---
tag: 09CBQB6V67
kind: statement
role: corollary
name: Area of a region as a line integral
aliases: [area-by-greens-theorem]
msc: [26B20]
uses: [0HR3N02317, 0YS8RX87F4, 0D032NSZKY]
sources:
  - url: https://en.wikipedia.org/wiki/Green%27s_theorem
    locator: Area calculation
---

## Statement

Let $C$ be a positively oriented, piecewise smooth, simple [[closed curve|0D032NSZKY]] in $\mathbb{R}^2$ that encloses the bounded region $D$.
Then the area $A$ of $D$ satisfies
$$A = \oint_C x \, dy = -\oint_C y \, dx = \frac{1}{2} \oint_C (x \, dy - y \, dx).$$

## Remarks

Proof: apply [[Green's theorem|0HR3N02317]] with $(L, M)$ equal to $(0, x)$, $(-y, 0)$ and $(-y/2, x/2)$. In each case $\partial M / \partial x - \partial L / \partial y = 1$.
Example: the ellipse $x = a \cos t$, $y = b \sin t$, $t \in [0, 2\pi]$, with $a, b > 0$, gives $A = \tfrac{1}{2} \int_0^{2\pi} a b \, dt = \pi a b$.
For a polygon with vertices $(x_1, y_1), \dots, (x_n, y_n)$ in counterclockwise order, the formula gives $A = \tfrac{1}{2} \sum_{k=1}^n (x_k y_{k+1} - x_{k+1} y_k)$, with $(x_{n+1}, y_{n+1}) = (x_1, y_1)$.

## Explanations

### L1 human

You can find the area of a region by going once around its boundary, without looking inside.
This is a direct use of Green's theorem with a field whose curl is $1$.