Chapter 14 - Surface Area - 14.5 Exercises - Page 1007: 8

$${\text{Surface area}} = \frac{{12\sqrt 3 - 8}}{8}$$

$$\eqalign{ & f\left( {x,y} \right) = 2 + \frac{2}{3}{y^{3/2}} \cr & {\text{Find }}{f_x}\left( {x,y} \right){\text{ and }}{f_y}\left( {x,y} \right) \cr & {f_x}\left( {x,y} \right) = \frac{\partial }{{\partial x}}\left[ {2 + \frac{2}{3}{y^{3/2}}} \right] = 0 \cr & {f_y}\left( {x,y} \right) = \frac{\partial }{{\partial y}}\left[ {2 + \frac{2}{3}{y^{3/2}}} \right] = {y^{1/2}} \cr & \cr & {\text{The limits of the region }}R{\text{ are:}} \cr & R = \left\{ {\left( {x,y} \right):0 \leqslant x \leqslant 2,{\text{ }}0 \leqslant y \leqslant 2 - x} \right\} \cr & \cr & {\text{The area of the surface }}S{\text{ is given by}} \cr & {\text{Surface area}} = \iint\limits_R {\sqrt {1 + {{\left[ {{f_x}\left( {x,y} \right)} \right]}^2} + {{\left[ {{f_y}\left( {x,y} \right)} \right]}^2}} dA} \cr & = \int_0^2 {\int_0^{2 - x} {\sqrt {1 + {{\left[ 0 \right]}^2} + {{\left[ {{y^{1/2}}} \right]}^2}} dydx} } \cr & = \int_0^2 {\int_0^{2 - x} {\sqrt {1 + y} dydx} } \cr & {\text{Integrate with respect to }}y \cr & = \int_0^2 {\left[ {\frac{{2{{\left( {1 + y} \right)}^{3/2}}}}{3}} \right]} _0^{2 - x}dx \cr & = \int_0^2 {\left[ {\frac{{2{{\left( {1 + 2 - x} \right)}^{3/2}}}}{3} - \frac{{2{{\left( {1 + 0} \right)}^{3/2}}}}{3}} \right]} dx \cr & = \int_0^2 {\left[ {\frac{{2{{\left( {3 - x} \right)}^{3/2}}}}{3} - \frac{2}{3}} \right]} dx \cr & {\text{Integrate}} \cr & = \left[ {\frac{{4{{\left( {3 - x} \right)}^{5/2}}}}{{15}} - \frac{2}{3}x} \right]_0^2 \cr & = \left[ { - \frac{{4{{\left( {3 - 2} \right)}^{5/2}}}}{{15}} - \frac{2}{3}\left( 2 \right)} \right] - \left[ { - \frac{{4{{\left( {3 - 0} \right)}^{5/2}}}}{{15}} - \frac{2}{3}\left( 0 \right)} \right] \cr & {\text{Simplify}} \cr & = \left[ { - \frac{4}{{15}} - \frac{4}{3}} \right] - \left[ { - \frac{{4{{\left( 3 \right)}^{5/2}}}}{{15}} - 0} \right] \cr & = - \frac{8}{5} + \frac{{12\sqrt 3 }}{5} \cr & = \frac{{12\sqrt 3 - 8}}{8} \cr} $$