Calculus: Early Transcendentals 8th Edition

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

Chapter 6 - Review - Exercises: 9

Answer

$ V=\frac{1656 \pi }{5} $

Work Step by Step

{Step 1 of 5}: First, find the points of intersection of the curves $ x=0 $ and$ x=9-x^{2} $ To do this, set $ 9-x^{2}=0 $ or $ y^{2}=9 $ $ y= \pm 3 $ The y-coordinates of the points of intersection are -3 and 3. {Step 2 of 5}: Sketch the curves. {Step 3 of 5}: Rotate this shaded region about $ x=-1 $ Sketch the solid and the washer. {Step 4 of 5}: The outer radius of the washer is $ \left( 9-y^{2} \right) +1=10-y^{2} $ The inner radius of the washer is 1. The cross-sectional area of the washer $ A \left( y \right) = \pi \left[ \left( 10-y^{2} \right) ^{1}-1^{2} \right] $ $ = \pi \left[ 100+y^{4}-20y^{2}-1 \right] $ $ = \pi \left[ 99-20y^{2}+y^{4} \right] $ {Step 5 of 5}: The volume of the solid $ V= \int _{-3}^{3}A \left( x \right) dy $ $ = \int _{-3}^{3} \pi \left( 99-20y^{2}+y^{4} \right) dy $ $ = \pi \int _{-3}^{3} \left( 99-20y^{2}+y^{4} \right) dy $ $ = \pi \left[ 99y-\frac{1}{3} \left( 20y^{3} \right) +\frac{1}{3}y^{5} \right] _{-3}^{3} $ [By the fundamental theorem of calculus, part 2] $ = \pi \left[ \left( 99 \times 3-9 \times 20+\frac{243}{5} \right) - \left( -99 \times 3+9 \times 20-\frac{243}{5} \right) \right] $ $ = \pi \left[ 297-180+\frac{243}{5}+297-180+\frac{243}{5} \right] $ $ = \pi \left[ 594-360+\frac{486}{5} \right] $ $ = \pi \left[ 234+\frac{486}{5} \right] $ $ V=\frac{1656 \pi }{5} $
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