Chapter 5 - Applications Of The Definite Integral In Geometry, Science, And Engineering - 5.5 Area Of A Surface Of Revolution - Exercises Set 5.5 - Page 380: 9

\[S = \frac{{16}}{9}\pi \]

$$\eqalign{ & y = \sqrt x - \frac{1}{3}{x^{3/2}},{\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} 1 \leqslant x \leqslant 3 \cr & {\text{Calculate }}\frac{{dy}}{{dx}} \cr & \frac{{dy}}{{dx}} = \frac{1}{{2\sqrt x }} - \frac{1}{2}{x^{1/2}} \cr & \frac{{dy}}{{dx}} = \frac{1}{2}{x^{ - 1/2}} - \frac{1}{2}{x^{1/2}} \cr & {\text{Use the formula }}S = \int_a^b {2\pi y\sqrt {1 + {{\left( {\frac{{dy}}{{dx}}} \right)}^2}} } dx,{\text{ then}} \cr & S = \int_1^3 {2\pi \left( {\sqrt x - \frac{1}{3}{x^{3/2}}} \right)\sqrt {1 + {{\left( {\frac{1}{2}{x^{ - 1/2}} - \frac{1}{2}{x^{1/2}}} \right)}^2}} } dx \cr & {\text{Simplifying}} \cr & S = 2\pi \int_1^3 {\left( {\sqrt x - \frac{1}{3}{x^{3/2}}} \right)\sqrt {1 + {{\left( {\frac{1}{2}{x^{ - 1/2}}} \right)}^2} - \frac{1}{2} + {{\left( { - \frac{1}{2}{x^{1/2}}} \right)}^2}} } dx \cr & S = 2\pi \int_1^3 {\left( {\sqrt x - \frac{1}{3}{x^{3/2}}} \right)\sqrt {{{\left( {\frac{1}{2}{x^{ - 1/2}} + \frac{1}{2}{x^{1/2}}} \right)}^2}} } dx \cr & S = 2\pi \int_1^3 {\left( {{x^{1/2}} - \frac{1}{3}{x^{3/2}}} \right)\left( {\frac{1}{2}{x^{ - 1/2}} + \frac{1}{2}{x^{1/2}}} \right)} dx \cr & {\text{Distribute and reduce like terms}} \cr & S = 2\pi \int_1^3 {\left( {\frac{1}{2} + \frac{1}{2}x - \frac{1}{6}x - \frac{1}{6}{x^2}} \right)} dx \cr & S = 2\pi \int_1^3 {\left( {\frac{1}{2} + \frac{1}{3}x - \frac{1}{6}{x^2}} \right)} dx \cr & {\text{Integrate}} \cr & S = 2\pi \left[ {\frac{1}{2}x + \frac{1}{6}{x^2} - \frac{1}{{18}}{x^3}} \right]_1^3 \cr & S = 2\pi \left[ {\frac{3}{2} + \frac{3}{2} - \frac{3}{2}} \right] - 2\pi \left[ {\frac{1}{2} + \frac{1}{6} - \frac{1}{{18}}} \right] \cr & S = 2\pi \left( {\frac{3}{2}} \right) - 2\pi \left( {\frac{{11}}{{18}}} \right) \cr & S = \frac{{16}}{9}\pi \cr} $$