Riemann area between two continuous graphs and the disc as a vertically simple region

Let $a<b$, and let $g,f:[a,b]\to \mathbb{R}$ be continuous with $g(x)\le f(x)$ for every $x$. The Riemann area between their graphs is

$\mathrm{area}\{(x,y):a\le x\le b,g(x)\le y\le f(x)\}:={\int }_a^b(f(x)-g(x))dx\mathrm{.}$

The integral exists by A continuous function on $[a,b]$ is Riemann integrable, by Heine-Cantor and Riemann's criterion and is the Darboux integral of The lower and upper Darboux integrals of a bounded $f$ on $[a,b]$ as ${\sup }_{P}L(f,P)$ and ${\inf }_{P}U(f,P)$, Darboux integrability as their equality, and the notation ${\int }_a^{b}f$. The square map is continuous by Sums, scalar multiples, products, absolute values, maxima, minima and quotients with nonvanishing denominator of continuous functions are continuous, as are constants, the identity and every polynomial function and strictly increasing on the nonnegative reals by Monotonicity of $x\mapsto x^n$ and of $n\mapsto a^n$, so Continuous inverse theorem: a continuous injective $f$ on an interval $I$ is a bijection onto the order-convex set $f[I]$, and the inverse $g:f[I]\to I$ is continuous and strictly monotone in the same sense as $f$ makes its inverse square-root function continuous; existence and uniqueness are Square roots exist: a unique $\sqrt{a}\ge 0$ with $(\sqrt{a}{)}^2=a$; the positives are $\{x^2:x\mathrm{\ne }0\}$. Hence, for $r>0$, the functions $x\mapsto \pm \sqrt{r^2-x^2}$ are well defined and continuous on $[-r,r]$, and the closed disc of radius $r$ is the region between them.

This is a local convention for regions between continuous graphs. It does not assign an area to an arbitrary bounded planar set.