Answer
$$V=\frac{15}{2}\pi$$
Work Step by Step
r=$\sqrt x$
$V=\pi\int\limits_1^4(\sqrt x)^2 dx$
$V=\pi\int\limits_1^4(x)dx$
$V=\pi\frac{x^2}{2}│_1^4$
$V=\pi (\frac{4^2}{2}-\frac{1^2}{2})$
$V=\pi(\frac{16}{2}-\frac{1}{2})$
$V=\frac{15}{2}\pi$
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 453: 3
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