Elementary Technical Mathematics

Published by Brooks Cole
ISBN 10: 1285199197
ISBN 13: 978-1-28519-919-1

Chapter 12 - Section 12.8 - Cylinders - Exercise - Page 427: 7

Answer

$56.5 \; in^3$.

Work Step by Step

The given values are Inner radius $r_1=4.00\;in$. Outer radius $r_2=5.00 \;in$. Height $h=2.00\;in$. The formula for volume is: $V=\pi r^2 h$ $V=\pi r_1^2h$ Plug values into the formula. $V=\pi (4.00\; in)^2(2.00\; in)$ Simplify. $V=32\pi \; in^3$ $V=\pi r_2^2h$ Plug values into the formula. $V=\pi (5.00\; in)^2(2.00\; in)$ Simplify. $V=50\pi \; in^3$ $V=$ outer volume $-$ inner volume. $=50\pi \; in^3- 32\pi \; in^3 $ $=(50-32)\pi \; in^3$ $=18\pi \; in^3$ Simplify (rounded to one decimal place). $=56.5 \; in^3$.
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