$\int_\gamma P\,dx+Q\,dy$ as the area of a domain

Question

Let $A$ be a domain in $\mathbb R^2$ whose boundary $\gamma $ is a smooth positively oriented curve.

Find two functions $P,Q:\mathbb R^2\to \mathbb R$ such that $\int_\gamma P\,dx+Q\,dy$ is the area of $A$.

Is it sufficient to apply Green's theorem and find $P$ and $Q$ with $Q_x-P_y=1$? So, will $Q(x,y)=x, P(x,y)=x$ do the job?

Answer 1

Given the line integral over a smooth, closed, positively oriented curve $\gamma$, Green's Theorem relates this line integral to the double integral over the region $D$ bound by the closed curve $\gamma$.

Green's Theorem: $$\int_\gamma P\ dx+Q\ dy=\iint_D\bigg(\frac{\partial{Q}}{\partial{x}}-\frac{\partial{P}}{\partial{y}}\bigg)dA$$

By definition, we see $$\iint_D1 \, dA=\text{area of region $D$ enclosed by } \gamma$$

All that is required is to select two functions $P(x,y)$ and $Q(x,y)$ such that $$\frac{\partial{Q}}{\partial{x}}-\frac{\partial{P}}{\partial{y}}=1$$

As Frederico points out, there are infinitely many functions to select!

Answer 2

Yes, that's right. You can choose $P(x,y) = x$ and $Q(x,y) = x$. Note that you can actually make many different choices, depending on the path. For example, also $$Q(x,y) = \frac{x}{2}, \, P(x,y) = -\frac{y}{2}$$ would do the job.