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: 12


$ 94579.5 \ J$

Work Step by Step

The volume of one layer is equal to $0.64 \Delta y \mathrm{m}^{3}$. The force weight of one layer is equal to $3763.2 \pi \Delta y \ N$ Therefore, the work done against can be computed as: $ W=\int_{0}^{4} 3763.2 \pi y \ d y\\=3763.2 \pi \int_{0}^{4} y \ d y\\ =3763.2 \ \pi [\dfrac{y^2}{2}]_0^4 \\=30105.6 \pi \\ \approx 94579.5 \ J$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.