Chapter 14 - Surface Area - 14.5 Exercises - Page 1007: 11

$$\sqrt 2 \pi $$

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