# Arbitrary area function

If I wish to express the area of any closed curve in $$R^2$$ as a function, is there a more efficient way to do so than to use integrals. I ask this because, in the case where this closed curve is something like the one below:

Then I feel as if it would simply be too cumbersome to describe the area of this curve using the antiderivatives of it's piecewise functions. So, is it formally acceptable to simply define an area function for such a closed, simple curve using the function $$A: C \rightarrow R$$. If it is formally acceptable to define such a function, then can this function be assumed to be continuous (when $$C$$ is continuous)?

• I think to define continuity of $A$ you need to define what an arbitrarily small change in $C$ means—continuity of $A$ is the ability to make $A(C)$ change by as small an amount aa we like by making a small enough change to $C$. – timtfj Jan 7 '19 at 19:09

There are various theorems and formulas of this kind, e.g., the following: If $$\gamma:\quad t\mapsto \gamma(t)=\bigl(x(t),y(t)\bigr)\qquad(0\leq t\leq T)$$ is a simply closed curve in the $$(x,y)$$-plane that bounds a certain shape $$\Omega\subset{\mathbb R}^2$$ counterclockwise then $${\rm area}(\Omega)=\int_0^T x(t)\,y'(t)\>dt\ ,\tag{1}$$ or $${\rm area}(\Omega)={1\over2}\int_0^T \bigl(x(t)\,y'(t)-x'(t)\,y(t)\bigr)\>dt\ .$$ In order to prove $$(1)$$ one applies Green's theorem to the vector field $${\bf F}(x,y):=(0,x)$$. This field has $${\rm curl}({\bf F})=Q_x-P_y\equiv1$$. It follows that \eqalign{{\rm area}(\Omega)&=\int_\Omega 1\>{\rm d}(x,y)=\int_\Omega {\rm curl}({\bf F})\>{\rm d}(x,y)\cr &=\int_{\partial\Omega}{\bf F}({\bf z})\cdot d{\bf z}=\int_{\partial \Omega}{\bf F}\bigl(\gamma(t)\bigr)\cdot\gamma'(t)\>dt\cr &=\int_0^T x(t)\,y'(t)\>dt\ .\cr}