Answer
$${\text{Surface area}} = \frac{9}{2}\sqrt {37} + \frac{3}{4}\ln \left( {\sqrt {37} + 6} \right)$$
Work Step by Step
$$\eqalign{
& f\left( {x,y} \right) = {y^2} \cr
& {\text{Find }}{f_x}\left( {x,y} \right){\text{ and }}{f_y}\left( {x,y} \right) \cr
& {f_x}\left( {x,y} \right) = \frac{\partial }{{\partial x}}\left[ {{y^2}} \right] = 0 \cr
& {f_y}\left( {x,y} \right) = \frac{\partial }{{\partial y}}\left[ {{y^2}} \right] = 2y \cr
& {\text{Let the region }}R: \cr
& {\text{Square with vertices }}\left( {0,0} \right),\left( {3,0} \right),\left( {0,3} \right){\text{,}}\left( {3,3} \right){\text{ }}\left( {{\text{Graph below}}} \right) \cr
& {\text{The limits of the region }}R{\text{ are:}} \cr
& R = \left\{ {\left( {x,y} \right):0 \leqslant y \leqslant 3,{\text{ }}0 \leqslant x \leqslant 3} \right\} \cr
& \cr
& {\text{The area of the surface }}S{\text{ is given by}} \cr
& {\text{Surface area}} = \iint\limits_R {\sqrt {1 + {{\left[ {{f_x}\left( {x,y} \right)} \right]}^2} + {{\left[ {{f_y}\left( {x,y} \right)} \right]}^2}} dA} \cr
& = \int_0^3 {\int_0^3 {\sqrt {1 + {{\left[ 0 \right]}^2} + {{\left[ {2y} \right]}^2}} dxdy} } \cr
& = \int_0^3 {\int_0^3 {\sqrt {1 + 4{y^2}} dxdy} } \cr
& {\text{Integrate with respect to }}x \cr
& {\text{Surface area}} = \int_0^3 {3\sqrt {1 + 4{y^2}} dy} \cr
& = 3\int_0^3 {\sqrt {1 + 4{y^2}} dy} \cr
& {\text{Integrate by tables usimg formula 26}} \cr
& = 3\left[ {\frac{1}{2}y\sqrt {4{y^2} + 1} + \frac{1}{4}\ln \left| {\sqrt {4{y^2} + 1} + 2y} \right|} \right]_0^3 \cr
& = 3\left[ {\frac{1}{2}\left( 3 \right)\sqrt {4{{\left( 3 \right)}^2} + 1} + \frac{1}{4}\ln \left| {\sqrt {4{{\left( 3 \right)}^2} + 1} + 2\left( 3 \right)} \right|} \right] - 3\left[ 0 \right] \cr
& = \frac{9}{2}\sqrt {37} + \frac{3}{4}\ln \left( {\sqrt {37} + 6} \right) \cr} $$
