Why does Hausdorff measure go to zero as diameter power increases? For example, why does the Hausdorff measure of a flat disc go to zero when the power that the diameter is raised to (in the definition of the Hausdorff measure) reaches 3?

A flat disc $D$ of radius $1$ has area $\pi.$ For $n\in \Bbb N$ let $S_n$ be a set of open discs, each of diameter $1/n$ or less, with $\cup S_n \supset D$ and  $$\sum_{t\in S_n}A(t)<2\pi,$$ where $A(t)$ is the area of $t.$ 
For $t\in S_n$ let $d(t)$ be the diameter of $t.$ Then $A(t)=\pi d(t)^2/4.$ 
Let $r>0.$ Then  $$\sum_{t\in S_n} \pi d(t)^{2+r}/4=\sum_{t\in S_n}A(t)d(t)^r\leq \sum_{t\in S_n}A(t)(1/n)^r<2\pi(1/n)^r.$$ As $n\to \infty$ we have $2\pi (1/n)^r\to 0.$
