University Calculus: Early Transcendentals (3rd Edition)

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

Chapter 6 - Section 6.1 - Volumes Using Cross-Sections - Exercises - Page 357: 60


$Volume= \pi (\dfrac{36}{5})$ and $\space Weight =192 \space g$

Work Step by Step

We need to integrate the integral to compute the volume. We have: $$Volume = \int_{0}^{6} (36-x^2) \dfrac{x^2}{144} dx \times \pi \\= \dfrac{ \pi }{144} [12x^3 - \dfrac{x^{5}}{5}]_{0}^{6} \\= \pi (\dfrac{36}{5})$$ Now, $$Weight =\pi (\dfrac{36}{5}) \approx 192 \space g$$
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