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: 17


$$ 5652.37 \mathrm{N}$$

Work Step by Step

Since the fluid surface is at height $y = 1$, the horizontal strip at height y is at a depth of $1 − y.$ Moreover, this strip has a width of $e − e^y$. Thus \begin{align*} F&=\rho g \int_{0}^{1}(1-y)\left(e-e^{y}\right) d y\\ &= e\rho g \int_{0}^{1}(1-y) d y-\rho g \int_{0}^{1}(1-y) e^{y} d y\\ &=e\rho g \left.\left(y-\frac{1}{2} y^{2}\right)\right|_{0} ^{1}-\rho g \left.\left((1-y) e^{y}+e^{y}\right)\right|_{0} ^{1}\\ &=\rho g\left(\frac{1}{2} e-(e-2)\right)\\ &=\rho g\left(2-\frac{1}{2} e\right)\\ &=900 \cdot 9.8\left(2-\frac{1}{2} e\right)\\ & \approx 5652.37 \mathrm{N} \end{align*}
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