Area bounded between two hyperbolas

It's easiest to picture this hyperbolic figure as being inscribed in a square of side length $6 \Rightarrow x, y \in [-3,3]$ . Knowing that $y^2-x^2=1$ intersects the line $y=3 \Rightarrow x = \pm 2\sqrt{2}$ , and the chamfered ends are right-isosceles triangles of side length $3-2\sqrt{2}$ , the required (and highly-symmetric) area computes according to:

$A = 6^2 - 4\int_{-2\sqrt{2}}^{2\sqrt{2}} 3 - \sqrt{x^2+1} dx - 4[\frac{1}{2}(3-2\sqrt{2})^2] \approx \boxed{9.05}.$