Calculus: Early Transcendentals (2nd Edition)

Published by Pearson
ISBN 10: 0321947347
ISBN 13: 978-0-32194-734-5

Chapter 6 - Applications of Integration - 6.6 Surface Area - 6.6 Exercises - Page 457: 9

Answer

$$S = \frac{{53}}{9}\pi $$

Work Step by Step

$$\eqalign{ & y = {x^{3/2}} - \frac{{{x^{1/2}}}}{3}{\text{ on }}\left[ {1,2} \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) = {x^3}{\text{ and }}\left[ {1,2} \right] \to a = 1{\text{ and }}b = 2.{\text{ then}} \cr & f'\left( x \right) = \frac{d}{{dx}}\left[ {{x^{3/2}} - \frac{{{x^{1/2}}}}{3}} \right] \cr & f'\left( x \right) = \frac{3}{2}{x^{1/2}} - \frac{1}{2}\left( {\frac{{{x^{ - 1/2}}}}{3}} \right) \cr & f'\left( x \right) = \frac{3}{2}{x^{1/2}} - \frac{1}{6}{x^{ - 1/2}} \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_1^2 {2\pi \left( {{x^{3/2}} - \frac{{{x^{1/2}}}}{3}} \right)\sqrt {1 + {{\left( {\frac{3}{2}{x^{1/2}} - \frac{1}{6}{x^{ - 1/2}}} \right)}^2}} } dx \cr & {\text{expand use }}{\left( {a - b} \right)^2} = {a^2} - 2ab + {b^2} \cr & S = \int_1^2 {2\pi \left( {{x^{3/2}} - \frac{{{x^{1/2}}}}{3}} \right)\sqrt {1 + \frac{9}{4}x - \frac{1}{2} + \frac{1}{{36}}{x^{ - 1}}} } dx \cr & S = \int_1^2 {2\pi \left( {{x^{3/2}} - \frac{{{x^{1/2}}}}{3}} \right)\sqrt {\frac{9}{4}x + \frac{1}{2} + \frac{1}{{36}}{x^{ - 1}}} } dx \cr & S = \int_1^2 {2\pi \left( {{x^{3/2}} - \frac{{{x^{1/2}}}}{3}} \right)\sqrt {{{\left( {\frac{3}{2}{x^{1/2}}} \right)}^2} + \frac{1}{2}\left( {\frac{3}{2}{x^{1/2}}} \right)\left( {\frac{1}{6}{x^{ - 1/2}}} \right) + {{\left( {\frac{1}{6}{x^{ - 1/2}}} \right)}^2}} } dx \cr & {\text{factoring}} \cr & S = \int_1^2 {2\pi \left( {{x^{3/2}} - \frac{{{x^{1/2}}}}{3}} \right)\sqrt {{{\left( {\frac{3}{2}{x^{1/2}} + \frac{1}{6}{x^{ - 1/2}}} \right)}^2}} } dx \cr & S = \int_1^2 {2\pi \left( {{x^{3/2}} - \frac{{{x^{1/2}}}}{3}} \right)\left( {\frac{3}{2}{x^{1/2}} + \frac{1}{6}{x^{ - 1/2}}} \right)} dx \cr & {\text{multiply by using the distributive property}} \cr & S = 2\pi \int_1^2 {\left( {\frac{3}{2}{x^2} + \frac{1}{6}x - \frac{1}{2}x - \frac{1}{{18}}} \right)} dx \cr & S = 2\pi \int_1^2 {\left( {\frac{3}{2}{x^2} - \frac{1}{3}x - \frac{1}{{18}}} \right)} dx \cr & {\text{integrating}} \cr & S = 2\pi \left( {\frac{1}{2}{x^3} - \frac{1}{6}{x^2} - \frac{1}{{18}}x} \right)_1^2 \cr & S = 2\pi \left( {\frac{1}{2}{{\left( 2 \right)}^3} - \frac{1}{6}{{\left( 2 \right)}^2} - \frac{1}{{18}}\left( 2 \right)} \right) - 2\pi \left( {\frac{1}{2}{{\left( 1 \right)}^3} - \frac{1}{6}{{\left( 1 \right)}^2} - \frac{1}{{18}}\left( 1 \right)} \right) \cr & {\text{simplifying}} \cr & S = 2\pi \left( {4 - \frac{2}{3} - \frac{1}{9}} \right) - 2\pi \left( {\frac{1}{2} - \frac{1}{6} - \frac{1}{{18}}} \right) \cr & S = 2\pi \left( {\frac{{29}}{9}} \right) - 2\pi \left( {\frac{5}{{18}}} \right) \cr & S = \frac{{53}}{9}\pi \cr} $$
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