Calculus (3rd Edition)

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

Chapter 9 - Further Applications of the Integral and Taylor Polynomials - 9.2 Fluid Pressure and Force - Exercises - Page 475: 21


$$5.460393400 \times 10^{11} \mathrm{N}$$

Work Step by Step

Let $y=0$ be at the bottom of the dam, so that the top of the dam is at $y=185 .$ Then the width of the dam at height $y$ is $$2000+\frac{1000 y}{185} .$$ The dam is inclined at an angle of $55^{\circ}$ to the horizontal, so the height of a horizontal strip is $$ \frac{\Delta y}{\sin 55^{\circ}} \approx 1.221 \Delta y $$ so that the area of such a strip is $$ 1.221\left(2000+\frac{1000 y}{185}\right) \Delta y $$ Then \begin{aligned} F &=\rho g \int_{0}^{185} 1.221 y\left(2000+\frac{1000 y}{185}\right) d y\\ &=\rho g \int_{0}^{185} 2442 y+6.6 y^{2} d y\\ &=\left.\rho g\left(1221 y^{2}+2.2 y^{3}\right)\right|_{0} ^{185} \\ &=55,718,300 \rho g\\ &=55,718,300 \cdot 9800\\ &=5.460393400 \times 10^{11} \mathrm{N} \end{aligned}
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