# Green's Theorem: Statement, Orientation and Worked Examples

> Green's theorem turns a line integral around a closed curve into a double integral over the region inside. Statement, hypotheses, area formula, examples.

Source: https://practicecalculus.com/theorems/greens-theorem
Updated: 2026-11-03

$$ \oint_C P\,dx + Q\,dy = \iint_D \left(\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}\right) dA $$

## At a glance

- **Hypotheses:** $C$ is a positively oriented, piecewise smooth, simple closed curve bounding a region $D$, and $P$, $Q$ have continuous partial derivatives on an open region containing $D$
- **Positive orientation:** Counterclockwise, so the region stays on your left
- **Integrand:** $Q_x - P_y$, the scalar curl of $\langle P, Q \rangle$
- **Area formula:** $A = \frac{1}{2}\oint_C x\,dy - y\,dx$
- **Generalizes to:** Stokes' theorem in three dimensions
- **Course:** Calculus 3

## Key takeaways

- Green's theorem converts a closed line integral into a double integral of $Q_x - P_y$ over the enclosed region.
- The curve must be traversed counterclockwise; clockwise flips the sign.
- $P$ and $Q$ need continuous partial derivatives on the whole region, including any holes. A singularity inside breaks the theorem.
- Choosing $Q_x - P_y = 1$ turns a line integral into an area formula.

**Green's theorem** relates a line integral around a closed curve in the plane to a double integral over the region it encloses. It says the total circulation around the boundary equals the sum of all the small circulations inside.

## Statement of Green's theorem

Let

1. $C$ be a **positively oriented** (counterclockwise), **piecewise smooth**, **simple closed** curve in the plane, and
2. $D$ be the region bounded by $C$, and
3. $P(x, y)$ and $Q(x, y)$ have **continuous partial derivatives** on an open region that contains $D$.

Then

$$\oint_C P\,dx + Q\,dy = \iint_D \left(\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}\right) dA$$

In vector form, with $\mathbf{F} = \langle P, Q \rangle$, the left side is the circulation $\oint_C \mathbf{F}\cdot d\mathbf{r}$ and the integrand on the right is the scalar curl $Q_x - P_y$.

## What Green's theorem means

Chop $D$ into tiny squares and add up the circulation of $\mathbf{F}$ around each one. Along every interior edge, two neighboring squares traverse it in opposite directions, so those contributions cancel. Only the outer boundary survives. Meanwhile, the circulation around a tiny square is approximately $(Q_x - P_y)$ times its area. Adding up both ways gives Green's theorem.

```graph
title: 'Green''s theorem region: between y = x² and y = x for 0 ≤ x ≤ 1'
caption: 'The boundary $C$ runs counterclockwise: along the parabola $y = x^2$ from $(0, 0)$ to $(1, 1)$, then back along the line $y = x$. The region $D$ stays on the left.'
x: [-0.2, 1.3]
y: [-0.2, 1.3]
functions:
  - expr: x
    label: y = x (top, traversed right to left)
    domain: [0, 1]
  - expr: x^2
    label: y = x² (bottom, traversed left to right)
    domain: [0, 1]
shade: {expr: x, from: 0, to: 1, below: x^2}
points:
  - [0, 0, '(0, 0)']
  - [1, 1, '(1, 1)']
```

## Why each hypothesis matters

**Orientation.** Traversing $C$ clockwise reverses the sign of the line integral, while the double integral does not change. Always check the direction.

**Continuity everywhere inside.** The vortex field

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

has $Q_x - P_y = 0$ at every point except the origin, where it is undefined. Around the unit circle $x = \cos t$, $y = \sin t$, the line integral is $\int_0^{2\pi} (\sin^2 t + \cos^2 t)\,dt = 2\pi$, not $0$. Green's theorem does not apply because the origin, inside $D$, is a point where $P$ and $Q$ are not even defined.

**Closed curve.** For a curve that does not close up, there is no enclosed region, so the theorem says nothing.

## Worked examples

**Example 1: compute $\oint_C xy\,dx + x^2\,dy$ around the region in the graph.**

**Step 1.** $P = xy$ and $Q = x^2$, so $Q_x - P_y = 2x - x = x$.

**Step 2.** The region is $x^2 \le y \le x$ for $0 \le x \le 1$:

$$\iint_D x\,dA = \int_0^1 \int_{x^2}^{x} x\,dy\,dx = \int_0^1 (x^2 - x^3)\,dx = \frac{1}{3} - \frac{1}{4} = \frac{1}{12}$$

**Check directly.** Along $y = x^2$ ($x = t$, $0 \to 1$) the integrand is $t^3\,dt + t^2(2t\,dt) = 3t^3\,dt$, contributing $\frac{3}{4}$. Along $y = x$ back from $(1, 1)$ to $(0, 0)$ it is $2t^2\,dt$ with $t$ from $1$ to $0$, contributing $-\frac{2}{3}$. The total is $\frac{3}{4} - \frac{2}{3} = \frac{1}{12}$. Same answer, more work.

**Example 2: area of an ellipse.** Choose $P = -\frac{y}{2}$ and $Q = \frac{x}{2}$, so $Q_x - P_y = 1$ and the double integral is the area:

$$A = \frac{1}{2}\oint_C x\,dy - y\,dx$$

For the ellipse $x = a\cos t$, $y = b\sin t$, $0 \le t \le 2\pi$: $x\,dy - y\,dx = (ab\cos^2 t + ab\sin^2 t)\,dt = ab\,dt$. So $A = \frac{1}{2}\cdot 2\pi ab = \pi ab$.

**Example 3: a messy field, an easy answer.** Evaluate $\oint_C (x^2 - y)\,dx + (x + y^3)\,dy$ around the unit circle, counterclockwise. Here $Q_x - P_y = 1 - (-1) = 2$, so the integral is $2 \cdot \text{area} = 2\pi$.

## Common mistakes

1. **Swapping the order in the integrand.** It is $Q_x - P_y$, not $P_y - Q_x$ or $P_x - Q_y$.
2. **Ignoring orientation.** A clockwise curve gives the negative of the double integral.
3. **Using it across a singularity.** Check that $P$ and $Q$ are smooth on all of $D$, not just on $C$.

## Related theorems and guides

Green's theorem is the flat case of [Stokes' theorem](/theorems/stokes-theorem), and its flux form is the two-dimensional [divergence theorem](/theorems/divergence-theorem). Since $Q_x - P_y = 0$ for gradient fields by [Clairaut's theorem](/theorems/clairauts-theorem), conservative fields have zero circulation. For the double integrals on the right side, see [double integrals](/blog/double-integrals).

## Frequently asked questions

### What does Green's theorem say?

For a positively oriented, piecewise smooth, simple closed curve $C$ bounding a region $D$, and $P$, $Q$ with continuous partial derivatives near $D$, $\oint_C P\,dx + Q\,dy = \iint_D (Q_x - P_y)\,dA$.

### What is positive orientation in Green's theorem?

Counterclockwise around the region, or more generally the direction in which the region stays on your left as you walk along the curve. Going clockwise multiplies the line integral by $-1$.

### When should I use Green's theorem?

Use it when a line integral around a closed curve is tedious, especially when the curve has several pieces, and $Q_x - P_y$ is simple. Also use it in reverse to compute areas with a line integral.

### Why does Green's theorem fail for the vortex field around the origin?

The field $\left\langle \frac{-y}{x^2 + y^2}, \frac{x}{x^2 + y^2} \right\rangle$ has $Q_x - P_y = 0$ away from the origin, but is undefined at the origin. If the curve encloses the origin, the hypothesis fails, and in fact the line integral around the unit circle is $2\pi$, not $0$.