Calculus: Early Transcendentals 8th Edition

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

Chapter 6 - Review - Exercises: 7

Answer

$ V=\frac{64 \pi }{15} $

Work Step by Step

{Step 1 of 5}: Find, find the points of intersection of $ y=2x $ and $ y=x^{2} $ $ x^{2}=2x $ $ x^{2}-2x=0 $ $ x \left( x-2 \right) =0 $ Or $ x=0 $ or $ x=2 $ The points of Intersection are (0, 0) and (2, 4). {Step 2 of 5}: Sketch the curves. {Step 3 of 5}: Rotate the shaded region about the x-axis. Sketch the solid obtained by rotating the shaded region about the x-axis and the washer. {Step 4 of 5}: The outer radius of the washer is 2x The inner radius of the washer is x2 The cross-sectional area of the washer $ A \left( x \right) = \pi \left[ \left( 2x \right) ^{2}- \left( x^{2} \right) ^{2} \right] $ $ = \pi \left[ 4x^{2}-x^{4} \right] $ {Step 5 of 5}: The volume of the solid obtained by rotating the shaded region about the x-axis $ V= \int _{0}^{2}A \left( x \right) dx $ $ = \int _{0}^{2} \pi \left[ 4x^{2}-x^{4} \right] dx $ $ = \pi \int _{0}^{2} \left( 4x^{2}-x^{4} \right) dx $ $ = \pi \left[ \frac{4}{3}x^{3}-\frac{1}{5}x^{5} \right] _{0}^{2} $ [By the fundamental theorem of calculus, part 2] $ = \pi \left[ \frac{32}{3}-\frac{32}{5} \right] $ $ V=\frac{64 \pi }{15} $
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.