Answer
$6 \pi$
Work Step by Step
We integrate the integral to calculate the volume as follows:
$V= (9\pi/4) \times \int_{0}^{2} y^2 dy$
Now, $V=(9\pi/4) (\dfrac{y^3}{3}]_{0}^{2}$
or, $=(9\pi/4) (\dfrac{8}{3})$
or, $=6 \pi$
Thomas' Calculus 13th Edition
by Thomas Jr., George B.
Published by
Pearson
ISBN 10:
0-32187-896-5
ISBN 13:
978-0-32187-896-0
Chapter 6: Applications of Definite Integrals - Section 6.1 - Volumes Using Cross-Sections - Exercises 6.1 - Page 322: 16
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