Answer
$2 \pi$
Work Step by Step
Area $=R^2\pi- r^2 \pi=\pi (4-4x)$
We integrate the integral to calculate the volume as follows:
$V= \pi \times \int_{0}^{1} (4-4x) dx$
Now, $V= \pi [4x-2x^2]_{0}^{1}$
or, $= \pi (4-2-0+0)$
or, $=2 \pi$
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: 36
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