jmc

The two circles intersect at $(0,0)$ and $(2,2)$, as shown.



Half of the region described is formed by removing an isosceles right triangle of leg length 2 from a quarter of one of the circles. Because the quarter-circle has area $(1/4)\pi (2{)}^2=\pi$ and the triangle has area $(1/2)(2{)}^2=2$, the area of the region is $2(\pi -2)$, or $\boxed{2\pi -4}$.