Answer
$${\text{Surface area}} = 2\sqrt {17} + \frac{1}{2}\ln \left( {4 + \sqrt {17} } \right)$$
Work Step by Step
$$\eqalign{
& f\left( {x,y} \right) = 9 - {x^2} \cr
& {\text{Find }}{f_x}\left( {x,y} \right){\text{ and }}{f_y}\left( {x,y} \right) \cr
& {f_x}\left( {x,y} \right) = \frac{\partial }{{\partial x}}\left[ {9 - {x^2}} \right] = - 2x \cr
& {f_y}\left( {x,y} \right) = \frac{\partial }{{\partial y}}\left[ {9 - {x^2}} \right] = 0 \cr
& {\text{Let the region R be:}} \cr
& {\text{Square with vertices }}\left( {0,0} \right),\left( {2,0} \right),\left( {0,2} \right){\text{,}}\left( {2,2} \right){\text{ }}\left( {{\text{Graph below}}} \right) \cr
& {\text{The limits of the region }}R{\text{ are:}} \cr
& R = \left\{ {\left( {x,y} \right):0 \leqslant y \leqslant 2,{\text{ }}0 \leqslant x \leqslant 2} \right\} \cr
& \cr
& {\text{The area of the surface }}S{\text{ is given by}} \cr
& {\text{Surface area}} = \iint\limits_R {\sqrt {1 + {{\left[ {{f_x}\left( {x,y} \right)} \right]}^2} + {{\left[ {{f_y}\left( {x,y} \right)} \right]}^2}} dA} \cr
& = \int_0^2 {\int_0^2 {\sqrt {1 + {{\left[ { - 2x} \right]}^2} + {{\left[ 0 \right]}^2}} dydx} } \cr
& = \int_0^2 {\int_0^2 {\sqrt {4{x^2} + 1} dydx} } \cr
& {\text{Integrate with respect to }}y \cr
& {\text{Surface area}} = 2\int_0^2 {\sqrt {4{x^2} + 1} dx} \cr
& {\text{Surface area}} = \int_0^2 {\sqrt {{{\left( {2x} \right)}^2} + 1} \left( 2 \right)dx} \cr
& {\text{Integrate by tables use formula 26}} \cr
& {\text{Surface area}} = 2\left[ {\frac{1}{2}x\sqrt {{{\left( {2x} \right)}^2} + 1} + \frac{1}{4}\ln \left| {2x + \sqrt {{{\left( {2x} \right)}^2} + 1} } \right|} \right]_0^2 \cr
& {\text{Surface area}} = \left[ {x\sqrt {4{x^2} + 1} + \frac{1}{2}\ln \left| {2x + \sqrt {4{x^2} + 1} } \right|} \right]_0^2 \cr
& = \left[ {2\sqrt {17} + \frac{1}{2}\ln \left| {4 + \sqrt {17} } \right|} \right] - \left[ 0 \right] \cr
& = 2\sqrt {17} + \frac{1}{2}\ln \left( {4 + \sqrt {17} } \right) \cr} $$
