Algebra 2 (1st Edition)

Published by McDougal Littell
ISBN 10: 0618595414
ISBN 13: 978-0-61859-541-9

Chapter 1, Equations and Inequalities - 1.4 Rewrite Equations and Formulas - 1.4 Exercises - Problem Solving - Page 32: 39

Answer

$$V_1= \frac{l^2}{4\pi} W$$ $$V_2= \frac{W^2}{4\pi} l$$

Work Step by Step

Generally, we know that the volume of a cylinder is defined as: $V=\pi r^2h$ We have to define $r$, and we will use the formula of circumference for that: $C=2\pi r$ $(1)$Let's first calculate the first scenario, where: $Circumference(C) = l$ $Height(h)=W$ Define $r$ in terms of the circumference $l=2\pi r$ $r=\frac{l}{2\pi}$ So we get the following: $V_1=\pi (\frac{l}{2\pi})^2 W$ Simplify: $V_1= \frac{l^2}{4\pi} W$ --- $(1)$Let's calculate the second scenario, where: $Circumference(C) = W$ $Height(h)=l$ Define $r$ in terms of the circumference $W=2\pi r$ $r=\frac{W}{2\pi}$ So we get the following: $V_2=\pi (\frac{W}{2\pi})^2 l$ Simplify: $V_2= \frac{W^2}{4\pi} l$
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