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: 35


$$16\sqrt 2 \ \pi.$$

Work Step by Step

Since $y=x$ then $y'=1$. Hence the surface area is given by $$2\pi\int_0^4 y\sqrt{1+(y')^2}dx =2\pi\int_0^4 x\sqrt{1+1}dx\\ =2\sqrt 2\pi\int_0^4 x dx=\sqrt 2 \pi x^2|_0^4=16\sqrt 2 \ \pi.$$
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