Answer
$$A = 2\sqrt 3 - \frac{{2\pi }}{3}$$
Work Step by Step
$$\eqalign{
& {\text{Let }}y = 2{\text{ and }}y = \frac{1}{{\sqrt {1 - {x^2}} }} \cr
& 2 = \frac{1}{{\sqrt {1 - {x^2}} }} \cr
& 4 = \frac{1}{{1 - {x^2}}} \cr
& 4 - 4{x^2} = 1 \cr
& 4{x^2} - 3 = 0 \cr
& x = \pm \frac{{\sqrt 3 }}{2} \cr
& 2 \geqslant \frac{1}{{\sqrt {1 - {x^2}} }}{\text{ on the interval }}\left[ { - \frac{{\sqrt 3 }}{2},\frac{{\sqrt 3 }}{2}} \right] \cr
& {\text{The area is given by}} \cr
& A = \int_{ - \sqrt 3 /2}^{\sqrt 3 /2} {\left( {2 - \frac{1}{{\sqrt {1 - {x^2}} }}} \right)} dx \cr
& {\text{Integrating}} \cr
& A = \left[ {2x - {{\sin }^{ - 1}}x} \right]_{ - \sqrt 3 /2}^{\sqrt 3 /2} \cr
& A = 2\left( {\frac{{\sqrt 3 }}{2}} \right) - {\sin ^{ - 1}}\left( {\frac{{\sqrt 3 }}{2}} \right) - 2\left( { - \frac{{\sqrt 3 }}{2}} \right) + {\sin ^{ - 1}}\left( { - \frac{{\sqrt 3 }}{2}} \right) \cr
& {\text{Simplifying}} \cr
& A = \sqrt 3 - \frac{\pi }{3} + \sqrt 3 - \frac{\pi }{3} \cr
& A = 2\sqrt 3 - \frac{{2\pi }}{3} \cr} $$
