Answer
$2 \pi (4 -\pi) $
Work Step by Step
We need to integrate the integral as shown below:
$ Volume = \int_0^{\pi/4} (2 \pi) (4) \times (tan^2 x) dx \\ =8 \pi \int_0^{\pi/4} (\sec^2 x-1) dx \\=8 \pi [\tan x-x] _0^{\pi/4} \\= 2 \pi (4 -\pi) $
University Calculus: Early Transcendentals (3rd Edition)
by Hass, Joel R.; Weir, Maurice D.; Thomas Jr., George B.
Published by
Pearson
ISBN 10:
0321999584
ISBN 13:
978-0-32199-958-0
Chapter 6 - Practice Exercises - Page 391: 14
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