Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

Chapter 1 - Section 1.10 - Modeling with Functions - Exercise Set - Page 296: 66


The required volume is\[V=\frac{1}{12}\pi {{h}^{3}}\]

Work Step by Step

The radius of the cone is $r=6\text{ feet}$ and the height is $12\text{ feet}$. Equate the small cone to the bigger cone: $\frac{r}{h}=\frac{6}{12}$ So, $\begin{align} & r=\frac{6}{12}h \\ & r=\frac{1}{2}h \\ \end{align}$ Then the volume of cone is: $\begin{align} & V\left( h \right)=\frac{1}{3}\pi {{r}^{2}}h \\ & =\frac{1}{3}\pi {{\left( \frac{1}{2}h \right)}^{2}}h \\ & =\frac{1}{3}\pi \left( \frac{1}{4}{{h}^{2}} \right)h \\ & =\frac{1}{12}\pi {{h}^{3}} \end{align}$ Therefore, the required volume is $V=\frac{1}{12}\pi {{h}^{3}}$.
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