Constructing a planar shape with area $\pi^2$

By wrapping a thread around a circle, it's possible to mark off a segment of length $\pi$ as the circumference of a circle of unit diameter, then to construct a square of side $\pi$. For example:

Similarly, using an Archimedean spiral $r = b\theta$, we can mark off a segment of length $2\pi b$ (the radial separation of successive windings), and then construct a square of side $\pi$ (by taking $b = \frac{1}{2}$, for example).

Along different lines, the region $a_{1} + b\theta \leq r \leq a_{2} + b\theta$ for $0 \leq \theta \leq 2\pi$ (bounded by Archimedean spirals and segments whose endpoints are intersections of the spirals with the positive $x$-axis) has area \begin{align*} \frac{1}{2} \int_{0}^{2\pi} \bigl[(a_{2} + b\theta)^{2} - (a_{1} + b\theta)^{2}\bigr]\, d\theta &= \frac{1}{2} \int_{0}^{2\pi} (a_{2} - a_{1})(a_{2} + a_{1} + 2b\theta)\, d\theta \\ &= (a_{2} - a_{1}) \bigl[\pi(a_{2} + a_{1}) + 2b\pi^{2}\bigr]. \end{align*}

Taking $a_{1} = 0$, $a_{2} = 1$, and $b = \frac{1}{2}$, for example, gives a region of area $\pi + \pi^{2}$. The excess can be whittled away by removing disks, say, nine disks of radius $\frac{1}{3}$ centered on the the intersections of the Archimedean spiral $r = \frac{1}{2}(1 + \theta)$ with the rays making (constructible!) angles $\frac{\pi}{10}(2i+1)$, for $1 \leq i \leq 9$: