Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 5 - Applications of Integration - 5.3 Volumes by Cylindrical Shells - 5.3 Exercises - Page 382: 32

Answer

The region bounded by $y= 2x ,\ \ y=\sin x ,\ \ x=0,\ \ x=\pi/2 $ rotates about $x=-1$

Work Step by Step

Given $$ \int_0^{\pi / 2} 2 \pi(x+1)(2 x-\sin x) d x $$ Compare with $$V= \int_a^b 2\pi r(x) h(x)dx$$ Here $r(x)= x+1,\ \ h(x)=2x-\sin x$. This represents the volume of the generated solid when the region bounded by $$y= 2x ,\ \ y=\sin(x) ,\ \ x=0,\ \ x=\pi/2 $$ rotates about $x=-1$
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