Chapter 6 - Applications of Integration - 6.6 Surface Area - 6.6 Exercises - Page 457: 7

$$S = \frac{{2912}}{3}\pi $$

$$\eqalign{ & y = 8\sqrt x {\text{ on }}\left[ {9,20} \right] \cr & {\text{use the Definition of Area of a Surface of Revolution }}\left( {{\text{see page 454}}} \right) \cr & {\text{The area of the surface generated when the graph of }}f{\text{ on the interval }}\left[ {a,b} \right] \cr & {\text{is revolved about the }}x - {\text{axis}} \cr & S = \int_a^b {2\pi f\left( x \right)\sqrt {1 + f'{{\left( x \right)}^2}} } dx \cr & {\text{then}} \cr & y = f\left( x \right) = 8\sqrt x {\text{ and }}\left[ {9,20} \right] \to a = 9{\text{ and }}b = 20.{\text{ then}} \cr & f'\left( x \right) = \frac{d}{{dx}}\left[ {8\sqrt x } \right] \cr & f'\left( x \right) = 8\left( {\frac{1}{{2\sqrt x }}} \right) = \frac{4}{{\sqrt x }} \cr & {\text{substituting the values into the formula }}S = \int_a^b {2\pi f\left( x \right)\sqrt {1 + f'{{\left( x \right)}^2}} } dx \cr & S = \int_9^{20} {2\pi \left( {8\sqrt x } \right)\sqrt {1 + {{\left( {\frac{4}{{\sqrt x }}} \right)}^2}} } dx \cr & S = 16\pi \int_9^{20} {\sqrt x \sqrt {1 + \frac{{16}}{x}} } dx \cr & S = 16\pi \int_9^{20} {\sqrt x \sqrt {\frac{{x + 16}}{x}} } dx \cr & S = 16\pi \int_9^{20} {\sqrt x \frac{{\sqrt {x + 16} }}{{\sqrt x }}} dx \cr & S = 16\pi \int_9^{20} {\sqrt x \frac{{\sqrt {x + 16} }}{{\sqrt x }}} dx \cr & S = 16\pi \int_9^{20} {\sqrt {x + 16} } dx \cr & {\text{integrate and evaluate}} \cr & S = 16\pi \left( {\frac{{{{\left( {x + 16} \right)}^{3/2}}}}{{3/2}}} \right)_9^{20} \cr & S = \frac{{32}}{3}\pi \left( {{{\left( {x + 16} \right)}^{3/2}}} \right)_9^{20} \cr & S = \frac{{32}}{3}\pi \left( {{{\left( {20 + 16} \right)}^{3/2}} - {{\left( {9 + 16} \right)}^{3/2}}} \right) \cr & {\text{simplifying}} \cr & S = \frac{{32}}{3}\pi \left( {{{\left( {36} \right)}^{3/2}} - {{\left( {25} \right)}^{3/2}}} \right) \cr & S = \frac{{32}}{3}\pi \left( {216 - 125} \right) \cr & S = \frac{{32}}{3}\pi \left( {91} \right) \cr & S = \frac{{2912}}{3}\pi \cr} $$