Area with Green's Theorem

Green's theorem can be read backward: a double integral over a region can be computed from a line integral around its edge. The simplest double integral, $\iint_D 1\,dA$, is the area. So any field with $Q_x - P_y = 1$ turns the area of a region into a line integral around its boundary. Surveyors' instruments called planimeters measure areas on a map this way, by tracing the outline. You need the circulation form.

Key idea

For a region $D$ with counterclockwise boundary $C$: $$\text{Area}(D) = \oint_C x\,dy = -\oint_C y\,dx = \frac{1}{2}\oint_C x\,dy - y\,dx.$$ For a polygon with vertices $(x_1, y_1), \ldots, (x_n, y_n)$ in counterclockwise order, this gives the shoelace formula $A = \frac{1}{2}\sum (x_i y_{i+1} - x_{i+1} y_i)$, with $(x_{n+1}, y_{n+1}) = (x_1, y_1)$.

Why it works

Each choice has $Q_x - P_y = 1$: for $\langle 0, x \rangle$, ${}1 - 0$; for $\langle -y, 0 \rangle$, ${}0 - (-1)$; for $\frac{1}{2}\langle -y, x \rangle$, $\frac{1}{2} + \frac{1}{2}$. Green's theorem turns each line integral into $\iint_D 1\,dA$.
Shoelace: on the segment $(x_1, y_1) + t(x_2 - x_1, y_2 - y_1)$, $$x\,dy - y\,dx = [x_1(y_2 - y_1) - y_1(x_2 - x_1)]\,dt = (x_1 y_2 - x_2 y_1)\,dt,$$ so each side contributes $x_1 y_2 - x_2 y_1$.

The field $\frac{1}{2}\langle -y, x \rangle$ around the ellipse $\langle 2\cos t, \sin t \rangle$. The work done around the loop is the area inside: ${}2\pi \approx 6.283$, which is $\pi ab$ with $a = 2$, $b = 1$.

The same idea gives other quantities of a region from its boundary. With $\mathbf{F} = \left\langle 0, \frac{x^2}{2} \right\rangle$, $Q_x - P_y = x$, so $\oint_C \frac{x^2}{2}\,dy = \iint_D x\,dA$, the moment used to find the centroid $\bar{x}$. Software that measures shapes in images works this way: it stores an outline as a list of points and computes areas and centroids from sums over the edges, without ever filling in the inside.

Worked examples

Worked example: An ellipse

Find the area inside $\mathbf{r}(t) = \langle a\cos t, b\sin t \rangle$.

$$x\,dy - y\,dx = (a\cos t)(b\cos t)\,dt - (b\sin t)(-a\sin t)\,dt = ab\,dt.$$
$A = \frac{1}{2}\int_0^{2\pi} ab\,dt = \pi ab$.

Answer: $\pi ab$

Worked example: A triangle

Find the area of the triangle with vertices $(0, 0)$, $(4, 1)$, $(1, 3)$.

The vertices go counterclockwise. Side terms: ${}0 \cdot 1 - 4 \cdot 0 = 0$; ${}4 \cdot 3 - 1 \cdot 1 = 11$; ${}1 \cdot 0 - 0 \cdot 3 = 0$.
$A = \frac{1}{2}(0 + 11 + 0) = \frac{11}{2}$.

Answer: $\frac{11}{2}$

Worked example: A loop

The curve $\mathbf{r}(t) = \langle t^2 - 1, t^3 - t \rangle$, $-1 \le t \le 1$, starts and ends at the origin and makes a loop. Find the area inside.

$$x\,y' - y\,x' = (t^2 - 1)(3t^2 - 1) - (t^3 - t)(2t) = t^4 - 2t^2 + 1 = (t^2 - 1)^2.$$
$A = \frac{1}{2}\int_{-1}^1 (t^2 - 1)^2\,dt = \frac{1}{2} \cdot \frac{16}{15} = \frac{8}{15}$.
The integral came out positive, so this direction of travel is counterclockwise.

Answer: $\frac{8}{15}$

Area with Green's Theorem

Worked examples

Practice

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