Chapter 12.4, Problem 31PPS

To determine

To find:

The volume of the rubberized material that makes up the holder.

Answer to Problem 31PPS

The volume of the rubberized material is about $304.1{\text{ cm}}^3$ .

Explanation of Solution

Given information:

A can $\text{12 cm}$ tall fits into a rubberized holder that is $\text{11}\text{.5 cm}$ tall including $\text{1 cm}$ for the thickness of the base of the holder. The thickness of the rim holder is $\text{1 cm}$ .

Calculation:

The container has a radius of $\frac{6.5}{2}+1=4.25\text{ cm}$ and a height of $\text{11}\text{.5 cm}$ .

The empty space used to keep the can has a radius of $\frac{6.5}{2}=3.25\text{ cm}$ and a height of $11.5-1=10.5\text{ cm}$ .

The volume of the rubberized material is the difference between the volumes of the container and the space used for the can.

Volume of the rubberized material is:

  $V=\pi r_c{}^2{h}_c-\pi r_e{}^2{h}_e$

Substitute $r_c=4.25$ , $h_c=11.5$ , $r_e=3.25$ and $h_e=10.5$ in the above formula:

  $\begin{array}{c}V=\pi r_c{}^2{h}_c-\pi r_e{}^2{h}_e\\ =\pi {\left(4.25\right)}^2\left(11.5\right)-\pi {\left(3.25\right)}^2\left(10.5\right)\\ =\pi \left(207.71875\right)-\pi \left(15.84375\right)\\ \approx 304.1\end{array}$

Hence the volume of the rubberized material is about $304.1{\text{ cm}}^3$ .