2013 AMC 10A Problem 20

Consider one quarter of the swept region and multiply its area by $4$. The sector has angle $45^\circ$ and radius $\frac{\sqrt2}{2}$, so its area is $\frac18\pi\left(\frac{\sqrt2}{2}\right)^2=\frac{\pi}{16}$. The two right-triangle pieces in that quarter have areas $\frac12\cdot\frac12\cdot\frac12=\frac18$ and $\frac12\left(\frac{\sqrt2-1}{2}\right)^2$. Multiplying the quarter-area sum by $4$ gives $2-\sqrt2+\frac{\pi}{4}$. Thus, C is the correct answer.