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 476: 23


$$(15 b+30 a) h^{2} \text { lb } $$

Work Step by Step

Place the origin along the top edge of the trough with the positive $y$ -axis pointing downward. The width of the front side of the trough varies linearly from $b$ when $y=0$ to $a$ when $y=h ;$ thus, the width of the front side of the trough at depth $y$ feet is given by $$ b+\frac{a-b}{h} y $$ Then \begin{aligned} F&=w \int_{a}^{b} y f(y) d y\\ &=90 \int_{0}^{h} y\left(b+\frac{a-b}{h} y\right) d y \\ &=90 b \int_{0}^{h} y d y+\frac{90(a-b)}{h} \int_{0}^{h} y^{2} d y \\ &=90 b\left[\frac{y^{2}}{2}\right]_{0}^{h}+\frac{90(a-b)}{h}\left[\frac{y^{3}}{3}\right]_{0}^{h} \\ &=45 b h^{2}+30 \frac{(a-b)}{h} h^{3} \\ &=(45 b+30(a-b)) h^{2} \\ &=(45 b+30 a-30 b) h^{2} \\ &= (15 b+30 a) h^{2} \text { lb } \end{aligned}
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.