Calculus: Early Transcendentals (2nd Edition)

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

Chapter 6 - Applications of Integration - 6.3 Volume by Slicing - 6.3 Exercises - Page 431: 29

Answer

$\dfrac{5 \pi}{6}$

Work Step by Step

Our aim is to compute the volume of the revolution of the curve by using the Washer method. Washer method for computing the volume of the revolution of the curve: Let us consider two functions $f(x)$ and $g(x)$ (both are continuous functions) with $f(x) \geq g(x) \geq 0$ on the interval $[m, n]$ . Then the volume of the solid can be obtained by rotating the region under the graph about the x-axis and can be expressed as: $\ Volume, V=\pi \int_m^n [f(x)^2-g(x)^2] \ dx \\ =\pi \int_m^n [R_{outer}^2-R_{inner}^2] \ dx$ We are given that $f(x)=e^{x/2}; g(x)=e^{-x/2}$ and $x=\ln 2; x=\ln 3 \implies m=\ln 2; n=\ln 3$ Thus, $V= \pi \int_{\ln 2}^{\ln 3} [(e^{x/2})^2-(e^{-x/2})^2] \ dx$ or, $= \pi \int_{\ln 2}^{\ln 3} (e^x -e^{-x}) \ dx$ or, $=\pi [e^x]_{\ln 2}^{\ln 3}-\pi [e^{-x}]_{\ln 2}^{\ln 3}$ Hence, $V=\pi (e^{\ln 3}-e^{\ln 2}+e^{\ln (1/3)}-e^{\ln (1/2)})=\dfrac{5 \pi}{6}$
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