Chapter 14 - Surface Area - 14.5 Exercises - Page 1007: 2

$${\text{Surface area}} = 9\sqrt {14} $$

$$\eqalign{ & f\left( {x,y} \right) = 15 + 2x - 3y \cr & {\text{Find }}{f_x}\left( {x,y} \right){\text{ and }}{f_y}\left( {x,y} \right) \cr & {f_x}\left( {x,y} \right) = \frac{\partial }{{\partial x}}\left[ {15 + 2x - 3y} \right] = 2 \cr & {f_y}\left( {x,y} \right) = \frac{\partial }{{\partial y}}\left[ {15 + 2x - 3y} \right] = - 3 \cr & {\text{Let the region R be:}} \cr & {\text{Square with vertices }}\left( {0,0} \right),\left( {3,0} \right),\left( {0,3} \right){\text{,}}\left( {3,3} \right){\text{ }}\left( {{\text{Graph below}}} \right) \cr & {\text{The limits of the region }}R{\text{ are:}} \cr & R = \left\{ {\left( {x,y} \right):0 \leqslant y \leqslant 3,{\text{ }}0 \leqslant x \leqslant 3} \right\} \cr & \cr & {\text{The area of the surface }}S{\text{ is given by}} \cr & {\text{Surface area}} = \iint\limits_R {\sqrt {1 + {{\left[ {{f_x}\left( {x,y} \right)} \right]}^2} + {{\left[ {{f_y}\left( {x,y} \right)} \right]}^2}} dA} \cr & = \int_0^3 {\int_0^3 {\sqrt {1 + {{\left[ 2 \right]}^2} + {{\left[ { - 3} \right]}^2}} dydx} } \cr & = \int_0^3 {\int_0^3 {\sqrt {14} dydx} } \cr & = \sqrt {14} \int_0^3 {\int_0^3 {dydx} } \cr & {\text{Integrate with respect to }}y \cr & {\text{Surface area}} = \sqrt {14} \int_0^4 3 dx \cr & {\text{Integrate}} \cr & {\text{Surface area}} = 3\sqrt {14} \left[ x \right]_0^3 \cr & {\text{Surface area}} = 9\sqrt {14} \cr} $$