Let $D$ be the set under the graph of $f(x)=x^2$ for $0\leq x\leq 1$, that is, $D= \{ (x_1,x_2) : 0\leq x_1\leq 1, 0\leq x_2\leq x_1^2 \}$. Denote by $A_{r}(y)$ the $2$-dimensional cube of side length $r$ with center $y\in \mathbb{R}^2$.
I want to find a short and simple proof of the following result: There exist constants $c>0, r_0,s_0\in (0,1)$ such that for all $r\in (0,r_0), s\in (0,s_0)$ and every $y\in D$: $|A_{rs}(y) \cap D| \leq c s |A_r(y)\cap D|$.
Here $|\cdot |$ is the Lebesgue measure.
So far, I have a proof using a case analysis. I distinguish cases where parts of the cube are above the graph and parts are under the graph. But there are a lot of cases and I was wondering if anyone has an idea for a simple proof.