Answer
$$A = 6\sqrt 2 + \ln \left( {2\sqrt 2 + 3} \right)$$
Work Step by Step
$$\eqalign{
& {y^2} - {x^2} = 1,{\text{ }}y = 3 \cr
& {y^2} = 1 + {x^2} \cr
& y = \sqrt {{x^2} + 1} \cr
& {\text{Let }}y = y \cr
& \sqrt {{x^2} + 1} = 3 \cr
& {x^2} + 1 = 9 \cr
& {x^2} = 8 \cr
& x = \pm 2\sqrt 2 \cr
& {\text{From the graph, we can define the area as}} \cr
& A = \int_{ - 2\sqrt 2 }^{2\sqrt 2 } {\sqrt {{x^2} + 1} } dx \cr
& {\text{By symmetry}} \cr
& A = 2\int_0^{2\sqrt 2 } {\sqrt {{x^2} + 1} } dx \cr
& {\text{Integrate by tables}} \cr
& *\int {\sqrt {{u^2} + {a^2}} du} = \frac{1}{2}\left( {u\sqrt {{u^2} + {a^2}} + {a^2}\ln \left| {u + \sqrt {{u^2} + {a^2}} } \right|} \right) + C,{\text{ then}} \cr
& A = \left[ {x\sqrt {{x^2} + 1} + \ln \left| {x + \sqrt {{x^2} + 1} } \right|} \right]_0^{2\sqrt 2 } \cr
& {\text{Evaluating}} \cr
& A = \left[ {2\sqrt 2 \sqrt {{{\left( {2\sqrt 2 } \right)}^2} + 1} + \ln \left| {2\sqrt 2 + \sqrt {{{\left( {2\sqrt 2 } \right)}^2} + 1} } \right|} \right] - \left[ 0 \right] \cr
& A = 2\sqrt 2 \sqrt 9 + \ln \left| {2\sqrt 2 + \sqrt 9 } \right| \cr
& A = 6\sqrt 2 + \ln \left( {2\sqrt 2 + 3} \right) \cr} $$
