Calculus: Early Transcendentals 8th Edition

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

Chapter 6 - Review - Exercises: 18

Answer

a) sketch the curves $y=1-x^2$ and $ y=x^6-x+1$ with a computer. These points are near (0,1) and (0.75, 0.43). b) $A \approx 0.12 $ c) V $ \approx 0.543 $ d) V $ \approx 0.308 $

Work Step by Step

Step 1 A) First, sketch the curves $y=1-x^2$ and $ y=x^6-x+1$ with a computer. Move the cursor to the points of intersection of these curves to estimate. These points are near (0,1) and (0.75, 0.43). Step 2 B) The area of shaded region R A = $ \int _{0}^{0.75} \left[ \left( 1-x^{2} \right) - \left( x^{6}-x+1 \right) \right] dx $ =$ \int _{0}^{0.75} \left[ 1-x^{2}-x^{6}+x-1 \right] dx $ =$ \left[ \frac{x^{2}}{2}-\frac{x^{3}}{3}-\frac{x^{7}}{7} \right] _{0}^{0.75} $ =$ \left[ \frac{ \left( 0.75 \right) ^{2}}{2}-\frac{ \left( 0.75 \right) ^{3}}{3}-\frac{ \left( 0.75 \right) ^{7}}{7} \right] $ A $ \approx 0.121556$ or $A \approx 0.12 $ Step 3 C) use the slicing method. The outer radius of the washer is (1-x2) The inner radius of the washer is (x6-x+1) The area of the washer A(x) = $ \pi \left[ \left( 1-x^{2} \right) ^{2}- \left( x^{6}-x+1 \right) ^{2} \right] $ = $ \pi \left[ 1+x^{4}-2x^{2}- \left( x^{12}-2x^{7}+2x^{6}-2x+x^{2}+1 \right) \right] $ = $ \pi \left[ 1+x^{4}-2x^{2}-x^{12}+2x^{7}-2x^{6}+2x-x^{2}-1 \right] $ = $ \pi \left[ x^{4}-3x^{2}-x^{12}+2x^{7}-2x^{6}+2x \right] $ Step 4 The volume of the solid V= $ \pi \int _{0}^{0.75} \left[ x^{4}-3x^{2}-x^{12}+2x^{7}-2x^{6}+2x \right] dx $ = $ \pi \left[ \frac{1}{5}x^{5}-x^{3}-\frac{1}{13}x^{13}+\frac{2}{8}x^{8}-\frac{2}{7}x^{7}+\frac{2}{2}x^{2} \right] _{0}^{0.75} $ = $ \pi \left[ \frac{1}{5} \left( 0.75 \right) ^{5}- \left( 0.75 \right) ^{3}-\frac{1}{13} \left( 0.75 \right) ^{13}+\frac{2}{8} \left( 0.75 \right) ^{8}-\frac{2}{7} \left( 0.75 \right) ^{7}+ \left( 0.75 \right) ^{2} \right] $ V $ \approx 0.173 \pi $ or V $ \approx 0.543 $ Step 5 D) Use the cylindrical method: Radius of the shell is x Circumference of the shell is 2$ \pi x $ The height of the shell $(1- x^{2} ) - ( x^{6}-x+1 ) =1-x^{2}-x^{6}+x-1 $ $ = x-x^2-x^6$ Step 6 Volume V = $ \int _{0}^{0.75} \left( 2 \pi x \right) \left( x-x^{2}-x^{6} \right) dx $ = 2$ \pi \int _{0}^{0.75} \left( x^{2}-x^{3}-x^{7} \right) dx $ = 2$ \pi \left[ \frac{1}{3}x^{3}-\frac{1}{4}x^{4}-\frac{1}{8}x^{8} \right] _{0}^{0.75} $ =2$ \pi \left[ \frac{1}{3} \left( 0.75 \right) ^{3}-\frac{1}{4} \left( 0.75 \right) ^{4}-\frac{1}{8} \left( 0.75 \right) ^{8} \right] $ V $ \approx 0.308 $
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