3. As shown in Figure 4, the side of square $ABCD$ is $AB=1$, and $\overparen{BD}$ and $\overparen{AC}$ are both circular arcs with a radius of 1. Then the difference in the areas of the two unshaded parts is $\qquad$
3. $\frac{\pi}{2}-1$.

From Figure 4, we know $S_{1}+S=\frac{\pi}{4}$. Therefore, $S_{1}=\frac{\pi}{4}-S$. Also, $S_{2}+S=1-\frac{\pi}{4}$, so $S_{2}=1-S-\frac{\pi}{4}$. Thus, $S_{1}-S_{2}=\left(\frac{\pi}{4}-S\right)-\left(1-S-\frac{\pi}{4}\right)$ $=\frac{\pi}{2}-1$.