smc

We can draw the hexagon between the centers of the circles, and compute its area. The hexagon is made of $6$ equilateral triangles each with length $2$, so the area is: $$\frac{\sqrt{3}}{4}\cdot 2^2\cdot 6=6\sqrt{3}.$$ Then, we add the areas of the three sectors outside the hexagon: $$\frac{2}{3}\pi \cdot 3=2\pi .$$ We now subtract the areas of the three sectors inside the hexagon but outside the figure (which is $\pi$) to get the area enclosed in the curved figure: $$6\sqrt{3}+2\pi -\pi =\pi +6\sqrt{3}.$$ Hence, our answer is $\boxed{\pi +6\sqrt{3}},$ and we are done.