Chapter 14 - Surface Area - 14.5 Exercises - Page 1007: 12

$${\text{Surface area}} = \frac{{2\left( {5\sqrt 5 - 1} \right)\pi }}{3}$$

$$\eqalign{ & f\left( {x,y} \right) = xy \cr & {\text{Find }}{f_x}\left( {x,y} \right){\text{ and }}{f_y}\left( {x,y} \right) \cr & {f_x}\left( {x,y} \right) = \frac{\partial }{{\partial x}}\left[ {xy} \right] = y \cr & {f_y}\left( {x,y} \right) = \frac{\partial }{{\partial y}}\left[ {xy} \right] = x \cr & {\text{The limits of the region }}R{\text{ are:}} \cr & R = \left\{ {\left( {x,y} \right):{x^2} + {y^2} \leqslant 16} \right\} \cr & {\text{Changing the region to polar coordinates}} \cr & R = \left\{ {\left( {r,\theta } \right):0 \leqslant r \leqslant 4,{\text{ 0}} \leqslant \theta \leqslant 2\pi } \right\} \cr & \cr & {\text{The area of the surface }}S{\text{ is given by}} \cr & {\text{Surface area}} = \iint\limits_R {\sqrt {1 + {{\left[ {{f_x}\left( {x,y} \right)} \right]}^2} + {{\left[ {{f_y}\left( {x,y} \right)} \right]}^2}} dA} \cr & = \int_0^{2\pi } {\int_0^4 {\sqrt {1 + {{\left[ y \right]}^2} + {{\left[ x \right]}^2}} rdrd\theta } } \cr & = \int_0^{2\pi } {\int_0^4 {\sqrt {1 + {y^2} + {x^2}} rdrd\theta } } \cr & = \int_0^{2\pi } {\int_0^4 {\sqrt {1 + {r^2}} rdrd\theta } } \cr & = \frac{1}{2}\int_0^{2\pi } {\int_0^4 {\sqrt {1 + {r^2}} \left( {2r} \right)drd\theta } } \cr & {\text{Integrate with respect to }}r \cr & = \frac{1}{2}\int_0^{2\pi } {\left[ {\frac{{2{{\left( {1 + {r^2}} \right)}^{3/2}}}}{3}} \right]_0^2} d\theta \cr & = \frac{1}{3}\int_0^{2\pi } {\left[ {{{\left( {1 + {r^2}} \right)}^{3/2}}} \right]_0^2} d\theta \cr & = \frac{1}{3}\int_0^{2\pi } {\left[ {{{\left( {1 + {2^2}} \right)}^{3/2}} - {{\left( {1 + {0^2}} \right)}^{3/2}}} \right]} d\theta \cr & = \frac{1}{3}\int_0^{2\pi } {\left( {5\sqrt 5 - 1} \right)} d\theta \cr & = \frac{{5\sqrt 5 - 1}}{3}\int_0^{2\pi } {d\theta } \cr & {\text{Integrate}} \cr & {\text{Surface area}} = \frac{{5\sqrt 5 - 1}}{3}\left( {2\pi } \right) \cr & {\text{Surface area}} = \frac{{2\left( {5\sqrt 5 - 1} \right)\pi }}{3} \cr} $$