Answer
$2 \pi$
Work Step by Step
Area $=\pi r^2=\pi (\sqrt 5 y^2)^2=5 \pi y^4$
We integrate the integral to calculate the volume as follows:
$V= \pi \int_{-1}^{1} 5y^4 dy$
Now, $V=\pi [(y^5) ]_{-1}^1$
or, $= \pi [1-(-1)]$
So, $V=2 \pi$
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: 27
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