Calculus: Early Transcendentals (2nd Edition)

Published by Pearson
ISBN 10: 0321947347
ISBN 13: 978-0-32194-734-5

Chapter 6 - Applications of Integration - 6.7 Physical Applications - 6.7 Exercises - Page 469: 43

Answer

$6533.33 \ N $

Work Step by Step

The width of the dam at height $y$ is given by $-y=-2x \implies y=2x \implies x=0.5 y$ So, $w(y)=(2)(0.5)y=y$ We need to use the formula such as: $F=\int_b^t \rho g (a-y) w(y) \ dy$. Plug in the above formula the given values to obtain: $F=\int_b^t \rho g (a-y) w(y) \ dy \\= \int_0^{1} \rho g (2-y) y \ dy\\=\dfrac{2}{3} \times \rho g \\=(\dfrac{2}{3}) (1000) (9.81) \\=6533.33 \ N $
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