Area of a region as a line integral

Statement

Let $C$ be a positively oriented, piecewise smooth, simple closed curve 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 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 $1$.