Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 6 - Applications of the Integral - 6.5 Work and Energy - Exercises - Page 317: 14


$1891.6 \ J$

Work Step by Step

The volume of one layer is equal to: $(0.64-y^2) \pi \Delta y \mathrm{m}^{3}$ The force of one layer is equal to: $5880 \pi (0.64-y^2) \Delta y \ N$ Therefore, the work done against can be computed as: $ W=\int_{0}^{0.8} 5880 \pi (0.64-y^2) \Delta y \ d y\\=5880 \pi \int_{0}^{0.8} (0.64-y^2) y \ d y\\ = 5880 \pi [0.32 y^2-\dfrac{y^4}{4}]_0^{0.8} \\=5880 \pi [\dfrac{64}{625}] \\ \approx 1891.6 \ J$
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