Answer
$\frac{\pi}{2}$
Work Step by Step
$$ V = \pi \int_{0}^{1}[{(1)}^2-{(\sqrt{y})}^2]dy$$
$$ V = \pi \int_{0}^{1}[1-y]dy$$
$$ V = \pi [y-\frac{y^2}{2}]_{0}^{1}dx$$
$$ V = \pi[1-\frac{1}{2}]=\frac{\pi}{2}$$
Calculus 10th Edition
by Larson, Ron; Edwards, Bruce H.
Published by
Brooks Cole
ISBN 10:
1-28505-709-0
ISBN 13:
978-1-28505-709-5
Chapter 7 - Applications of Integration - 7.2 Exercises - Page 454: 45
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