Answer
$${\text{Surface area}} = \frac{4}{{27}}\left( {31\sqrt {31} - 8} \right)$$
Work Step by Step
$$\eqalign{
& f\left( {x,y} \right) = 3 + {x^{3/2}} \cr
& {\text{Find }}{f_x}\left( {x,y} \right){\text{ and }}{f_y}\left( {x,y} \right) \cr
& {f_x}\left( {x,y} \right) = \frac{\partial }{{\partial x}}\left[ {3 + {x^{3/2}}} \right] = \frac{3}{2}{x^{1/2}} \cr
& {f_y}\left( {x,y} \right) = \frac{\partial }{{\partial y}}\left[ {3 + {x^{3/2}}} \right] = 0 \cr
& {\text{Let the region R be:}} \cr
& {\text{Square with vertices }}\left( {0,0} \right),\left( {0,4} \right),\left( {3,4} \right){\text{,}}\left( {3,0} \right){\text{ }}\left( {{\text{Graph below}}} \right) \cr
& {\text{The limits of the region }}R{\text{ are:}} \cr
& R = \left\{ {\left( {x,y} \right):0 \leqslant y \leqslant 4,{\text{ }}0 \leqslant x \leqslant 3} \right\} \cr
& \cr
& {\text{The area of the surface }}S{\text{ is given by}} \cr
& {\text{Surface area}} = \iint\limits_R {\sqrt {1 + {{\left[ {{f_x}\left( {x,y} \right)} \right]}^2} + {{\left[ {{f_y}\left( {x,y} \right)} \right]}^2}} dA} \cr
& = \int_0^3 {\int_0^3 {\sqrt {1 + {{\left[ {\frac{3}{2}{x^{1/2}}} \right]}^2} + {{\left[ 0 \right]}^2}} dydx} } \cr
& = \int_0^4 {\int_0^3 {\sqrt {1 + \frac{9}{4}x} dxdy} } \cr
& = \frac{4}{9}\int_0^4 {\int_0^3 {\frac{9}{4}\sqrt {1 + \frac{9}{4}x} dxdy} } \cr
& {\text{Integrate with respect to }}x \cr
& = \frac{4}{9}\int_0^4 {\left[ {\frac{2}{3}{{\left( {1 + \frac{9}{4}x} \right)}^{3/2}}} \right]_0^3} dy \cr
& = \frac{8}{{27}}\int_0^4 {\left[ {{{\left( {1 + \frac{9}{4}\left( 3 \right)} \right)}^{3/2}} - {{\left( {1 + \frac{9}{4}\left( 0 \right)} \right)}^{3/2}}} \right]} dy \cr
& = \frac{8}{{27}}\int_0^4 {\left[ {{{\left( {\frac{{31}}{4}} \right)}^{3/2}} - 1} \right]} dy \cr
& {\text{Integrate}} \cr
& = \frac{8}{{27}}\left[ {{{\left( {\frac{{31}}{4}} \right)}^{3/2}} - 1} \right]\left( 4 \right) \cr
& = \frac{{32}}{{27}}\left( {\frac{{31\sqrt {31} }}{8} - 1} \right) \cr
& = \frac{4}{{27}}\left( {31\sqrt {31} - 8} \right) \cr} $$
