Answer
False
Work Step by Step
False.
Graphically, the curve $y=\sqrt{x}$ is above $y=x$ for $0\leq x\leq 1$.
Then, the volume of the resulting solid by revolving the region bounded by both curves should be $V=\int_0^1\pi (\sqrt{x}^2-x^2)dx$.
Calculus: Early Transcendentals 9th Edition
by Stewart, James
Published by
Cengage Learning
ISBN 10:
1337613924
ISBN 13:
978-1-33761-392-7
Chapter 6 - Review - True-False Quiz - Page 479: 3
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