∂Q/∂x = e^{x²}. 
Then Q = ∫ e^{x²} dx — but this can't be expressed in elementary functions!

Alternative: Let Q = 0 and −∂P/∂y = e^{x²}. So ∂P/∂y = −e^{x²}.
Then P = −y e^{x²} (treating x constant).

So: ∬_D e^{x²} dA = ∮_C (−y e^{x²}) dx + 0 dy = −∮_C y e^{x²} dx.

Now C is the triangle boundary:
C₁: (0,0)→(1,0): y=0. −∫₀¹ 0·e^{x²} dx = 0.
C₂: (1,0)→(0,1): r(t)=⟨1−t,t⟩, t∈[0,1]. dx=−dt. y=t, x=1−t.
    −∫₀¹ t·e^{(1−t)²}·(−dt) = ∫₀¹ t e^{(1−t)²} dt.
C₃: (0,1)→(0,0): x=0. −∫₁⁰ y·e⁰·0 = 0.

So ∬_D e^{x²} dA = ∫₀¹ t e^{(1−t)²} dt.
Let u = 1−t, du = −dt. = ∫₁⁰ (1−u) e^{u²} (−du) = ∫₀¹ (1−u) e^{u²} du.
= ∫₀¹ e^{u²} du − ∫₀¹ u e^{u²} du.
= ∫₀¹ e^{u²} du − [e^{u²}/2]₀¹
= ∫₀¹ e^{u²} du − (e−1)/2.

Unfortunately, ∫₀¹ e^{u²} du is the imaginary error function — not elementary.
But either representation (the double integral or the line integral) is valid.
$

Quiz

Q1: Green's Theorem relates a line integral around a closed curve C to:

A) A triple integral over the solid bounded by C B) A double integral over the region D enclosed by C C) A surface integral over the surface bounded by C D) A line integral along any other curve

Correct: B)

• If you chose B: ∮_C P dx + Q dy = ∬_D (∂Q/∂x − ∂P/∂y) dA — a line integral around a closed curve equals a double integral over the enclosed region. Correct!
• If you chose A: That's the Divergence Theorem (Gauss), relating flux to triple integrals.
• If you chose C: That's Stokes' Theorem.
• If you chose D: Green's theorem relates a line integral to an area integral, not to another line integral.

Q2: To apply Green's Theorem, the curve C must be:

A) Any curve, open or closed B) A simple closed curve, oriented counterclockwise (positively) C) A straight line D) Parametrized by arc length

Correct: B)

• If you chose B: Green's Theorem requires a positively oriented (counterclockwise), piecewise-smooth, simple closed curve. Correct!
• If you chose A: The curve must be closed for the theorem to apply.
• If you chose C: A straight line is not closed — can't use Green's Theorem.
• If you chose D: Arc length parametrization is not required.

Q3: Green's Theorem can be used to compute area by setting:

A) P = 0, Q = 0 B) P = −y/2, Q = x/2 C) P = 1, Q = 1 D) P = x, Q = y

Correct: B)

• If you chose B: With P = −y/2, Q = x/2, ∂Q/∂x − ∂P/∂y = 1/2 − (−1/2) = 1, so ∬_D 1 dA = Area = ∮_C (−y/2 dx + x/2 dy). Correct!
• If you chose A: Gives zero — useless.
• If you chose C: ∂Q/∂x − ∂P/∂y = 0 − 0 = 0 — gives zero area.
• If you chose D: ∂Q/∂x − ∂P/∂y = 1 − 1 = 0 — gives zero.

Q4: For F = ⟨−y, x⟩ around the unit circle (counterclockwise), ∮_C F · dr equals:

A) 0 B) π C) 2π D) 1

Correct: C)

• If you chose C: By Green's: P = −y, Q = x. ∂Q/∂x − ∂P/∂y = 1 − (−1) = 2. ∬_D 2 dA = 2·Area(disk) = 2π. Or directly: r(t) = ⟨cos t, sin t⟩, ∫₀²π (−sin t)(−sin t) + (cos t)(cos t) dt = ∫₀²π 1 dt = 2π. Correct!
• If you chose A: The field has circulation — integral is nonzero.
• If you chose B: Half the correct value.
• If you chose D: Not the correct value.

Q5: If ∮_C F · dr = 0 for every closed curve C in a region, then in that region:

A) F = 0 everywhere B) ∂Q/∂x − ∂P/∂y = 0 everywhere (F is irrotational) C) F is constant D) The region must be simply connected

Correct: B)

• If you chose B: By Green's Theorem, ∮_C F · dr = ∬_D (∂Q/∂x − ∂P/∂y) dA = 0 for all D implies the integrand is zero: ∂Q/∂x = ∂P/∂y. Correct!
• If you chose A: F could be nonzero but still have zero curl (e.g., F = ⟨x, y⟩ fails this test, but F = ⟨y, x⟩ has curl 0? No: ∂P/∂y=1, ∂Q/∂x=1 — wait, that does have zero curl. Actually F = ⟨x, 0⟩ has ∂Q/∂x=0, ∂P/∂y=0 — zero curl but nonzero.)
• If you chose C: Not required.
• If you chose D: Simple connectivity isn't required for this implication.

Q6: Green's Theorem fails for a region with a hole because:

A) The integral is always zero B) The curve is not simple closed C) The region is not simply connected — you need to add cuts or use the extended version D) The theorem only works for circles

Correct: C)

• If you chose C: For multiply connected regions (with holes), you must decompose the region or add extra boundary components. The standard Green's theorem assumes a simply connected region. Correct!
• If you chose A: The integral is not necessarily zero for regions with holes.
• If you chose B: The outer boundary can still be a simple closed curve.
• If you chose D: Green's theorem works for any nice region, not just circles.

Practice Problems

(Answers are below. Try each problem before checking.)

Problem 1: Verify Green's theorem for F(x, y) = ⟨x²y, xy²⟩ over the square [0, 1] × [0, 1].

Problem 2: Use Green's theorem to evaluate ∮_C (xy dx + x² dy) where C is the triangle with vertices (0,0), (1,0), (0,2).

Problem 3: Use Green's theorem to evaluate ∮_C (y² dx − x² dy) where C is the circle x² + y² = 1.

Problem 4: Find the area of one arch of the cycloid x = t − sin t, y = 1 − cos t for t ∈ [0, 2π] and the x-axis.

Problem 5: Use Green's theorem to evaluate ∬_D (x² + y²) dA where D is the region bounded by the cardioid r = 1 + cos θ. (Hint: find a line integral representation.)

Problem 6: Evaluate ∮_C ⟨−y³, x³⟩ · dr where C is the circle x² + y² = 4.

Problem 7: Use the area formula (1/2)∮(x dy − y dx) to find the area inside the ellipse x²/16 + y²/9 = 1.

Summary

1. Green's theorem ∮_C P dx + Q dy = ∬_D (∂Q/∂x − ∂P/∂y) dA connects the circulation around a closed curve to the integral of the curl over the enclosed region — it's the 2D analog of the Fundamental Theorem of Calculus
2. The theorem transforms difficult line integrals into simple double integrals (especially when ∂Q/∂x − ∂P/∂y is constant) and can also work in reverse to compute double integrals via line integrals
3. Three area formulas emerge as special cases: Area = ∮ x dy = −∮ y dx = (1/2)∮ (x dy − y dx); the last form is particularly useful for parametric curves
4. Green's theorem requires P and Q to have continuous partial derivatives on an open region containing D; if the field has a singularity inside D, the theorem does not apply
5. The flux-divergence form ∮ F · n ds = ∬ div F dA measures net outward flow and is the 2D precursor to the Divergence Theorem

Pitfalls

1. Applying Green's theorem when the field has a singularity inside the region. If P or Q is undefined at any point within D, the theorem's hypotheses are violated. The classic case is ⟨−y, x⟩/(x² + y²) around the unit circle — curl is zero everywhere except at the origin, yet the line integral is 2π, not 0.

2. Using the wrong orientation. Green's theorem in its standard form requires counterclockwise (positive) orientation. If the curve is traversed clockwise, the result is the negative. Always check orientation before applying the formula.

3. Forgetting to verify that ∂Q/∂x − ∂P/∂y simplifies before integrating. The power of Green's theorem comes when the curl simplifies to a constant or simple function. Students sometimes set up the double integral without first computing and simplifying the integrand.

4. Confusing the circulation-curl and flux-divergence forms. The circulation form uses ∂Q/∂x − ∂P/∂y (the 2D curl); the flux form uses ∂P/∂x + ∂Q/∂y (the 2D divergence). Using one when the problem calls for the other will produce a wrong answer.

5. Misapplying the area formulas. The formulas Area = (1/2)∮(x dy − y dx), Area = ∮ x dy, and Area = −∮ y dy work for any simple closed curve, but only when the integrand is correctly paired with the differential. Using, say, ∮ x dx — which always gives zero around a closed curve — is a common mistake.

Next Steps

Move on to 07-07 — Curl and Divergence to explore the physical meaning and computation of these two fundamental differential operators in 3D, and to understand how they characterize vector fields.