13. 1. 5 * Given a $2 \times 3$ rectangle, with vertices at $(0,0),(2,0),(0,3),(2,3)$. Rotate it $90^{\circ}$ clockwise around the point $(2,0)$, then $90^{\circ}$ clockwise around the point $(5,0)$, then $90^{\circ}$ clockwise around the point $(7,0)$, and finally $90^{\circ}$ clockwise around the point $(10,0)$ (at this point, the side that was originally on the $x$-axis is back on the $x$-axis). A point on the rectangle initially at $(1,1)$ traces out a curve during these rotations. Find the area of the region below this curve and above the $x$-axis.
As shown in the figure, the required area includes four right-angled triangles with an area of $\frac{1}{2}$ each, of size $1 \times 1$; four right-angled triangles with an area of 1 each, of size $1 \times 2$; two quarter-circles with an area of $\frac{\pi}{4}(\sqrt{5})^{2}=\frac{5 \pi}{4}$ each; and two quarter-circles with an area of $\frac{\pi}{4}(\sqrt{2})^{2}=\frac{\pi}{2}$ each. Therefore, the total area of the required region is $4 \times \frac{1}{2}+4 \times 1$ $+2 \times \frac{\pi}{2}+2 \times \frac{5 \pi}{4}=6+\frac{7}{2} \pi$