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: 10

$$S = \frac{{515}}{{64}}\pi $$

$$\eqalign{ & y = \frac{1}{3}{x^3} + \frac{1}{4}{x^{ - 1}},\,\,\,\,1 \leqslant x \leqslant 2 \cr & {\text{Calculate }}\frac{{dy}}{{dx}} \cr & \frac{{dy}}{{dx}} = {x^2} - \frac{1}{4}{x^{ - 2}} \cr & {\text{Use }}S = \int_a^b {2\pi y\sqrt {1 + {{\left( {\frac{{dy}}{{dx}}} \right)}^2}} } dx \cr & S = \int_1^2 {2\pi \left( {\frac{1}{3}{x^3} + \frac{1}{4}{x^{ - 1}}} \right)\sqrt {1 + {{\left( {{x^2} - \frac{1}{4}{x^{ - 2}}} \right)}^2}} } dx \cr & S = 2\pi \int_1^2 {\left( {\frac{1}{3}{x^3} + \frac{1}{4}{x^{ - 1}}} \right)\sqrt {1 + {{\left( {{x^2}} \right)}^2} - 2\left( {\frac{1}{4}} \right) + {{\left( { - \frac{1}{4}{x^{ - 2}}} \right)}^2}} } dx \cr & S = 2\pi \int_1^2 {\left( {\frac{1}{3}{x^3} + \frac{1}{4}{x^{ - 1}}} \right)\sqrt {1 + {{\left( {{x^2}} \right)}^2} + \frac{1}{2} + {{\left( { - \frac{1}{4}{x^{ - 2}}} \right)}^2}} } dx \cr & S = 2\pi \int_1^2 {\left( {\frac{1}{3}{x^3} + \frac{1}{4}{x^{ - 1}}} \right)\sqrt {{{\left( {{x^2} + \frac{1}{4}{x^{ - 2}}} \right)}^2}} } dx \cr & S = 2\pi \int_1^2 {\left( {\frac{1}{3}{x^3} + \frac{1}{4}{x^{ - 1}}} \right)\left( {{x^2} + \frac{1}{4}{x^{ - 2}}} \right)} dx \cr & {\text{Distribute and simplify}} \cr & S = 2\pi \int_1^2 {\left( {\frac{1}{3}{x^5} + \frac{1}{{12}}x + \frac{1}{4}x + \frac{1}{{16}}{x^{ - 3}}} \right)} dx \cr & S = 2\pi \int_1^2 {\left( {\frac{1}{3}{x^5} + \frac{1}{3}x + \frac{1}{{16}}{x^{ - 3}}} \right)} dx \cr & {\text{Integrate}} \cr & S = 2\pi \left[ {\frac{1}{{18}}{x^6} + \frac{1}{6}{x^2} - \frac{1}{{32}}{x^{ - 2}}} \right]_1^2 \cr & S = 2\pi \left[ {\frac{1}{{18}}{{\left( 2 \right)}^6} + \frac{1}{6}{{\left( 2 \right)}^2} - \frac{1}{{32}}{{\left( 2 \right)}^{ - 2}}} \right] - 2\pi \left[ {\frac{1}{{18}}{{\left( 1 \right)}^6} + \frac{1}{6}{{\left( 1 \right)}^2} - \frac{1}{{32}}{{\left( 1 \right)}^{ - 2}}} \right] \cr & S = 2\pi \left[ {\frac{{4855}}{{1152}}} \right] - 2\pi \left[ {\frac{{55}}{{288}}} \right] \cr & S = \frac{{515}}{{64}}\pi \cr} $$