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 - 9.1 Arc Length and Surface Area - Exercises - Page 469: 44


$$ 3.43653$$

Work Step by Step

Since $$ y=x^{4}\to y'=4x^{3}$$ Then \begin{align*} \text{Surface area =}& 2\pi \int_{a}^{b} f(x) \sqrt{1+f'^2(x)}dx\\ &=2 \pi \int_{0}^{1}x^4 \sqrt{1+16x^6 } dx\\ & \text{Using a calculator gives:} \\ &\approx 3.43653 \end{align*}
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