## Calculus 8th Edition

Published by Cengage

# Chapter 5 - Applications of Integration - 5.2 Volumes - 5.2 Exercises - Page 375: 30

#### Answer

$\displaystyle{V=\frac{4\pi}{15} }$

#### Work Step by Step

$\displaystyle{A(x)=\pi\left(1-x\right)^2-\pi\left(1-x^{\frac{1}{4}}\right)^2}\\ \displaystyle{A(x)=\pi\left(x^2-2x-x^{\frac{1}{2}}+2x^{\frac{1}{4}}\right)}$ \begin{aligned} V &=\int_{0}^{1} A(x) \ d x \\ V &=\int_{0}^{1} \pi\left(x^2-2x-x^{\frac{1}{2}}+2x^{\frac{1}{4}}\right) \ d x \\ V &=\pi \int_{0}^{1} x^2-2x-x^{\frac{1}{2}}+2x^{\frac{1}{4}}\ dx \\ V &=\pi\left[\frac{1}{3}x^3-x^2-\frac{2}{3} x^{\frac{3}{2}}+\frac{8}{5} x^{\frac{5}{4}}\right]_{0}^{1} \\ V &=\pi\left(\left(\frac{1}{3}(1)^3-(1)^2-\frac{2}{3} (1)^{\frac{3}{2}}+\frac{8}{5} (1)^{\frac{5}{4}}\right)-(0)\right) \\ V &=\frac{4\pi}{15} \end{aligned}

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