jmc

The center of the disk lies in a region $R$, consisting of all points within 1 unit of both $A$ and $B$. Let $C$ and $D$ be the points of intersection of the circles of radius 1 centered at $A$ and $B$. Because $△ABC$ and $△ABD$ are equilateral, arcs $CAD$ and $CBD$ are each $120^{\circ }$. Thus the sector bounded by $\overline{BC}$, $\overline{BD}$, and arc $CAD$ has area $\pi /3$, as does the sector bounded by $\overline{AC}$, $\overline{AD}$, and arc $CBD$. The intersection of the two sectors, which is the union of the two triangles, has area $\sqrt{3}/2$, so the area of $R$ is $$\frac{2\pi }{3}-\frac{\sqrt{3}}{2}.$$

The region $S$ consists of all points within 1 unit of $R$. In addition to $R$ itself, $S$ contains two $60^{\circ }$ sectors of radius 1 and two $120^{\circ }$ annuli of outer radius 2 and inner radius 1. The area of each sector is $\pi /6$, and the area of each annulus is $$\frac{\pi }{3}(2^2-1^2)=\pi .$$Therefore the area of $S$ is $$\left(\frac{2\pi }{3}-\frac{\sqrt{3}}{2}\right)+2\left(\frac{\pi }{6}+\pi \right)=\boxed{3\pi -\frac{\sqrt{3}}{2}}.$$