Green’s Theorem, multivariable and vector calculus lesson

Walk once around the edge of a pond and add up how much the current pushes you along. That total looks like it depends only on the edge. Green’s theorem says it equals the sum of all the tiny swirls inside the pond. Small loops inside cancel along shared edges, and only the outer edge survives.

Key terms

Positively oriented piecewise smooth boundary

Definition A curve $C$ given by a continuous $\mathbf{r}(t)$ for $a \le t \le b$ is a simple closed curve if $\mathbf{r}(a) = \mathbf{r}(b)$ and $\mathbf{r}$ is one-to-one on $[a, b)$ . It is piecewise smooth if there are finitely many points $a = t_0 < t_1 < \dots < t_k = b$ such that on each closed piece $[t_{i-1}, t_i]$ the map $\mathbf{r}$ is $C^1$ , with one-sided derivatives at the two ends, and

\mathbf{r}'(t) \neq \mathbf{0} \quad \text{for } t_{i-1} < t < t_i.

So the velocity may vanish or jump only at the finitely many joining points $t_i$ . A line integral is an ordinary integral in $t$ , and finitely many values of $t$ do not affect it, so these points do no harm. This is slightly more permissive than the piecewise regular curves of the lesson on scalar line integrals, which ask for a nonzero velocity at the ends of each piece too, and every piecewise regular curve is piecewise smooth in this sense. A simple closed curve splits the plane into a bounded inside and an unbounded outside, and $D$ denotes the inside together with $C$ . The curve is positively oriented if $D$ lies to the left as you travel along $C$ . For a single closed curve this means counterclockwise.

In plain words The boundary may have finitely many corners or cusps, where the smooth pieces join. Walk along it with the region always on your left hand.

Example The boundary of the unit square $[0, 1] \times [0, 1]$ is positively oriented when traversed

\begin{gathered} (0, 0) \to (1, 0) \to (1, 1) \\ \to (0, 1) \to (0, 0). \end{gathered}

It is four straight segments joined at four corners, so it is piecewise smooth but not smooth.

A C1 field on a neighborhood of the region

Definition The theorem asks for $P$ and $Q$ to have continuous first partial derivatives on an open set $U$ that contains the whole closed region $D$ together with its boundary $C$ . Being defined and smooth only on the inside of $D$ is not enough, and a single point of $D$ where $P$ or $Q$ is undefined breaks the hypothesis.

In plain words The field must be smooth everywhere on the region and a little beyond its edge, with no bad points hiding inside.

Example The field $\langle x^2 y, e^{x} \rangle$ is $C^1$ on all of $\mathbb{R}^2$ , so any region is fine. The field

\mathbf{F} = \left\langle \frac{-y}{x^2 + y^2},\ \frac{x}{x^2 + y^2} \right\rangle

is $C^1$ only on the plane minus the origin. It meets the hypothesis on the annulus $1 \le x^2 + y^2 \le 4$ , but not on the unit disk.

Circulation form

Definition Let $C$ be a positively oriented, piecewise smooth, simple closed curve in the plane, let $D$ be the closed region made of $C$ and its inside, and let $P$ and $Q$ be $C^1$ on an open set containing $D$ . Then

\begin{aligned} &\oint_C P \, dx + Q \, dy \\[0.4em] &\quad = \iint_D \left( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \right) dA. \end{aligned}

The left side is the circulation of $\mathbf{F} = \langle P, Q \rangle$ around $C$ , and the integrand on the right is the scalar curl.

In plain words Total circulation around the edge equals the total curl inside. Reverse the direction of $C$ and the left side changes sign, so the orientation is part of the theorem.

Example Take $\mathbf{F} = \langle xy, x^2 \rangle$ on the unit square. Here $Q_x - P_y = 2x - x = x$ , so

\iint_D x \, dA = \int_0^1 \!\! \int_0^1 x \, dx \, dy = \frac{1}{2}.

Directly, the bottom edge has $y = 0$ and gives $0$ . The right edge has $x = 1$ , $dx = 0$ and gives $\int_0^1 1 \, dy = 1$ . The top edge has $y = 1$ and runs from $x = 1$ to $x = 0$ , giving

\int_1^0 x \, dx = -\frac{1}{2}.

The left edge has $x = 0$ and gives $0$ . The total is $\frac{1}{2}$ , as promised.

Flux form

Definition Under the same hypotheses, let $\mathbf{n}$ be the outward unit normal to $C$ , which for positive orientation is the unit tangent turned clockwise by a right angle. Then

\begin{aligned} \oint_C \mathbf{F} \cdot \mathbf{n} \, ds &= \oint_C P \, dy - Q \, dx \\[0.4em] &= \iint_D \left( P_x + Q_y \right) dA. \end{aligned}

The left side is the outward flux of $\mathbf{F}$ across $C$ , and the integrand on the right is the divergence.

In plain words Net flow out through the edge equals the total source strength inside. Circulation and flux are different measurements of the same field, so keep them apart.

Example Let $\mathbf{F} = \langle x, y \rangle$ and let $D$ be the disk of radius $2$ . The divergence is $2$ , so the right side is $2 \cdot 4\pi = 8\pi$ . Directly, on the circle of radius $2$ the outward normal is $\langle x, y \rangle / 2$ , so $\mathbf{F} \cdot \mathbf{n} = 2$ , and the circle has length $4\pi$ . The flux is $2 \cdot 4\pi = 8\pi$ .

Simple regions and regions with holes

Definition The proof first handles simple regions, which are both of type I, lying between two graphs $y = g_1(x)$ and $y = g_2(x)$ , and of type II, lying between $x = h_1(y)$ and $x = h_2(y)$ . It extends to any region that can be cut into finitely many simple regions, because the line integrals along the cuts are traversed once in each direction and cancel.

A region with holes is handled the same way. Let $C_0$ be a piecewise smooth simple closed curve, and let $C_1, \dots, C_m$ be piecewise smooth simple closed curves inside $C_0$ that do not meet $C_0$ or each other, none of them inside another. Let $D$ be the closed region inside $C_0$ and outside every $C_j$ . Its boundary is the union of all these curves, and the positive orientation keeps $D$ on the left, which means $C_0$ counterclockwise and each inner curve $C_j$ clockwise. If $P$ and $Q$ are $C^1$ on an open set containing $D$ , then both forms of the theorem hold with the line integral taken as the sum over all boundary curves. With every curve written counterclockwise, the circulation form reads

\begin{aligned} &\oint_{C_0} \mathbf{F} \cdot d\mathbf{r} - \sum_{j=1}^{m} \oint_{C_j} \mathbf{F} \cdot d\mathbf{r} \\[0.4em] &\quad = \iint_D \left( Q_x - P_y \right) dA. \end{aligned}

The field may be undefined inside the holes, since the holes are not part of $D$ .

In plain words Holes are allowed as long as the field is smooth on what remains, including the inner edges. Traverse the inner curves clockwise and add them in.

Example Let $\mathbf{F} = \langle -y, x \rangle / (x^2 + y^2)$ and let $D$ be the annulus $1 \le x^2 + y^2 \le 4$ . On the plane minus the origin $Q_x - P_y = 0$ . Let $C_1$ and $C_2$ be the circles of radius $1$ and $2$ about the origin, both counterclockwise. Green’s theorem gives

\oint_{C_2} \mathbf{F} \cdot d\mathbf{r} - \oint_{C_1} \mathbf{F} \cdot d\mathbf{r} = 0.

The minus sign comes from traversing the inner circle clockwise. Each counterclockwise circle of radius $R$ gives

\int_0^{2\pi} \left( \sin^2 t + \cos^2 t \right) dt = 2\pi,

and indeed $2\pi - 2\pi = 0$ . The same argument, with a small circle about the origin as the hole, shows that every positively oriented, piecewise smooth, simple closed curve with the origin inside gives circulation $2\pi$ .

Area formula

Definition Let $D$ satisfy the hypotheses of Green’s theorem with positively oriented boundary $C$ . Choosing fields with $Q_x - P_y = 1$ gives

\begin{aligned} \operatorname{area}(D) &= \oint_C x \, dy = -\oint_C y \, dx \\[0.4em] &= \frac{1}{2} \oint_C x \, dy - y \, dx. \end{aligned}

In plain words You can find an area by walking around its edge. This is how a planimeter works.

Example The ellipse $x = 3\cos t$ , $y = 2\sin t$ for $0 \le t \le 2\pi$ is traversed counterclockwise. The third formula gives

\frac{1}{2} \int_0^{2\pi} \left( 6\cos^2 t + 6\sin^2 t \right) dt = 6\pi,

which matches the formula $\pi a b$ with $a = 3$ and $b = 2$ .

Check yourself

Use Green’s theorem to find the circulation of $\mathbf{F} = \langle y^2, 3xy \rangle$ around the triangle with vertices $(0, 0)$ , $(1, 0)$ , $(0, 1)$ , oriented counterclockwise.

Show the answer

The field is $C^1$ everywhere and the boundary is positively oriented and piecewise smooth. The scalar curl is $3y - 2y = y$ , so

\int_0^1 \!\! \int_0^{1 - x} y \, dy \, dx = \frac{1}{6}.

Find the area enclosed by the astroid $x = \cos^3 t$ , $y = \sin^3 t$ , $0 \le t \le 2\pi$ .

Show the answer

The astroid is a simple closed curve traversed counterclockwise. It is piecewise smooth with joining points $t = 0, \frac{\pi}{2}, \pi, \frac{3\pi}{2}, 2\pi$ . On each quarter the velocity $\langle -3\cos^2 t \sin t, 3\sin^2 t \cos t \rangle$ is continuous and nonzero inside, and it vanishes at the joining points, which are the four cusps. This is exactly what the definition allows. Here $x y' - y x' = 3\sin^2 t \cos^2 t$ , so the area is

\frac{1}{2} \int_0^{2\pi} 3 \sin^2 t \cos^2 t \, dt = \frac{3\pi}{8}.

For $\mathbf{F} = \langle -y, x \rangle / (x^2 + y^2)$ the scalar curl is $0$ , yet the circulation around the unit circle is $2\pi$ . Which hypothesis of Green’s theorem fails on the unit disk?

Show the answer

The field must be $C^1$ on an open set containing the whole disk. It is undefined at the origin, which lies inside, so the theorem does not apply. On regions that exclude the origin, such as an annulus, it applies and gives consistent results.

Green’s Theorem

Key terms

Positively oriented piecewise smooth boundary

A C1 field on a neighborhood of the region

Circulation form

Flux form

Simple regions and regions with holes

Area formula

Check yourself

How this lesson connects

Resources for this lesson

Vector field explorer