Calculus: Early Transcendentals 8th Edition

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

Chapter 6 - Review - Exercises: 16

Answer

a) A=$ \frac{5}{12} $ b)V = $ \frac{41 \pi }{105} $ c) V= $ \frac{13 \pi }{30} $

Work Step by Step

{Step 1 of 9} A) First, find the points of intersection of the curves y=x3 and y=2x-x2 $x^3=2x-x^2$ $x^3+x^2-2x=0$ $x(x^2+x-2)=0$ $x=0$ or $ x2+x-2=0$ $x=0$ or $x2+2x-x-2=0$ $x=0$ or $(x+2)(x-1)=0$ $x=0$ or $x=-2 $ or $x=1$ The points of intersection are (0,0), (-2,-8), and (1,1). Consider the two points (0,0) and (1,1), which are in the first quadrant. {Step 2 of 9} Sketch the curves. {Step 3 of 9} Find the area of the shaded region, which is labelled R. A=$ \int _{0}^{1} \left[ \left( 2x-x^{2} \right) -x^{3} \right] dx $ =$ \left[ x^{2}-\frac{1}{3}x^{3}-\frac{1}{4}x^{4} \right] _{0}^{1} $ =1-$ \frac{1}{3} $ -$ \frac{1}{4} $ A=$ \frac{5}{12} $ {Step 4 of 9} B) Rotate this region about the x-axis. Use the slicing method. Consider a strip in this region parallel to the y-axis. Rotate this region about the x-axis to produce a washer. {Step 5 of 9} The outer radius of the washer is $2x-x^2$ The inner radius of the washer is $x^3$ The cross-sectional area of the washer A(x) =$ \pi \left[ \left( 2x-x^{2} \right) ^{2}- \left( x^{3} \right) ^{2} \right] $ =$ \pi \left[ 4x^{2}+x^{4}-4x^{3}-x^{6} \right] $ {Step 6 of 9} Then the volume of the solid V=$ \int _{0}^{1}A \left( x \right) dx $ =$ \pi \int _{0}^{1} \left[ 4x^{2}+x^{4}-4x^{3}-x^{6} \right] $dx =$ \pi \left[ \frac{4}{3}x^{3}+\frac{1}{5}x^{5}-x^{4}-\frac{1}{7}x^{7} \right] _{0}^{1} $ =$ \pi \left[ \frac{4}{3}+\frac{1}{5}-1-\frac{1}{7} \right] $ V = $ \frac{41 \pi }{105} $ {Step 7 of 9} ( C ) rotate region R about the y-axis and use the cylindrical shell method. Consider a vertical strip in region R. rotate the region about the y-axis to produce a cylindrical shell with a radius of x and a height of $(2x-x^2-x^3)$. {Step 8 of 9} Draw a Figure {Step 9 of 9} The circumference of the shell is 2$ \pi x $ The volume of the solid V = $ \int _{0}^{1} \left( 2 \pi x \right) \left( 2x-x^{2}-x^{3} \right) dx $ V= 2$ \pi \int _{0}^{1} \left( 2x^{2}-x^{3}-x^{4} \right) dx $ =2$ \pi \left[ \frac{2}{3}x^{3}-\frac{1}{4}x^{4}-\frac{1}{5}x^{5} \right] _{0}^{1} $ =2$ \pi \left[ \frac{2}{3}-\frac{1}{4}-\frac{1}{5} \right] $ V= $ \frac{13 \pi }{30} $
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