Chapter 12.4, Problem 22PPS

To determine

Volume of given figure.

Answer to Problem 22PPS

Volume $\left(V\right)$ of given figure is $500298.6f{t}^3$

Explanation of Solution

Given:

   Diameter $\left(d_1\right)$ of cylindrical portion $=70ft$

Radius $\left(r_1\right)$ of cylindrical portion $=\frac{\left(d_1\right)}{2}=35ft$

Diameter $\left(d_2\right)$ of cone $=70ft$

Radius $\left(r_2\right)$ of cone $=\frac{\left(d_1\right)}{2}=35ft$

Height $\left(h_1\right)$ of cylindrical portion $=55ft$

Height $\left(h_2\right)$ of cone $=15ft$

Concept Used:

Volume of cylindrical part is calculated as,

  $V=\pi r^{2}h$

Where,

  $V=$ volume of cylindrical part.

  $r=$ radius of cylindrical part.

  $h=$ height of cylindrical part.

Volume of cone is calculated as,

  $V=\frac{1}{3}\pi r^2{h}^2$

Where,

  $V=$ volume of cone.

  $r=$ radius of cone.

  $h=$ height of cone.

Calculation:

For the given composite figure volume is calculated as,

  $\begin{array}{l}V=volumeofcone+volumeofcylindricalportion\\ V=\frac{1}{3}\pi r_1{}^2{h}_1{}^2+\pi r_2{}^2{h}_2\end{array}$

Put the values in the given equation,

  $\begin{array}{l}V=volumeofcone+volumeofcylindricalportion\\ V=\frac{1}{3}\pi r_1{}^2{h}_1{}^2+\pi r_2{}^2{h}_2\\ V=\frac{1}{3}\pi {\left(35\right)}^2{\left(15\right)}^2+\pi {\left(35\right)}^2\left(55\right)\\ V=288633.8+211664.8\\ V=500298.6f{t}^3\end{array}$

Conclusion:

Thus, volume $\left(V\right)$ of given figure is $500298.6f{t}^3$