Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 6 - Review - Exercises: 10

Answer

$ V=256 \pi $

Work Step by Step

{Step 1 of 5}: First, find the points of intersection of the curves $ y=x^{2}+1 $ and $ y=9-x^{2} $ $ x^{2}+1=9-x^{2} $ $ 2x^{2}=8 $ $ x^{2}=4 $ $ x= \pm 2 $ The points of intersection are (-2, 5) and (2, 5). {Step 2 of 5}: Sketch the curves. {Step 3 of 5}: Rotate this shaded region about $ y=-1 $ Sketch the solid obtained after rotation and a washer. {Step 4 of 5}: The outer radius of the washer $ 1+ \left( 9-x^{2} \right) =10-x^{2} $ The inner radius of the washer $ 1+ \left( 1+x^{2} \right) =2+x^{2} $ The cross-sectional area of the washer $ A \left( x \right) = \pi \left[ \left( 10-x^{2} \right) ^{2}- \left( 2+x^{2} \right) ^{2} \right] $ $ = \pi \left[ 100+x^{4}-20x^{2}-4-x^{4}-4x^{2} \right] $ $ = \pi \left[ 96-24x^{2} \right] $ {Step 5 of 5}: The volume of solid $ V= \int _{-2}^{2} \pi \left[ 96-24x^{2} \right] dx $ $ = \pi \int _{-2}^{2} \left( 96-24x^{2} \right) dx $ $ = \pi \left[ 96x-\frac{24}{3}x^{3} \right] _{-2}^{2} $ [By the fundamental theorem of calculus, part 2] $ = \pi \left[ 96 \times 2-8 \times 8+96 \times 2-8 \times 8 \right] $ $ = \pi \left[ 384-128 \right] $ $ V=256 \pi $
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