Calculus (3rd Edition)

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

Chapter 6 - Applications of the Integral - Chapter Review Exercises - Page 320: 53


$9800 \ \pi (h^3+ 2h^2-\dfrac{h^4}{4}) \ J$

Work Step by Step

The volume of one layer is equal to: $ \pi (4y-y^2) \Delta y \mathrm{m}^{3}$ . The force required to lift the layer is equal to: $1000 (9.8) \pi (4y-y^2) \Delta y \ N$ Therefore, the work done to fill the tank can be computed as: $ W=9800 \pi \int_{0}^{h} (y+1) (4y-y^2) \ d y\\= 9800 \pi \int_{0}^{h} (3y^2+4y-y^3) \ d y\\ =9800 \ \pi [y^3+2y^2-\dfrac{y^4}{4}]_0^-\dfrac{y^4}{4} \\=9800 \ \pi (h^3+ 2h^2-\dfrac{h^4}{4}) \ J$
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