Calculus: Early Transcendentals 8th Edition

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

Chapter 6 - Review - Exercises: 13

Answer

$ V=2 \pi \int _{\frac{- \pi }{3}}^{\frac{ \pi }{3}} \left( \frac{ \pi }{2}-x \right) \left( cos^{2}x-\frac{1}{4} \right) dx $

Work Step by Step

{Step 1 of 3} Consider the functions $ y=cos^{2}x, \vert x \vert \leq \frac{ \pi }{2},y=\frac{1}{4} $ is shown below {Step 2 of 3} The shaded region p when rotated about $ x=\frac{ \pi }{2} $ will have a radius $ \left( \frac{ \pi }{2}-x \right) $ and its height is $ \left( cos^{2}x-\frac{1}{4} \right) $,like this the solid will form and will integrate it from $ x=-\frac{ \pi }{3} $ to $ x=\frac{ \pi }{3} $ {Step 3 of 3} Since the solid lies between $ x=-\frac{ \pi }{3} $ and $ x=\frac{ \pi }{3} $ , its integral for the volume is $ V=2 \pi \int _{\frac{- \pi }{3}}^{\frac{ \pi }{3}} \left( \frac{ \pi }{2}-x \right) \left( cos^{2}x-\frac{1}{4} \right) dx $
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