Calculus: Early Transcendentals 8th Edition

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

Chapter 6 - Review - Exercises: 15

Answer

a) $ V=\frac{2 \pi }{15} $ b) $ V=\frac{ \pi }{6} $ c) $ V=\frac{8 \pi }{15} $

Work Step by Step

{Step 1 of 10} First, find the points of intersection of the curves $ y=x $ and $ y=x^{2} $ For this, $ x=x^{2} $ $ x^{2}-x=0 $ $ x \left( x-1 \right) =0 or \left( x=0 or x=1 \right) $ The point of intersection are $ \left( 0,0 \right) $ and $ \left( 1,1 \right) $ . {Step 2 of 10} Sketch the curves. {Step 3 of 10} Rotate this shaded region about the x-axis. Consider a strip parallel to the y-axis in this shaded region. Rotation about the x-axis produces a washer. The outer radius of the washer is $ x $ The inner radius of the washer is $ x^{2} $ {Step 4 of 10} The cross-sectional area of the washer $ A \left( x \right) = \pi \left[ x^{2}- \left( x^{2} \right) ^{2} \right] $ $ = \pi \left[ x^{2}-x^{4} \right] $ The volume of the solid $ V= \int _{0}^{1} \pi \left( x^{2}-x^{4} \right) dx $ $ V= \pi \int _{0}^{1} \left( x^{2}-x^{4} \right) dx $ $ = \pi \left[ \frac{x^{3}}{3}-\frac{x^{5}}{5} \right] _{0}^{1} $ $ = \pi \left[ \frac{1}{3}-\frac{1}{5} \right] $ $ V=\frac{2 \pi }{15} $ {Step 5 of 10} Consider the solid obtained by rotating this region about the y-axis. It is easier to use the cylindrical shell method to find the volume of this solid. Consider a strip at a distance of x from the origin. Rotation produces a cylindrical shell with radius x. {Step 6 of 10} Draw the Figure {Step 7 of 10} The circumference of the shell is $ 2 \pi x $, and the height of the shell is $ x-x^{2} $ The volume of the solid $ V= \int _{0}^{1} \left( 2 \pi x \right) \left( x-x^{2} \right) dx $ $ =2 \pi \int _{0}^{1} \left( x^{2}-x^{3} \right) dx $ $ =2 \pi \left[ \frac{1}{3}x^{3}-\frac{1}{4}x^{4} \right] _{0}^{1} $ $ =2 \pi \left[ \frac{1}{3}-\frac{1}{4} \right] $ $ V=\frac{ \pi }{6} $ {Step 8 of 10} Rotate this shaded region about $ y=2 $Use the slicing method {Step 9 of 10} The outer radius of the washer is $ 2-x^{2} $ The inner radius of the washer is $ 2-x $ The cross-sectional area of washer is $ A \left( x \right) = \pi \left[ \left( 2-x^{2} \right) ^{2}- \left( 2-x \right) ^{2} \right] $ $ = \pi \left[ 4+x^{4}-4x^{2}-4-x^{2}+4x \right] $ $ = \pi \left[ x^{4}-5x^{2}+4x \right] $ {Step 10 of 10} The volume of the solid obtained by rotating the bounded region about $ y=2 $ $ V= \int _{0}^{1}A \left( x \right) dx $ $ = \int _{0}^{1} \pi \left[ x^{4}-5x^{2}+4x \right] dx $ $ = \pi \int _{0}^{1} \left( x^{4}-5x^{2}+4x \right) dx $ $ = \pi \left[ \frac{1}{5}x^{5}-\frac{5}{3}x^{3}+4\frac{x^{2}}{2} \right] _{0}^{1} $ $ = \pi \left[ \frac{1}{5}-\frac{5}{3}+2 \right] $ $ V=\frac{8 \pi }{15} $
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