Can i use Green's theorem to calculate the area of an abstract triangle on a plane?

Question

I want to see some examples of Green's theorem used to calculate the area of some simple 2D shapes, but i haven't encountered a lot of them. My goal is to find (or study) a general procedure for achieving this. I proposed myself a simple question:

Let $p_i=(a_i,b_i)$, with $\{p_1,p_2,p_3\}\in \mathbb{R}^2$ be points on a 2-dimensional vector space (you may use Euclidean instead, if you like), and let $\mathcal{C}$ be the only curve (up to isomorphisms) that describes a triangle with $p_i$ as vertices. Also, let $D$ be the region enclosed by $\mathcal{C}$ (i.e $\mathcal{C}=\partial D$). Then, by Green's theorem, there may be some functions $\{L,M\}:\mathbb{R^2\rightarrow R^2}$ such that: $$A=\iint_D dD=\oint_{\mathcal{C}}(L\:dx+M\:dy)$$ Where $A$ is the area of the triangle.

My question is: can i obtain $L$ and $M$ functions for arbitrary points in the plane $p_i$?

Answer 1

Sure. There are several choices, the simplest one being $L(x,y)=-y/2$ and $M(x,y)=x/2$. According to Green's theorem, your line integral is equal to the double integral of $\partial M/\partial x-\partial L/\partial y=1$ over $D$, which is precisely your desired area.

Answer 2

The following generic formula for $m-$dimensional functional $f:\mathbb{R^m}\to\mathbb{R}$ defined as the divergence of $n$-th power of the radial vector: \begin{align} \nabla\cdot(r^n\mathbf{\hat r})&=\nabla\cdot(r^{n-1}\mathbf{r})=r^{n-1}\nabla\cdot\mathbf{r}+\mathbf{r}\cdot\nabla r^{n-1}=(m+n-1)r^{n-1} \end{align} can be used for a 2D polygon with $N$ vertices $\mathbf{r}=(x_i,y_i)=[(x_1,y_1)\ldots,(x_i,y_i),\ldots,(x_N,y_N)]$., where $i=1,\ldots,N$ \begin{align} \nabla\cdot\mathbf{r}&=2\\ \nabla\cdot(r^n\mathbf{\hat r})&=(n+1)r^{n-1} \end{align} where $\mathbf{\hat r}$ is unit radius vector. The area of the polygon can be defined as \begin{align} A=\iint_Sr^0\text{d}{S}=\frac{1}{2}\iint_S\nabla\cdot (r^{}\mathbf{\hat r})\text{d}{S}=\frac{1}{2}\oint_C\mathbf{\hat n}\cdot (r^{}\mathbf{\hat r})\text{d}{\ell} \end{align} where $\mathbf{\hat n}$ is the outward pointing unit normal of the segment and $\mathbf{r}$ is the tangential of the segment. Let $\text{d}{\mathbf{r}}=(\text{d} x,\text{d} y)$ be the tangential vector along the curve for a positively oriented (i.e. anticlockwise) polygon along the segment, an outward normal would be a vector which points 90° to the right of segment, which can be $(\text{d} y,-\text{d} x)$. The length of this vector or the increment of arclength is $\sqrt{\text{d} x^2 + \text{d} y^2}=\text{d}\ell.$ So $\mathbf{r}=(x,y)$ and $(\text{d} y,-\text{d} x)=\mathbf{\hat n}\text{d}\ell$. The formula is actually the area of the polygon since \begin{align} A&=\frac{1}{2}\oint_C\mathbf{r}\cdot(\mathbf{\hat n}\text{d}{\ell})\\ &=\frac{1}{2}\oint_C(x,y)\cdot(\text{d} y,-\text{d} x)\\ &=\tfrac 12 \oint_C(-y\text{d} x + x\text{d} y)\\ &=\oint_C x\text{d} y=-\oint_C y\text{d} x \end{align} where in the last step the Green's theorem is used and this formula is related to the shoelace formula.

If found the derivation is loose, please do correct me (since I just copied from my earlier notes). Your comments are welcome.