Calculus 10th Edition

Published by Brooks Cole
ISBN 10: 1-28505-709-0
ISBN 13: 978-1-28505-709-5

Chapter 7 - Applications of Integration - 7.2 Exercises: 21

Answer

$\frac{832\pi}{15}$

Work Step by Step

Setup the integration using washer method about the line x=5. $ 2\pi \int_0^2 [(5-y^2)^2 - (1)^1]dy$, It is easier to use the interval [0,2] because the graph is symmetric and then just multiply the integration by two. $2\pi \int_0^2 (25-10y^2 +y^4 -1)dy$ $2\pi \int_0^2 (y^4 -10y^2 +24)dy$ $2\pi [\frac{1}{5}y^5 - \frac{10}{3}y^3 +24y]_0^2$ $2\pi(\frac{32}{5}- \frac{80}{3}+48)-0$ $\frac{832\pi}{15}$
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