Green's theorem — proof sketch, both forms: Exercises

Reminders we will use constantly (both from the parent): $\oint_{C} P d x + Q d y = \iint_{D} (\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}) d A (circulation)$ $\oint_{C} F \cdot n d s = \iint_{D} (\frac{\partial P}{\partial x} + \frac{\partial Q}{\partial y}) d A (flux)$

Here $\oint_{C}$ means "add up along the whole closed loop $C$ ", and $\iint_{D}$ means "add up over the whole flat region $D$ that the loop fences in". The little $\partial Q / \partial x$ is the partial derivative: how fast $Q$ changes if you nudge only $x$ and freeze $y$ .

Related toolkit: Line Integrals, Double Integrals, Curl and Divergence, Conservative Vector Fields.

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

Level 1 — Recognition

L1.1

State which form of Green's theorem you would use, and write its integrand, for $\oint_{C} 3 x^{2} y d x + x^{3} d y .$

Recall Solution

This is a line integral written as $P d x + Q d y$ , so it is the circulation form. Read off $P = 3 x^{2} y$ , $Q = x^{3}$ . The integrand is $Q_{x} - P_{y}$ : $Q_{x} = \frac{\partial}{\partial x} (x^{3}) = 3 x^{2}, P_{y} = \frac{\partial}{\partial y} (3 x^{2} y) = 3 x^{2} .$ So the integrand is $3 x^{2} - 3 x^{2} = 0$ . The loop integral is $0$ for any closed curve (the field is conservative — see Conservative Vector Fields).

L1.2

For $F = (P, Q) = (x^{2}, y^{2})$ , write the flux-form integrand $\nabla \cdot F$ .

Recall Solution

Flux uses the sum $P_{x} + Q_{y}$ : $P_{x} = \frac{\partial}{\partial x} (x^{2}) = 2 x, Q_{y} = \frac{\partial}{\partial y} (y^{2}) = 2 y .$ $\nabla \cdot F = 2 x + 2 y .$

Level 2 — Application

L2.1

Compute $\oint_{C} (y - x) d x + (2 x) d y$ where $C$ is the boundary of the square $0 \leq x \leq 1$ , $0 \leq y \leq 1$ , counterclockwise.

Recall Solution

Circulation form: $P = y - x$ , $Q = 2 x$ . $Q_{x} - P_{y} = \frac{\partial}{\partial x} (2 x) - \frac{\partial}{\partial y} (y - x) = 2 - 1 = 1.$ So the loop integral collapses to the plain area of the square: $\iint_{D} 1 d A = Area = 1 \times 1 = 1.$ Answer: $1$ .

L2.2

Use the boundary–area formula $Area = \frac{1}{2} \oint_{C} (x d y - y d x)$ to find the area enclosed by the ellipse $x = 3 cos t, y = 2 sin t$ , $0 \leq t \leq 2 π$ .

Recall Solution

Why this formula? Choosing $P = - y /2, Q = x /2$ makes $Q_{x} - P_{y} = 1$ , so the double integral becomes the raw area — proved in the parent. We just plug in the parametrisation. $d x = - 3 sin t d t, d y = 2 cos t d t .$ $x d y - y d x = (3 cos t) (2 cos t d t) - (2 sin t) (- 3 sin t d t) = 6 cos^{2} t d t + 6 sin^{2} t d t = 6 d t .$ $Area = \frac{1}{2} \int_{0}^{2 π} 6 d t = \frac{1}{2} \cdot 12 π = 6 π .$ Answer: $6 π$ . (Sanity check: an ellipse with semi-axes $a = 3, b = 2$ has area $π ab = 6 π$ . ✓)

L2.3

Compute the outward flux $\oint_{C} F \cdot n d s$ of $F = (x^{2}, y^{2})$ across the boundary of the unit square $0 \leq x \leq 1, 0 \leq y \leq 1$ .

Recall Solution

Flux form turns the loop into a double integral of the divergence (from L1.2, $\nabla \cdot F = 2 x + 2 y$ ): $\iint_{D} (2 x + 2 y) d A = \int_{0}^{1} \int_{0}^{1} (2 x + 2 y) d y d x .$ Inner integral in $y$ : $\int_{0}^{1} (2 x + 2 y) d y = [2 x y + y^{2}]_{0}^{1} = 2 x + 1.$ Outer integral in $x$ : $\int_{0}^{1} (2 x + 1) d x = [x^{2} + x]_{0}^{1} = 1 + 1 = 2.$ Answer: flux $= 2$ .

Level 3 — Analysis

L3.1

Compute $\oint_{C} (e^{x} - y^{3}) d x + (cos y + x^{3}) d y$ where $C$ is the unit circle $x^{2} + y^{2} = 1$ , counterclockwise.

Recall Solution

The line integral looks brutal ( $e^{x}$ , $cos y$ ...). Green's theorem rescues us — those "own-variable" pieces vanish under $Q_{x} - P_{y}$ . $P = e^{x} - y^{3}, Q = cos y + x^{3} .$ $Q_{x} = 3 x^{2}, P_{y} = - 3 y^{2}, Q_{x} - P_{y} = 3 x^{2} + 3 y^{2} = 3 (x^{2} + y^{2}) .$ Convert to polar coordinates $x = r cos θ, y = r sin θ$ , where $x^{2} + y^{2} = r^{2}$ and $d A = r d r d θ$ . Why polar? The integrand and the region are both circular, so polar makes the limits trivial. $\iint_{D} 3 r^{2} d A = \int_{0}^{2 π} \int_{0}^{1} 3 r^{2} \cdot r d r d θ = \int_{0}^{2 π} \int_{0}^{1} 3 r^{3} d r d θ .$ $\int_{0}^{1} 3 r^{3} d r = [\frac{3}{4} r^{4}]_{0}^{1} = \frac{3}{4}, \int_{0}^{2 π} \frac{3}{4} d θ = \frac{3}{4} \cdot 2 π = \frac{3 π}{2} .$ Answer: $\frac{3 π}{2}$ .

L3.2

The vector field $F = \frac{( - y , x )}{x ^{2} + y ^{2}}$ has curl zero everywhere it is defined, yet $\oint_{C} F \cdot d r = 2 π$ around the unit circle. Explain the apparent contradiction and compute $Q_{x} - P_{y}$ for $(x, y) \neq = (0, 0)$ .

Recall Solution

Write $P = \frac{- y}{x ^{2} + y ^{2}}$ , $Q = \frac{x}{x ^{2} + y ^{2}}$ . Differentiate carefully: $Q_{x} = \frac{( x ^{2} + y ^{2} ) - x ( 2 x )}{( x ^{2} + y ^{2} ) ^{2}} = \frac{y ^{2} - x ^{2}}{( x ^{2} + y ^{2} ) ^{2}},$ $P_{y} = \frac{- ( x ^{2} + y ^{2} ) + y ( 2 y )}{( x ^{2} + y ^{2} ) ^{2}} = \frac{y ^{2} - x ^{2}}{( x ^{2} + y ^{2} ) ^{2}} .$ So $Q_{x} - P_{y} = 0$ wherever $F$ is defined. If Green's theorem applied, the loop integral would be $0$ — but it is $2 π$ . Resolution: Green's theorem requires $P, Q$ to have continuous partials everywhere inside $D$ . Here the field blows up at the origin, which lies inside the unit disk. The hypothesis fails, so we may not apply the theorem. The $2 π$ is real; the "contradiction" is only apparent because the theorem was never licensed.

Level 4 — Synthesis

L4.1

Compute $\oint_{C} F \cdot d r$ for $F = (y - sin x, cos x - y)$ where $C$ is the triangle with vertices $(0, 0)$ , $(2, 0)$ , $(0, 2)$ , traversed counterclockwise.

Recall Solution

$P = y - sin x, Q = cos x - y$ . $Q_{x} = - sin x, P_{y} = 1, Q_{x} - P_{y} = - sin x - 1.$ The triangle is the Type I region $0 \leq x \leq 2$ , $0 \leq y \leq 2 - x$ (bottom edge along the $x$ -axis, hypotenuse $y = 2 - x$ ). Why these limits? For a fixed $x$ the vertical strip runs from the $x$ -axis up to the slanted top. $\iint_{D} (- sin x - 1) d A = \int_{0}^{2} \int_{0}^{2 - x} (- sin x - 1) d y d x .$ Inner integral (integrand is constant in $y$ ): $\int_{0}^{2 - x} (- sin x - 1) d y = (- sin x - 1) (2 - x) .$ Now expand and integrate in $x$ : $\int_{0}^{2} (- sin x - 1) (2 - x) d x = \int_{0}^{2} [- 2 sin x + x sin x - 2 + x] d x .$ Term by term over $[0, 2]$ :

$\int_{0}^{2} - 2 sin x d x = 2 cos x_{0}^{2} = 2 cos 2 - 2.$
$\int_{0}^{2} x sin x d x = [sin x - x cos x]_{0}^{2} = sin 2 - 2 cos 2.$
$\int_{0}^{2} - 2 d x = - 4.$
$\int_{0}^{2} x d x = 2.$

Add them: $(2 cos 2 - 2) + (sin 2 - 2 cos 2) + (- 4) + 2 = sin 2 - 4.$ Answer: $sin 2 - 4 \approx 0.9093 - 4 = - 3.0907$ .

L4.2

A closed curve $C$ (counterclockwise) has the property that $\frac{1}{2} \oint_{C} (x d y - y d x) = 10$ . If you shrink the whole curve toward the origin by the factor $\frac{1}{2}$ (map every point $(x, y) \mapsto (\frac{1}{2} x, \frac{1}{2} y)$ ), what is the new value of $\frac{1}{2} \oint (x d y - y d x)$ ?

Recall Solution

That integral is the enclosed area (parent Example 1). Scaling every coordinate by a factor $k$ scales all lengths by $k$ and therefore areas by $k^{2}$ . Here $k = \frac{1}{2}$ , so area $\to (\frac{1}{2})^{2} \times 10 = \frac{1}{4} \times 10 = 2.5$ . Answer: $2.5$ .

Level 5 — Mastery

L5.1 (Full argument)

Let $D$ be an annulus: the region between the circle $x^{2} + y^{2} = 1$ (inner) and $x^{2} + y^{2} = 4$ (outer). Let $F = (- y, x)$ . Using Green's theorem correctly for a region with a hole, compute the total circulation, i.e. (outer boundary counterclockwise) $+$ (inner boundary clockwise).

Recall Solution

Why an annulus needs care. Green's theorem for a region with a hole says: sum the boundary integrals with the region always on your left. That means the outer circle counterclockwise and the inner circle clockwise. Then $\oint_{\partial D} F \cdot d r = \iint_{D} (Q_{x} - P_{y}) d A .$ Here $P = - y, Q = x$ , so $Q_{x} - P_{y} = 1 - (- 1) = 2$ (well-defined everywhere in the annulus — no singularity, since the origin is excluded). The annulus area is $π (2^{2}) - π (1^{2}) = 4 π - π = 3 π$ . $\iint_{D} 2 d A = 2 \cdot 3 π = 6 π .$ Answer: total circulation $= 6 π$ .

Cross-check the pieces directly. On the outer circle (radius $2$ , CCW), $\oint (- y d x + x d y) = 2 \times area = 2 π (2^{2}) = 8 π$ . On the inner circle (radius $1$ , clockwise = minus its CCW value) it is $- 2 π (1^{2}) = - 2 π$ . Sum $= 8 π - 2 π = 6 π$ . ✓

L5.2 (Full argument)

Prove that if $F = (P, Q)$ has $Q_{x} = P_{y}$ throughout a simply-connected region $D$ , then $\oint_{C} F \cdot d r = 0$ for every simple closed curve $C$ inside $D$ . Then use it to evaluate $\oint_{C} (2 x y d x + x^{2} d y)$ on any loop.

Recall Solution

The argument. Take any simple closed $C$ inside $D$ enclosing region $R \subseteq D$ . Since $D$ is simply connected (no holes), $R$ contains no forbidden points, so Green's theorem applies: $\oint_{C} F \cdot d r = \iint_{R} (Q_{x} - P_{y}) d A .$ By hypothesis $Q_{x} = P_{y}$ , so the integrand is $0$ everywhere, hence the double integral is $0$ , hence $\oint_{C} = 0$ . This is precisely the statement that $F$ is conservative (path-independent) on $D$ — see Conservative Vector Fields and the Fundamental Theorem of Calculus analogue for line integrals. Application. For $F = (2 x y, x^{2})$ : $Q_{x} = \frac{\partial}{\partial x} (x^{2}) = 2 x, P_{y} = \frac{\partial}{\partial y} (2 x y) = 2 x .$ They match everywhere in the plane (which is simply connected), so $\oint_{C} = 0$ for every closed loop. Answer: $0$ . (Indeed $F = \nabla (x^{2} y)$ , a gradient field.)

L5.3 (Synthesis with the flux form)

For $F = (x^{3}, y^{3})$ , compute the outward flux across the circle $x^{2} + y^{2} = a^{2}$ (radius $a$ ), leaving the answer in terms of $a$ .

Recall Solution

Flux form: $\nabla \cdot F = P_{x} + Q_{y} = 3 x^{2} + 3 y^{2} = 3 (x^{2} + y^{2})$ . In polar coordinates ( $x^{2} + y^{2} = r^{2}$ , $d A = r d r d θ$ ): $\iint_{D} 3 r^{2} d A = \int_{0}^{2 π} \int_{0}^{a} 3 r^{2} \cdot r d r d θ = \int_{0}^{2 π} \int_{0}^{a} 3 r^{3} d r d θ .$ $\int_{0}^{a} 3 r^{3} d r = [\frac{3}{4} r^{4}]_{0}^{a} = \frac{3}{4} a^{4}, \int_{0}^{2 π} \frac{3}{4} a^{4} d θ = \frac{3}{4} a^{4} \cdot 2 π = \frac{3 π a ^{4}}{2} .$ Answer: $\frac{3 π a ^{4}}{2}$ . (Check with $a = 1$ : gives $\frac{3 π}{2}$ , matching the shape of L3.1's radial computation.)

Recall One-line self-test

When is $\oint_{C} F \cdot d r = 0$ guaranteed for all loops? ::: When $Q_{x} = P_{y}$ on a simply-connected region (no singular points inside) — the field is conservative. For a region with a hole, which way does the inner boundary go? ::: Clockwise (so the region stays on your left).