## Calculus 8th Edition

Published by Cengage

# Chapter 8 - Further Applications of Integration - 8.3 Applications to Physics and Engineering - 8.3 Exercises: 1

#### Answer

(a) $187.5 lb/ft^{2}$ (b) $1875 lb$ (c) $562.5 lb$

#### Work Step by Step

Weight density of water is $\delta = 62.5 lb / ft^{3}$ (a) $P = \delta d \approx (62.5 lb/ft^{3})(3ft) = 187.5 lb/ft^{2}$ (b) $F = PA \approx (187.5 lb/ft^{2})(5ft)(2ft) = 1875 lb$ (c) $F = \int^{3}_{0} \delta x 2dx \approx (62.5)(2) \int^{3}_{0} x dx = 125 [\frac{1}{2} x^{2}]^{3}_{0} = 562.5 lb$

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