Chapter 11.3, Problem 54PPE

To determine

What fraction of the can is empty.

Answer to Problem 54PPE

  $3-\frac{2\pi r^2}{2\pi r^3+\pi r^2}$

Explanation of Solution

Given:

Three tennis balls with radius r are packed into a cylindrical can with radius r and height $6r+1$ .

Concept Used:

  $\begin{array}{l}\text{Volume of Sphere}=\frac{4}{3}\pi r^3\\ \text{Volume of Cylinder}=\pi r^{2}h\end{array}$

Calculation:

In order to find the fraction of the can that is empty, first find the volume of the balls and the volume of the can.

Volume of 1 tennis ball is given by $\frac{4}{3}\pi r^3$ , so the volume of 3 tennis balls would be $3\left(\frac{4}{3}\pi r^3\right)=4\pi r^3$ .

And the volume of the cylindrical can is $=\pi r^{2}h=\pi r^2\left(6r+1\right)=6\pi r^3+\pi r^2$ .

So, the empty space in the can after packing the three tennis balls is

  $\begin{array}{l}\text{Total volume of can}-\text{Volume of 3 balls}\\ \text{=}\left(6\pi r^3+\pi r^2\right)-\left(4\pi r^3\right)\\ =2\pi r^3+\pi r^2\end{array}$

Now to find the fraction of the can which is empty, divide the total volume of can by the empty space as shown below

  $2\pi r^3+\pi r^2\begin{array}{c}\hfill 3\\ \hfill \overline{)\begin{array}{l}6\pi r^3+\pi r^2\\ \underset{\_}{-\left(6\pi r^3+3\pi r^2\right)}\\ -2\pi r^2\end{array}}\end{array}$

Thus, the fraction of the can that is empty is $3-\frac{2\pi r^2}{2\pi r^3+\pi r^2}$ .