Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 9 - Further Applications of the Integral and Taylor Polynomials - Chapter Review Exercises - Page 496: 8



Work Step by Step

Given $$ y=\frac{2}{3} x^{3 / 4}-\frac{2}{5} x^{5 / 4}, \quad[0,1]$$ \begin{align*} y'&= \frac{1}{2} x^{-1 / 4}-\frac{1}{2} x^{1 / 4}\\ 1+y'^2&= 1+\left(\frac{1}{2} x^{-1 / 4}-\frac{1}{2} x^{1 / 4}\right)^2\\ &=\left(\frac{1}{2} x^{-1 / 4}+\frac{1}{2} x^{1 /4}\right)^2 \end{align*} Then the surface area is given by \begin{align*} S&=2\pi \int_{a}^{b} y\sqrt{1+y'^2}dx\\ &=2\pi \int_{0}^{1}\left(\frac{2}{3} x^{3 / 4}-\frac{2}{5} x^{5 / 4}\right)\sqrt{\left(\frac{1}{2} x^{-1 / 4}+\frac{1}{2} x^{1 /4}\right)^2} dx\\ &=2\pi \int_{0}^{1}\left(\frac{2}{3} x^{3 / 4}-\frac{2}{5} x^{5 / 4}\right) \left(\frac{1}{2} x^{-1 / 4}+\frac{1}{2} x^{1 /4}\right) dx\\ &=2\pi \int_0^1 [\frac{1}{3} x^{1/2} +\frac{1}{3}x -\frac{1}{5} x -\frac{1}{5} x^{3/2}]dx\\ &=2\pi[ \frac{2}{9} x^{3/2} +\frac{1}{15}x^2 -\frac{2}{25} x^{5/2 }]\bigg|_{0}^{1}\\ &=\frac{94\pi}{225} \end{align*}
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