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: 24

Answer

$2.36163$

Work Step by Step

a) Given $$y=x, y=2 x /\left(1+x^3\right) ; \quad \text { about } x=-1$$ First, we find the intersection points \begin{aligned} x&= \frac{2x}{1+x^3}\\ x+ x^4 &= 2x\\ x^4-x&=0\\ x(x^3-1)&=0\\ x(x-1)(x^2+x+1)&=0 \end{aligned} Then $x=0,\ \ x= 1 $. Since the volume of the generated solid given by \begin{aligned} V&=2\pi \int_a^b r(x)h(x)dx \end{aligned} Since $\frac{2x}{1+x^3}\geq x,\ \ 0\leq x\leq 1$ , then $$ h(x) =\frac{2x}{1+x^3}-x,\ \ r(x)= x+1$$ Then \begin{aligned} V&=2\pi \int_a^b r(x)h(x)dx\\ &= 2\pi \int_{0}^{1} \left(x+1\right)\left(\frac{2x}{1+x^3}-x \right)dx \end{aligned} b) Using the calculator, we get \begin{aligned} V &= 2\pi \int_{0}^{1} \left(x+1\right)\left(\frac{2x}{1+x^3}-x \right)dx \\ &=2\pi \left(\frac{-5}{6}+ \frac{2\sqrt{3}\pi}{9}\right)\\ &\approx 2.36163 \end{aligned}
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