---
title: Green's Theorem, Curl, and Divergence
module: Multiple Integrals and Vector Calculus
moduleNumber: 12
lessonNumber: 4
order: 1204
summary: >
  Green's Theorem equates the line integral of a field around a positively
  oriented closed curve with a double integral over the enclosed region, turning
  a boundary computation into an area computation and vice versa. Curl measures
  local circulation and divergence measures local outflow; the two vector forms
  of Green's Theorem express the boundary integral as the integrated curl or
  divergence, the planar case of Stokes' and the Divergence Theorem.
topics: [Multiple Integrals and Vector Calculus]
draft: false
sources:
  - book: Stewart
    ref: "Ch. 16 — Vector Calculus; §16.4 Green's Theorem"
  - book: Stewart
    ref: "§16.5 Curl and Divergence"
---

The [Fundamental Theorem for Line Integrals](/calculus/multiple-integrals-and-vector-calculus/vector-fields-and-line-integrals)
handled conservative fields by reducing a line integral to endpoint values.
Green's Theorem handles every field but requires a closed curve: it equates the
circulation of $\vec{F}$ around a closed curve $C$ with a double integral of
a derivative quantity over the region $C$ encloses. That single equation
explains the [component test](/calculus/multiple-integrals-and-vector-calculus/vector-fields-and-line-integrals)
for conservative fields, computes areas from boundary data, and, once its two
derivative quantities are named **curl** and **divergence**, generalizes into
Stokes' Theorem and the Divergence Theorem.

## Green's Theorem

Fix a region $D$ in the plane bounded by a simple closed curve $C$. The boundary
gets an orientation.

> **Definition (Positive orientation).** The **positive orientation** of a simple
> closed curve $C$ bounding a region $D$ is the counterclockwise traversal — the
> one that keeps $D$ on the left as the curve is traced. It is written $\partial D$.

With that convention the theorem reads as follows.

> **Theorem (Green).** Let $C$ be a positively oriented, piecewise-smooth, simple
> closed curve bounding a region $D$, and let $P, Q$ have continuous partial
> derivatives on an open set containing $D$. Then
> $$
> \oint\limits_C P\,\d x + Q\,\d y = \iint\limits_D \left( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \right) \d A .
> $$

The left side integrates the field around the boundary; the right side integrates
a derivative of the field over the interior. It is the two-dimensional counterpart
of the Fundamental Theorem of Calculus $\int_a^b F'(x)\,\d x = F(b) - F(a)$: an
integral of a derivative over a region equals an integral of the field over the
oriented boundary of that region.

$$
% caption: Green's Theorem: the circulation of $\mathbf{F}$ around the
% counterclockwise boundary $C$ equals the double integral of $Q_x-P_y$ over the
% enclosed region $D$.
\begin{tikzpicture}[scale=1.0]
\definecolor{acc}{HTML}{4A6FA5}
% region D
\fill[acc!10] (1.0,0.6) .. controls (2.6,-0.3) and (4.4,0.4) .. (4.6,1.8)
  .. controls (4.7,3.0) and (2.4,3.3) .. (1.0,2.6)
  .. controls (0.2,2.2) and (0.3,1.1) .. (1.0,0.6) -- cycle;
\draw[acc, very thick] (1.0,0.6) .. controls (2.6,-0.3) and (4.4,0.4) .. (4.6,1.8)
  .. controls (4.7,3.0) and (2.4,3.3) .. (1.0,2.6)
  .. controls (0.2,2.2) and (0.3,1.1) .. (1.0,0.6) -- cycle;
\node[font=\small] at (2.5,1.5) {$D$};
% orientation arrows on boundary (counterclockwise)
\draw[acc, very thick, ->] (3.9,0.55) -- (4.35,1.0);
\draw[acc, very thick, ->] (2.1,3.15) -- (1.5,3.0);
\draw[acc, very thick, ->] (0.35,1.4) -- (0.45,1.9);
\node[acc, font=\small, anchor=west] at (4.4,1.0) {boundary $C$};
\node[font=\small, anchor=north] at (2.5,-0.5) {circulation equals area integral};
\end{tikzpicture}
$$

**Green's Theorem and the component test.** For a conservative field $Q_x = P_y$, so the
integrand on the right vanishes and every closed-loop integral is zero — exactly
the closed-loop characterization of conservative fields. Green's Theorem is the
reason the cross-partial test detects conservativeness on a simply connected
domain.

> **Worked example.** Evaluate $\oint\limits_C (x^4 + y)\,\d x + (2x - y^3)\,\d y$ where $C$
> is the positively oriented unit circle. Here $P = x^4 + y$ and $Q = 2x - y^3$, so
> $Q_x - P_y = 2 - 1 = 1$, and
>
> $$
> \oint\limits_C P\,\d x + Q\,\d y = \iint\limits_D 1\,\d A = \text{area of the unit disk} = \pi .
> $$

Parametrizing the circle and integrating directly would involve four unpleasant
trigonometric integrals; Green's Theorem replaces them with the area of the disk.

### Area from the boundary

Run the theorem backward. Choosing $P, Q$ so that $Q_x - P_y = 1$ turns the double
integral into the area of $D$, computable from the boundary alone. Three standard
choices:

$$
A(D) = \oint\limits_C x\,\d y = -\oint\limits_C y\,\d x = \tfrac{1}{2}\oint\limits_C x\,\d y - y\,\d x .
$$

The symmetric third form is what a planimeter mechanizes: trace the boundary and
read off the enclosed area. For the ellipse $\vec{r}(t) = \langle a\cos t, b\sin t\rangle$,
$0 \le t \le 2\pi$,

$$
A = \tfrac{1}{2}\oint\limits_C x\,\d y - y\,\d x = \tfrac{1}{2}\int_0^{2\pi}\big(a\cos t \cdot b\cos t - b\sin t\cdot(-a\sin t)\big)\,\d t = \tfrac{1}{2}\int_0^{2\pi} ab\,\d t = \pi ab .
$$

### The proof idea

Green's Theorem holds for a simple region (both type I and type II) by two direct
computations, one matching $\oint\limits_C P\,\d x$ to $-\iint\limits_D P_y\,\d A$ and the other
matching $\oint\limits_C Q\,\d y$ to $\iint\limits_D Q_x\,\d A$. A general region is cut into simple
pieces; along each internal cut the boundary is traversed twice in opposite
directions, so those line integrals cancel and only the outer boundary remains.

$$
% caption: A region that is not simple is split by internal cuts into simple
% pieces; each cut is traversed once in each direction, so the interior line
% integrals cancel and only the outer boundary remains.
\begin{tikzpicture}[scale=1.0]
\definecolor{acc}{HTML}{4A6FA5}
% an L-shaped / non-simple region split into two
\fill[acc!10] (0,0) -- (4,0) -- (4,1.6) -- (2,1.6) -- (2,3) -- (0,3) -- cycle;
\draw[acc, very thick] (0,0) -- (4,0) -- (4,1.6) -- (2,1.6) -- (2,3) -- (0,3) -- cycle;
% internal cut
\draw[black, very thick, dashed] (2,0) -- (2,1.6);
\node[font=\small] at (0.95,0.8) {$D_1$};
\node[font=\small] at (3.0,0.8) {$D_2$};
% cancelling arrows on the cut
\draw[acc, thick, ->] (1.82,0.4) -- (1.82,1.2);
\draw[acc, thick, ->] (2.18,1.2) -- (2.18,0.4);
\node[font=\small, anchor=west] at (2.25,1.5) {cuts cancel};
\end{tikzpicture}
$$

## Curl

The quantity $Q_x - P_y$ that appears in Green's Theorem measures how much the
field circulates near a point. In three dimensions the corresponding object is a
vector, the **curl**, whose $\hat{k}$-component is exactly $Q_x - P_y$.

> **Definition (Curl).** For a vector field $\vec{F} = P\,\hat\imath + Q\,\hat\jmath + R\,\hat{k}$
> with differentiable components, the **curl** is
> $$
> \operatorname{curl}\vec{F} = \left( \frac{\partial R}{\partial y} - \frac{\partial Q}{\partial z} \right)\hat\imath + \left( \frac{\partial P}{\partial z} - \frac{\partial R}{\partial x} \right)\hat\jmath + \left( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \right)\hat{k} .
> $$

Introduce the **del** operator
$\nabla = \hat\imath\,\partial_x + \hat\jmath\,\partial_y + \hat{k}\,\partial_z$.
Then the curl is the formal cross product

$$
\operatorname{curl}\vec{F} = \nabla \times \vec{F} = \begin{vmatrix} \hat\imath & \hat\jmath & \hat{k} \\[1mm] \partial_x & \partial_y & \partial_z \\[1mm] P & Q & R \end{vmatrix} ,
$$

a mnemonic that reproduces all three components. Physically, if $\vec{F}$ is a
fluid velocity field, $\operatorname{curl}\vec{F}$ at a point is twice the
local angular velocity: drop a tiny paddle wheel into the flow and it spins about
the axis $\operatorname{curl}\vec{F}$ at a rate proportional to its magnitude.
Where the curl is zero, the flow is **irrotational** — the paddle wheel translates
but does not turn.

$$
% caption: The curl at a point is the axis and rate of local rotation: a paddle
% wheel in the flow spins about $\operatorname{curl}\mathbf{F}$, fast where the
% tangential circulation is strong.
\begin{tikzpicture}[scale=1.0]
\definecolor{acc}{HTML}{4A6FA5}
% swirling field arrows
\foreach \a in {0,45,...,315} {
  \draw[acc, thick, ->] ({1.6*cos(\a)},{1.6*sin(\a)}) -- ({1.6*cos(\a)-0.6*sin(\a)},{1.6*sin(\a)+0.6*cos(\a)});
}
% paddle wheel at center
\draw[black, thick] (-0.55,0) -- (0.55,0);
\draw[black, thick] (0,-0.55) -- (0,0.55);
\fill[black] (0,0) circle (1.6pt);
\draw[black, thick, ->] (0.35,0.62) arc (60:120:0.7);
% curl vector out of page (drawn as circle-dot)
\draw[black, thick] (2.6,1.9) circle (0.22);
\fill[black] (2.6,1.9) circle (1.6pt);
\node[font=\small, anchor=west] at (2.95,1.9) {curl (out of page)};
\end{tikzpicture}
$$

The curl connects to conservative fields. Taking the curl of a gradient always
gives zero, because the mixed partials cancel in pairs by Clairaut's Theorem.

> **Theorem (Curl of a gradient).** If $f$ has continuous second partials, then
> $\operatorname{curl}(\nabla f) = \vec{0}$. Consequently a conservative field
> is irrotational. On a simply connected domain the converse holds:
> $\operatorname{curl}\vec{F} = \vec{0}$ implies $\vec{F}$ is
> conservative.

This is the three-dimensional component test. In the plane it reduces to
$Q_x - P_y = 0$, the planar condition on $P_y = Q_x$.

## Divergence

Curl measures rotation; **divergence** measures expansion. It is the formal dot
product of del with the field, a scalar.

> **Definition (Divergence).** For $\vec{F} = P\,\hat\imath + Q\,\hat\jmath + R\,\hat{k}$,
> the **divergence** is
> $$
> \operatorname{div}\vec{F} = \nabla\cdot\vec{F} = \frac{\partial P}{\partial x} + \frac{\partial Q}{\partial y} + \frac{\partial R}{\partial z} .
> $$

If $\vec{F}$ is a fluid velocity field, $\operatorname{div}\vec{F}$ at a
point is the net rate at which fluid flows out of a tiny box around the point, per
unit volume. Positive divergence marks a **source** (fluid produced, box
emptying); negative divergence marks a **sink**; zero divergence means
**incompressible** flow, whatever enters a region also leaves it.

$$
% caption: Divergence as net outflow: over a small box, the flux leaving through
% the faces minus the flux entering, per unit volume, is
% $\operatorname{div}\mathbf{F}$; positive marks a source.
\begin{tikzpicture}[scale=1.0]
\definecolor{acc}{HTML}{4A6FA5}
% small box (oblique)
\draw[black] (1.4,1.0) rectangle (2.8,2.4);
\draw[black] (1.4,1.0) -- (2.0,1.4);
\draw[black] (2.8,1.0) -- (3.4,1.4);
\draw[black] (2.8,2.4) -- (3.4,2.8);
\draw[black] (1.4,2.4) -- (2.0,2.8);
\draw[black] (2.0,1.4) -- (3.4,1.4) -- (3.4,2.8) -- (2.0,2.8) -- cycle;
% outward flux arrows
\draw[acc, thick, ->] (2.4,2.6) -- (2.4,3.3);
\draw[acc, thick, ->] (3.5,1.9) -- (4.2,1.9);
\draw[acc, thick, ->] (1.2,1.7) -- (0.5,1.7);
\draw[acc, thick, ->] (2.1,0.9) -- (2.1,0.2);
\node[font=\small, anchor=south] at (2.4,3.35) {out};
\node[font=\small, anchor=west] at (4.25,1.9) {out};
\node[font=\small] at (2.4,1.75) {source};
\end{tikzpicture}
$$

Divergence pairs with curl through a second identity: the divergence of any curl
is zero.

> **Theorem (Divergence of a curl).** If $\vec{F}$ has continuous second
> partials, then $\operatorname{div}(\operatorname{curl}\vec{F}) = 0$.

The six terms cancel in pairs by Clairaut's Theorem. A field that is itself a curl
therefore has no sources or sinks; this is why magnetic fields, which are curls of
a vector potential, are divergence-free.

## The two vector forms of Green's Theorem

Green's Theorem, written with curl and divergence, is the two-dimensional case of
Stokes' Theorem and the Divergence Theorem. Regard the planar field
$\vec{F} = P\,\hat\imath + Q\,\hat\jmath$ as living in space with zero
$\hat{k}$-component.

- **Circulation form.** Since $(\operatorname{curl}\vec{F})\cdot\hat{k} = Q_x - P_y$,
  Green's Theorem is
  $$
  \oint\limits_C \vec{F}\cdot \d\vec{r} = \iint\limits_D (\operatorname{curl}\vec{F})\cdot\hat{k}\,\d A .
  $$
  The circulation around the boundary equals the integrated curl over the region.
  Lifting $C$ and $D$ off the plane into space gives
  [Stokes' Theorem](/calculus/multiple-integrals-and-vector-calculus/stokes-and-the-divergence-theorem).

- **Flux form.** With the outward normal $\hat{n}$ on $C$, a parallel
  computation gives
  $$
  \oint\limits_C \vec{F}\cdot\hat{n}\,\d s = \iint\limits_D \operatorname{div}\vec{F}\,\d A .
  $$
  The outward flux across the boundary equals the integrated divergence over the
  region. Replacing the plane region by a solid and its boundary curve by a
  boundary surface gives
  [the Divergence Theorem](/calculus/multiple-integrals-and-vector-calculus/stokes-and-the-divergence-theorem).

| Operator | Formula | Type | Measures | Zero when |
| --- | --- | --- | --- | --- |
| gradient | $\nabla f$ | vector | steepest increase of $f$ | $f$ constant |
| curl | $\nabla\times\vec{F}$ | vector | local circulation | irrotational (conservative) |
| divergence | $\nabla\cdot\vec{F}$ | scalar | local outflow | incompressible |

The two identities $\operatorname{curl}(\nabla f) = \vec{0}$ and
$\operatorname{div}(\operatorname{curl}\vec{F}) = 0$ organize these three
operators into a chain: gradient, then curl, then divergence, with each
composition vanishing.[^stewart164]

[^stewart164]: Stewart, §16.4 — Green's Theorem; §16.5 — Curl and Divergence. Positive orientation, the equivalence of circulation with integrated curl and flux with integrated divergence, the identities $\operatorname{curl}(\nabla f)=\vec 0$ and $\operatorname{div}(\operatorname{curl}\vec F)=0$, and the area-from-boundary formulas.