Answer
$\dfrac{\pi (3 \sqrt 3-\pi)}{3}$
Work Step by Step
We need to integrate the integral as shown below:
$Volume = \pi \int_{0}^{\pi/3}(\tan^2 x) dx \\=\pi \int_{0}^{\pi/3}(\sec^2 x-1) dx \\=\dfrac{\pi}{3} (3 \sqrt 3-\pi)$
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: 11
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