Back in my (early) wannabe days, in the summer between sophomore and junior year, I was studying single-variable calculus. From the AoPS book, I found this extra sidenote.

Consider a positively oriented closed simple curve (a curve that is closed, non-self intersecting, and traced such that region enclosed by the curve is always on the left) parameterized by \((x(t),y(t))\) from \(t_0\) to \(t_1\). Then, the area of the enclosed region is given by: \[\frac{1}{2}\int_{t_0}^{t_1}{x(t)y’(t)-x’(t)y(t)\textrm{ d}t}.\] Back then, I could barely understand the statement, and once I did, I had no clue as to how it would be true. The sidenote said that it was a special case of Green’s Theorem from vector calculus.

Two years later, I can finally say that I know where this comes from.

Green’s theorem states that for a positively oriented closed simple curve \(C\) enclosing a region \(R\), and a vector field \(\mathbf{F}\), \[\oint_{C}{\mathbf{F}\cdot\textrm{d}\mathbf{r}}=\iint_{R}{\nabla\times\mathbf{F}\textrm{ d}A},\] where \(\nabla\times\mathbf{F}\) is the 2D curl (I know, it’s a 3D thing so I’m abusing the notation a little). This is a cool result. For instance, it immediately establishes all the stuff about path-independence of line integrals over conservative fields which have to be curl-free and blah, blah, blah.

If \(\nabla\times\mathbf{F}=1\), then the RHS evaluates to the area of \(R\). So we want a vector field, \(\mathbf{F}=\langle M,N\rangle\), such that \(\frac{\partial N}{\partial x}-\frac{\partial M}{\partial y}=1\). Perhaps the simplest \(\mathbf{F}\) satisfying this is \(\mathbf{F}=\left\langle -\frac{y}{2},\frac{x}{2}\right\rangle\). Then by Green’s theorem, we have \[\begin{split} A&=\oint_{C}{-\frac{y}{2}\textrm{ d}x+\frac{x}{2}\textrm{ d}y}
&=\frac{1}{2}\oint_{C}{\left(x\frac{\textrm{d}y}{\textrm{d}t}-y\frac{\textrm{d}x}{\textrm{d}t}\right)\textrm{d}t}
&=\frac{1}{2}\int_{t_0}^{t_1}{x(t)y’(t)-x’(t)y(t)\textrm{ d}t}, \end{split}\] as desired. \(\square\)

​Cool beans.