We draw a unit-radius quarter circle around each vertex of a unit-sided square, which connects the adjacent vertices. Calculate the area of the common part of the four circle segments.
The three types of segments divide the square into quarters. We denote the areas of the congruent segments with the letters $A, B$, and $T$ as shown in the figure.

# 1984-04-163-1.eps 

The following equations can be written for the areas:

The area of the square is 1, so $4 A+4 B+T=1$,

$$
\text { The area of a quarter circle is } \pi / 4 \text {, so } 3 A+2 B+T=\frac{\pi}{4} \text {. }
$$

Finally, the area of the "dome" bounded by a square side and two quarter-circle arcs drawn around its endpoints, which is outlined with a thick line in the figure, is obtained by subtracting the common part, the area of an equilateral triangle with a unit side, from the area of two $60^{\circ}$ central angle sectors, that is,

$$
2 A+B+T=\frac{2 \pi}{6}-\frac{\sqrt{3}}{4}
$$

The value of the desired $T$ can be obtained from equations (1), (2), and (3), for example, as follows. Multiply equation (2) by (-4) and equation (3) by 4

$$
\begin{gathered}
4 A+4 B+T=1 \\
-12 A-8 B-4 T=-\pi \\
8 A+4 B+4 T=\frac{4 \pi}{3}-\sqrt{3}
\end{gathered}
$$

Adding them together, we get

$$
T=1-\sqrt{3}+\frac{\pi}{3} \approx 0.3151
$$