Chapter 10 - Conics, Parametric Equations, and Polar Coordinates - Review Exercises - Page 742: 25

$$\eqalign{ & \left( {\text{a}} \right)\left( {0,50} \right) \cr & \left( {\text{b}} \right)S \approx 38294.488 \cr} $$

$$\eqalign{ & y = \frac{{{x^2}}}{{200}},{\text{ }} - 100 \leqslant x \leqslant 100 \cr & {\text{The equation of the parabola is on the form }}{x^2} = 4py \cr & {\text{With Vertex }}\left( {0,0} \right){\text{ and Focus }}\left( {0,p} \right) \cr & y = \frac{{{x^2}}}{{200}},{\text{ }}{x^2} = 4py \cr & \frac{{{x^2}}}{{200}} = \frac{{{x^2}}}{{4p}} \cr & 200 = 4p \cr & p = 50 \cr & \cr & \left( {\text{a}} \right){\text{ The coordinates of the focus are:}} \cr & \left( {0,50} \right) \cr & \cr & \left( {\text{b}} \right){\text{The surface of the antena is given by}} \cr & S = 2\pi \int_a^b {x\sqrt {1 + {{\left( {\frac{{dy}}{{dx}}} \right)}^2}} dx} \cr & y = \frac{{{x^2}}}{{200}} \to \frac{{dy}}{{dx}} = \frac{{2x}}{{200}} = \frac{x}{{100}},{\text{ so}} \cr & S = 2\pi \int_{ - 100}^{100} {x\sqrt {1 + {{\left( {\frac{x}{{100}}} \right)}^2}} dx} \cr & {\text{By symmetry and the property }}\int_{ - a}^a {f\left( x \right)} dx = 0,{\text{ }}f\left( x \right){\text{ odd}} \cr & {\text{We obtain }}S = 0,{\text{ because }}y = \frac{{{x^2}}}{{200}}{\text{ is symmetrical with}} \cr & {\text{respect to the }}y{\text{ - axis}}{\text{, then It is necessary only revolved}} \cr & {\text{the function for the interval }}\left[ {0,100} \right] \cr & S = \frac{{2\pi }}{{100}}\int_0^{100} {x\sqrt {10000 + {x^2}} dx} \cr & S = \frac{\pi }{{100}}\int_0^{100} {\left( {2x} \right)\sqrt {10000 + {x^2}} dx} \cr & {\text{Integrating}} \cr & S = \frac{\pi }{{100}}\left[ {\frac{2}{3}{{\left( {10000 + {x^2}} \right)}^{3/2}}} \right]_{ - 100}^{100} \cr & S = \frac{\pi }{{150}}\left[ {{{\left( {10000 + {{\left( {100} \right)}^2}} \right)}^{3/2}} - {{\left( {10000 + {{\left( 0 \right)}^2}} \right)}^{3/2}}} \right] \cr & {\text{Simplifying}} \cr & S \approx 38294.488 \cr} $$