Answer
$$16\sqrt 2 \ \pi.$$
Work Step by Step
Since $y=x$ then $y'=1$. Hence the surface area is given by
$$2\pi\int_0^4 y\sqrt{1+(y')^2}dx =2\pi\int_0^4 x\sqrt{1+1}dx\\
=2\sqrt 2\pi\int_0^4 x dx=\sqrt 2 \pi x^2|_0^4=16\sqrt 2 \ \pi.$$
Calculus (3rd Edition)
by Rogawski, Jon; Adams, Colin
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
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