University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 6 - Practice Exercises - Page 390: 6


$2 \sqrt 3$

Work Step by Step

We need to integrate the integral as shown below: Area of each cross-section triangle, $A=\pi \space (r)^2 =\dfrac{(2 \sqrt 4)^2 \sin 60^{\circ}}{2}=4 \sqrt 3 x$ Now, $Volume = \int_{0}^{1} 4 \sqrt 3 \space x \space dx \\=2 \sqrt 3 \times [x^2]_0^1 \\=2 \sqrt 3$
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