Chapter 14 - Surface Area - 14.5 Exercises - Page 1007: 1

$${\text{Surface area}} = 24$$

$$\eqalign{ & f\left( {x,y} \right) = 2x + 2y \cr & {\text{Find }}{f_x}\left( {x,y} \right){\text{ and }}{f_y}\left( {x,y} \right) \cr & {f_x}\left( {x,y} \right) = \frac{\partial }{{\partial x}}\left[ {2x + 2y} \right] = 2 \cr & {f_y}\left( {x,y} \right) = \frac{\partial }{{\partial y}}\left[ {2x + 2y} \right] = 2 \cr & {\text{Let the region R be:}} \cr & {\text{Triangle with vertices }}\left( {0,0} \right),\left( {4,0} \right),\left( {0,4} \right){\text{ }}\left( {{\text{Graph below}}} \right) \cr & \cr & m = \frac{{0 - 4}}{{4 - 0}} = - 1 \cr & y = mx + b \cr & y = - x + 4 \cr & {\text{The limits of the region }}R{\text{ are:}} \cr & R = \left\{ {\left( {x,y} \right):0 \leqslant y \leqslant 4 - x,{\text{ }}0 \leqslant x \leqslant 4} \right\} \cr & \cr & {\text{The area of the surface }}S{\text{ is given by}} \cr & {\text{Surface area}} = \iint\limits_R {\sqrt {1 + {{\left[ {{f_x}\left( {x,y} \right)} \right]}^2} + {{\left[ {{f_y}\left( {x,y} \right)} \right]}^2}} dA} \cr & = \int_0^4 {\int_0^{4 - x} {\sqrt {1 + {{\left[ 2 \right]}^2} + {{\left[ 2 \right]}^2}} dydx} } \cr & = \int_0^4 {\int_0^{4 - x} {\sqrt 9 dydx} } \cr & = 3\int_0^4 {\int_0^{4 - x} {dydx} } \cr & {\text{Integrate with respect to }}y \cr & {\text{Surface area}} = 3\int_0^4 {\left( {4 - x} \right)} dx \cr & {\text{Integrate}} \cr & {\text{Surface area}} = - 3\left[ {\frac{{{{\left( {4 - x} \right)}^2}}}{2}} \right]_0^4 \cr & {\text{Surface area}} = - 3\left[ {\frac{{{{\left( {4 - 4} \right)}^2}}}{2} - \frac{{{{\left( {4 - 0} \right)}^2}}}{2}} \right] \cr & {\text{Surface area}} = 24 \cr} $$