4. As shown in Figure 2, an equilateral triangle $\triangle A B C$ with side length 1, and circles with radius 1 are drawn with centers at vertices $A$, $B$, and $C$. Then the area of the region covered by these three circles is ( ).
(A) $\frac{3 \pi}{2}+\sqrt{3}$
(B) $\frac{5 \pi}{2}-\sqrt{3}$
(C) $\frac{7 \pi}{2}-2 \sqrt{3}$
(D) $3 \pi-2 \sqrt{3}$
4.A.

Given $S_{\triangle A B C}=S_{\triangle A B D}=\frac{\sqrt{3}}{4}$,
$S_{\text {sector }- \text { BC }}=\frac{\pi}{6}, S_{\text {sector } A}-\Omega C D=\frac{\pi}{3}$,
the area of the intersection of any two circles is
$$
2\left(\frac{\pi}{3}-\frac{\sqrt{3}}{4}\right)=\frac{2 \pi}{3}-\frac{\sqrt{3}}{2} \text {, }
$$

the area of the intersection of all three circles is
$$
3 \times \frac{\pi}{6}-2 \times \frac{\sqrt{3}}{4}=\frac{\pi}{2}-\frac{\sqrt{3}}{2} \text {. }
$$

Therefore, $S=3 \pi-3\left(\frac{2 \pi}{3}-\frac{\sqrt{3}}{2}\right)+\left(\frac{\pi}{2}-\frac{\sqrt{3}}{2}\right)=\frac{3 \pi}{2}+\sqrt{3}$.