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 - 7.3 Exercises: 25

Answer

$16\pi$

Work Step by Step

Setup the integration using shell method about the line x=4 $ 2\pi \int _0^2 (4-x)(4x-x^2-x^2)dx$ $ 4\pi \int _0^2(4-x)(2x-x^2)dx$, Take a factor of 2 outside the integration $4\pi \int_0^2 (8x-4x^2-2x^2+x^3)dx$ $4\pi \int_0^2 (x^3 -6x^2 +8x)dx$ $4\pi [\frac{1}{4}x^4 - 2x^3 +4x^2]_0^2$ $4\pi(4-16+16)$ $16\pi$
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