8. Let $S$ be the set of discs $D$ contained completely in the set $\{(x, y): y<0\}$ (the region below the $x$-axis) and centered (at some point) on the curve $y=x^{2}-\frac{3}{4}$. What is the area of the union of the elements of $S$ ?
Answer: $\sqrt[2 \pi]{3}+\frac{\sqrt{3}}{4}$ Solution 1. An arbitrary point $\left(x_{0}, y_{0}\right)$ is contained in $S$ if and only if there exists some $(x, y)$ on the curve $\left(x, x^{2}-\frac{3}{4}\right)$ such that $\left(x-x_{0}\right)^{2}+\left(y-y_{0}\right)^{2}0$, we find that we need $-2 x_{0}^{2}-2 y_{0}^{2}+\frac{3}{2}-2 y_{0}>$ $0 \Longleftrightarrow 1>x_{0}+\left(y_{0}+\frac{1}{2}\right)^{2}$.
$S$ is therefore the intersection of the lower half-plane and a circle centered at $\left(0,-\frac{1}{2}\right)$ of radius 1 . This is a circle of sector angle $4 \pi / 3$ and an isosceles triangle with vertex angle $2 \pi / 3$. The sum of these areas is $\frac{2 \pi}{3}+\frac{\sqrt{3}}{4}$.
Solution 2. Let $O=\left(0,-\frac{1}{2}\right)$ and $\ell=\{y=-1\}$ be the focus and directrix of the given parabola. Let $\ell^{\prime}$ denote the $x$-axis. Note that a point $P^{\prime}$ is in $S$ iff there exists a point $P$ on the parabola in the lower half-plane for which $d\left(P, P^{\prime}\right)<d\left(P, \ell^{\prime}\right)$. However, for all such $P, d\left(P, \ell^{\prime}\right)=1-d(P, \ell)=1-d(P, O)$, which means that $P^{\prime}$ is in $S$ iff there exists a $P$ on the parabola for which $d\left(P^{\prime}, P\right)+d(P, O)<1$. It is not hard to see that this is precisely the intersection of the unit circle centered at $O$ and the lower half-plane, so now we can proceed as in Solution 1.