Answer
$$2{a^2}\pi $$
Work Step by Step
$$\eqalign{
& f\left( x \right) = \sqrt {{a^2} - {x^2} - {y^2}} \cr
& {\text{Find }}{f_x}\left( {x,y} \right){\text{ and }}{f_y}\left( {x,y} \right) \cr
& {f_x}\left( {x,y} \right) = \frac{\partial }{{\partial x}}\left[ {\sqrt {{a^2} - {x^2} - {y^2}} } \right] = - \frac{x}{{\sqrt {{a^2} - {x^2} - {y^2}} }} \cr
& {f_y}\left( {x,y} \right) = \frac{\partial }{{\partial y}}\left[ {\sqrt {{a^2} - {x^2} - {y^2}} } \right] = - \frac{y}{{\sqrt {{a^2} - {x^2} - {y^2}} }} \cr
& {\text{The limits of the region }}R{\text{ are:}} \cr
& R = \left\{ {\left( {x,y} \right):{x^2} + {y^2} \leqslant {a^2}} \right\} \cr
& {\text{Changing the region to polar coordinates}} \cr
& R = \left\{ {\left( {r,\theta } \right):0 \leqslant r \leqslant a,{\text{ 0}} \leqslant \theta \leqslant 2\pi } \right\} \cr
& {\text{The area of the surface }}S{\text{ is given by}} \cr
& {\text{Surface area}} = \iint\limits_R {\sqrt {1 + {{\left[ {{f_x}\left( {x,y} \right)} \right]}^2} + {{\left[ {{f_y}\left( {x,y} \right)} \right]}^2}} dA} \cr
& = \int_0^{2\pi } {\int_0^a {\sqrt {1 + {{\left[ { - \frac{x}{{\sqrt {{a^2} - {x^2} - {y^2}} }}} \right]}^2} + {{\left[ { - \frac{y}{{\sqrt {{a^2} - {x^2} - {y^2}} }}} \right]}^2}} rdrd\theta } } \cr
& = \int_0^{2\pi } {\int_0^a {\sqrt {1 + \frac{{{x^2}}}{{{a^2} - {x^2} - {y^2}}} + \frac{{{y^2}}}{{{a^2} - {x^2} - {y^2}}}} rdrd\theta } } \cr
& = \int_0^{2\pi } {\int_0^a {\sqrt {\frac{{{a^2} - {x^2} - {y^2} + {x^2} + {y^2}}}{{{a^2} - {x^2} - {y^2}}}} rdrd\theta } } \cr
& = \int_0^{2\pi } {\int_0^b {\sqrt {\frac{{{a^2}}}{{{a^2} - \left( {{x^2} + {y^2}} \right)}}} rdrd\theta } } \cr
& = \int_0^{2\pi } {\int_0^a {\sqrt {\frac{{{a^2}}}{{{a^2} - {r^2}}}} rdrd\theta } } \cr
& = a\int_0^{2\pi } {\int_0^a {\frac{r}{{\sqrt {{a^2} - {r^2}} }}drd\theta } } \cr
& = - \frac{a}{2}\int_0^{2\pi } {\int_0^a {\frac{{ - 2r}}{{\sqrt {{a^2} - {r^2}} }}drd\theta } } \cr
& {\text{Integrate with respect to }}r \cr
& = - \frac{a}{2}\int_0^{2\pi } {\left[ {\frac{{{{\left( {{a^2} - {r^2}} \right)}^{1/2}}}}{{1/2}}} \right]} _0^ad\theta \cr
& = - a\int_0^{2\pi } {\left[ {\sqrt {{a^2} - {r^2}} } \right]} _0^ad\theta \cr
& = - a\int_0^{2\pi } {\left[ {\sqrt {{a^2} - {a^2}} - \sqrt {{a^2} - {0^2}} } \right]} d\theta \cr
& = - a\int_0^{2\pi } {\left[ {\sqrt 0 - a} \right]} d\theta \cr
& = {a^2}\int_0^{2\pi } {d\theta } \cr
& {\text{Integrate}} \cr
& = {a^2}\left[ {2\pi - 0} \right] \cr
& = 2{a^2}\pi \cr} $$
