Green's theorem — proof sketch, both forms: Worked examples

We use two objects throughout. Both were defined in the parent note; here they are again in one place so no symbol is unearned:

The scenario matrix

Every problem Green's theorem can pose falls into one of these cells. The examples below each carry a [cell] tag so you can see the coverage is complete.

Cell	What makes it distinct	Example(s)
A. Curl, positive	$Q_{x} - P_{y} > 0$ everywhere → nonzero circulation	Ex 1
B. Curl, zero by symmetry	integrand odd over a symmetric region → answer $0$	Ex 2
C. Wrong orientation	curve given clockwise → sign flip	Ex 3
D. Non-standard region	annulus / region with a hole	Ex 4
E. Singularity (hypothesis breaks)	field blows up inside $D$ → theorem misused	Ex 5
F. Flux form	using divergence, not curl	Ex 6
G. Degenerate / limiting	shrinking loop, curl as a limit	Ex 7
H. Word problem	real-world flux (fluid crossing a boundary)	Ex 8
I. Exam twist	non-circular boundary, area trick	Ex 9

Example 1 — Curl, positive `[cell A]`

Worked example Circulation of a genuine swirl

Compute $\oint_{C} (- y) d x + x d y$ where $C$ is the counterclockwise boundary of the disk of radius $R$ .

Forecast: this field $F = (- y, x)$ points along circles (a pure swirl). Walking counterclockwise you're always pushed forward, so the answer should be positive and grow with area. Guess before reading on.

Step 1. Identify $P = - y$ , $Q = x$ . Why this step? Green's theorem needs $P$ and $Q$ named before we can differentiate.

Step 2. Compute the curl integrand: $Q_{x} - P_{y} = \frac{\partial x}{\partial x} - \frac{\partial ( - y )}{\partial y} = 1 - (- 1) = 2.$ Why this step? The circulation form converts the loop integral into $\iint_{D} (Q_{x} - P_{y}) d A$ . We need that integrand.

Step 3. Integrate the constant over the disk: $\iint_{D} 2 d A = 2 \cdot (area of disk) = 2 π R^{2} .$ Why this step? Integrating the constant $2$ over $D$ is just $2 \times$ the area, and the disk has area $π R^{2}$ .

Result: $\oint_{C} (- y) d x + x d y = 2 π R^{2}$ .

Verify: parametrise $x = R cos t, y = R sin t$ . Then $- y d x + x d y = R^{2} sin^{2} t + R^{2} cos^{2} t = R^{2}$ , and $\int_{0}^{2 π} R^{2} d t = 2 π R^{2}$ . ✅ Matches, and it grew with area exactly as forecast.

Figure — Green's theorem — proof sketch, both forms

Look at the red arrows: they curl counterclockwise, aligned with our walking direction — that is why the circulation is positive.

Example 2 — Curl zero by symmetry `[cell B]`

Example 3 — Wrong orientation `[cell C]`

Worked example Clockwise loop

Compute $\oint_{C} (- y) d x + x d y$ where $C$ is the unit circle traversed clockwise.

Forecast: it's the same field as Ex 1 (with $R = 1$ ), but we now walk the wrong way. Green's theorem is stated for counterclockwise loops, so we expect the answer to be the negative of Ex 1.

Step 1. Green's theorem requires positive (counterclockwise) orientation. Our curve is clockwise, so let $C^{+}$ be the counterclockwise version and note $C = - C^{+}$ . Why this step? Reversing the direction of a line integral flips its sign; we must convert to the orientation the theorem assumes.

Step 2. Apply the theorem to $C^{+}$ (from Ex 1 with $R = 1$ ): $\oint_{C^{+}} (- y) d x + x d y = 2 π (1)^{2} = 2 π .$

Step 3. Flip the sign for the clockwise curve: $\oint_{C} (- y) d x + x d y = - \oint_{C^{+}} = - 2 π .$ Why this step? $C = - C^{+}$ , and $\int_{- C} = - \int_{C}$ .

Result: $- 2 π$ .

Verify: clockwise means $x = cos t, y = - sin t$ (angle decreasing). Then $- y d x + x d y = - sin^{2} t - cos^{2} t = - 1$ , and $\int_{0}^{2 π} (- 1) d t = - 2 π$ . ✅

The red arrow shows the walking direction now runs against the field's swirl — the region is on your right, not your left, which is the visual signature of a sign flip.

Example 4 — Region with a hole `[cell D]`

Worked example Annulus

Compute $\oint (- y) d x + x d y$ over the boundary of the annulus $1 \leq x^{2} + y^{2} \leq 4$ (radii $1$ and $2$ ). The boundary has two circles: the outer $C_{2}$ counterclockwise, the inner $C_{1}$ clockwise (so the region stays on your left everywhere).

Forecast: the field is the swirl again. The double integral only counts the ring's area, not the hole. Expect $2 \times (ring area)$ .

Step 1. The correctly oriented boundary is $C = C_{2} - C_{1}^{cc}$ where $C_{1}^{cc}$ is the inner circle counterclockwise. Why this step? For a region on your left, the outer curve runs counterclockwise and every inner hole runs clockwise. This is the multiply-connected version from Step 3 of the parent proof.

Step 2. Curl integrand is $2$ (same as Ex 1). Integrate over the ring: $\iint_{D} 2 d A = 2 (π \cdot 2^{2} - π \cdot 1^{2}) = 2 (4 π - π) = 6 π .$ Why this step? The ring's area is (outer disk area) − (inner disk area) $= 4 π - π = 3 π$ .

Result: total oriented circulation $= 6 π$ .

Verify (by hand, both circles): each circle of radius $R$ counterclockwise gives $2 π R^{2}$ (Ex 1). Outer counterclockwise $= 2 π (4) = 8 π$ ; inner clockwise $= - 2 π (1) = - 2 π$ . Sum $= 8 π - 2 π = 6 π$ . ✅

Example 5 — Singularity: the hypothesis breaks `[cell E]`

Worked example The trap field

For $F = (\frac{- y}{x ^{2} + y ^{2}}, \frac{x}{x ^{2} + y ^{2}})$ , a student computes $Q_{x} - P_{y} = 0$ and concludes $\oint_{C} F \cdot d r = 0$ for the unit circle. Where is the error, and what is the true value?

Forecast: something is wrong — this field is famous. Suspect the origin.

Step 1. Check the curl integrand away from the origin. Writing $Q_{x} - P_{y}$ and simplifying gives $0$ wherever the field is defined. Why this step? We must confirm the student's algebra before blaming it.

Step 2. Locate the flaw: $F$ is undefined at $(0, 0)$ because $x^{2} + y^{2} = 0$ there. Green's theorem demands $P, Q$ have continuous partial derivatives everywhere inside $D$ — the origin is inside the unit disk and violates this. Why this step? A theorem's conclusion is void if its hypothesis fails. No hypothesis, no $\iint$ -to- $\oint$ equality.

Step 3. Compute the honest line integral directly, $x = cos t, y = sin t$ (so $x^{2} + y^{2} = 1$ ): $\oint_{C} F \cdot d r = \int_{0}^{2 π} [(- sin t) (- sin t) + (cos t) (cos t)] d t = \int_{0}^{2 π} 1 d t = 2 π .$ Why this step? When the theorem is unavailable, fall back to the definition of a line integral.

Result: the true value is $2 π$ , not $0$ . The double-integral shortcut was illegal.

Fix: exclude the origin with a tiny circle; on the resulting hole-punched region the theorem does apply, and it correctly reports the leftover $2 π$ living at the puncture. This is exactly the mistake flagged in the parent note.

Verify: $(- sin t) (- sin t) + (cos t) (cos t) = sin^{2} t + cos^{2} t = 1$ , integrating to $2 π$ . ✅

Example 6 — Flux form `[cell F]`

Every red arrow crosses the fence pointing outward — no inflow anywhere, which is why the total is a clean positive number.

Example 7 — Degenerate limit: curl as a shrinking loop `[cell G]`

Worked example Circulation per area

For $F = (- y, x)$ , show that the circulation around a small circle of radius $ε$ , divided by its area, approaches the curl integrand as $ε \to 0$ .

Forecast: the parent note called $Q_{x} - P_{y}$ the "local spin." A limit should recover the local value $2$ from the global loop integral.

Step 1. From Ex 1, circulation around radius $ε$ is $2 π ε^{2}$ . Why this step? We reuse the closed-form we already trust.

Step 2. Divide by the enclosed area $π ε^{2}$ : $\frac{circulation}{area} = \frac{2 π ε ^{2}}{π ε ^{2}} = 2.$ Why this step? "Circulation per unit area" is the physical definition of scalar curl.

Step 3. Take the limit: $lim_{ε \to 0} \frac{1}{π ε ^{2}} \oint_{C_{ε}} F \cdot d r = 2 = (Q_{x} - P_{y}) .$ Why this step? It shows the double-integral integrand really is a pointwise limit of loop integrals — the microscopic meaning behind Green's theorem, echoing the Fundamental Theorem of Calculus idea that a boundary encodes an interior rate.

Result: the ratio is exactly $2$ for every $ε$ , so the limit is $2$ . ✅ (A rare "no limit surprise" — the field's curl is constant.)

Verify: $2 π ε^{2} / (π ε^{2}) = 2$ for all $ε > 0$ , hence the limit is $2$ . ✅

Example 8 — Word problem: fluid across a channel gate `[cell H]`

Worked example How much water leaves the pond?

Water in a shallow pond has velocity field $F = (x, y)$ metres/second (a spring at the origin pushing outward). The pond boundary is the circle of radius $3$ m. What is the net outward volume flow rate (per unit depth) through the boundary?

Forecast: water flows outward everywhere, so net outflow is positive; by Ex 6 it should be $2 π R^{2}$ with $R = 3$ .

Step 1. Net outflow rate $= \oint_{C} F \cdot n d s$ — the flux form. Why this step? "Volume crossing the boundary per second" is precisely a flux integral.

Step 2. $\nabla \cdot F = 1 + 1 = 2$ (per second), integrate over the pond: $\iint_{D} 2 d A = 2 \cdot π \cdot 3^{2} = 18 π .$ Why this step? Divergence $= 2$ means each square metre "creates" $2$ m³/s of outflow; multiply by area $9 π$ .

Result: $18 π \approx 56.5 m^{3} / s$ per unit depth of pond.

Verify (units & value): $[divergence] = \frac{m/s}{m} = 1/ s$ ; times area $m^{2}$ gives $m^{2} / s$ (area flow per unit depth) ✅. Numerically $18 π = 56.548...$ , positive as forecast. ✅

Example 9 — Exam twist: area of a triangle from its edges `[cell I]`

Worked example Area with no double integral

Use $Area = \frac{1}{2} \oint_{C} (x d y - y d x)$ to find the area of the triangle with vertices $(0, 0)$ , $(4, 0)$ , $(0, 3)$ , walking counterclockwise.

Forecast: classic $3$ - $4$ - $5$ -ish right triangle; area $= \frac{1}{2} \cdot 4 \cdot 3 = 6$ . The boundary formula must reproduce this.

Step 1. The area formula comes from choosing $P = - y /2, Q = x /2$ , giving $Q_{x} - P_{y} = 1$ , so $\iint_{D} 1 d A = Area$ . Why this step? We engineer the integrand to be $1$ so the double integral literally is the area — no antiderivative in the plane needed, only edge sums.

Step 2. For a straight segment from $(x_{1}, y_{1})$ to $(x_{2}, y_{2})$ , one shows $\frac{1}{2} \int (x d y - y d x) = \frac{1}{2} (x_{1} y_{2} - x_{2} y_{1}) .$ Why this step? On a line segment the integral of $x d y - y d x$ is the signed twice-area of the triangle the segment makes with the origin — the "shoelace" building block.

Step 3. Sum over the three edges $(0, 0) \to (4, 0) \to (0, 3) \to (0, 0)$ : $\frac{1}{2} [(0 \cdot 0 - 4 \cdot 0) + (4 \cdot 3 - 0 \cdot 0) + (0 \cdot 0 - 0 \cdot 3)] = \frac{1}{2} [0 + 12 + 0] = 6.$ Why this step? Each bracket is one edge's contribution; adding them walks the whole boundary once.

Result: Area $= 6$ .

Verify: elementary geometry $\frac{1}{2} \cdot base \cdot height = \frac{1}{2} \cdot 4 \cdot 3 = 6$ . ✅

Each red edge contributes the signed area of the wedge it sweeps from the origin; the wedges add up (with cancellation) to the triangle. This is the line-integral way to measure area — the boundary knowing the interior once more.

Recall Did the matrix get fully covered?

Positive curl ::: Ex 1 Zero by symmetry ::: Ex 2 Wrong orientation (sign flip) ::: Ex 3 Region with a hole (annulus) ::: Ex 4 Singularity breaks the hypothesis ::: Ex 5 Flux / divergence form ::: Ex 6 Degenerate shrinking loop (curl as a limit) ::: Ex 7 Real-world flux word problem ::: Ex 8 Exam twist (area from boundary) ::: Ex 9

Related deep structure: Green's theorem is the flat-plane case of both Stokes' Theorem (curl form) and the Divergence Theorem (flux form); when $Q_{x} - P_{y} = 0$ everywhere the field is often a conservative field with zero circulation.

Flashcards

Circulation of

(- y, x)

around radius

R

(CCW)?

2 π R^{2}

Why is

\oint y^{2} d x + 3 x y d y = 0

on the unit disk?

Curl integrand is

y

, odd over a disk symmetric about

y = 0

, so it cancels.

Clockwise loop — what do you do?

Flip the sign; apply the theorem to the CCW version.

Annulus boundary orientation?

Outer CCW, inner CW (region on your left).

Why does

(- y, x) / (x^{2} + y^{2})

give

2 π

not

0

Singularity at origin breaks the continuous-partials hypothesis.

Flux of

(x, y)

across radius

R

2 π R^{2}

, since divergence

= 2

Area from the boundary?

Area = \frac{1}{2} \oint_{C} (x d y - y d x)