University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 3 - Section 3.11 - Linearization and Differentials - Exercises - Page 199: 53


$dV=180\pi=565.4867 in^3$

Work Step by Step

We are given: cylinder height(h)=30 in, radius=6in, and thickness=0.5in We need to find the change in volume of the cylinder. The thickness is changing at a rate of (dr)=0.5in The height is not changing, so dh=0 Volume(V)=$\pi r^2 h$ on differentiating the above equation: $dV=2\pi rh dr+\pi r^2 dh$ $dV=2\pi 6\times30\times0.5 dr+0$ $dV=180\pi=565.4867 in^3$ Thus, the change in volume of the cylinder is: $180\pi=565.4867 in^3$
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