Calculus: Early Transcendentals (2nd Edition)

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

Chapter 6 - Applications of Integration - 6.4 Volume by Shells - 6.4 Exercises - Page 443: 57

Answer

$$V = \frac{{608}}{3}\pi $$

Work Step by Step

$$\eqalign{ & {\text{Let the functions }}y = x,{\text{ }}y = 2x + 2,{\text{ and the lines }}x = 2,{\text{ }}x = 6 \cr & {\text{revolved about the }}y{\text{ - axis}} \cr & {\text{Using the shell method about the }}y{\text{ - axis}} \cr & V = \int_a^b {2\pi x\left[ {f\left( x \right) - g\left( x \right)} \right]} dx \cr & {\text{From the graph of the region shown below}} \cr & f\left( x \right) = 2x + 2{\text{ and }}g\left( x \right) = x{\text{ on the interval }}\left[ {2,6} \right],{\text{ then}} \cr & V = \int_2^6 {2\pi x\left[ {\left( {2x + 2} \right) - x} \right]} dx \cr & V = 2\pi \int_2^6 {x\left( {x + 2} \right)} dx \cr & V = 2\pi \int_2^6 {\left( {{x^2} + 2x} \right)} dx \cr & {\text{Integrating}} \cr & V = 2\pi \left[ {\frac{1}{3}{x^3} + {x^2}} \right]_2^6 \cr & V = 2\pi \left[ {\frac{1}{3}{{\left( 6 \right)}^3} + {{\left( 6 \right)}^2}} \right] - 2\pi \left[ {\frac{1}{3}{{\left( 2 \right)}^3} + {{\left( 2 \right)}^2}} \right] \cr & V = 2\pi \left( {108} \right) - 2\pi \left( {\frac{{20}}{3}} \right) \cr & V = \frac{{608}}{3}\pi \cr} $$
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