Chapter 12.6, Problem 4CYP

To determine

Find the volume of air that was added to the balloon.

Answer to Problem 4CYP

Volume of the sphere that was added to the balloon is $55.7i{n}^3$

Explanation of Solution

Given:

R inflates a spherical balloon to a circumference of about 14 inches . He then adds more air to the balloon until the circumference is about 18 inches.

Formula Used:

Volume of a sphere:

  $V=\frac{4}{3}\pi r^3$ , where r is the radius of the sphere

Circumference of a sphere:

  $C=2\pi r$ , where r is the radius of the sphere

Calculation:

Let the radius of the balloon initially and finally be $r_1$ and $r_2$ respectively.

Given , circumference of the balloon at initial inflation = $\begin{array}{l}{C}_1=2\pi r_1=14inches\\ \Rightarrow r_1=\frac{14}{2\pi }\mathrm{................}[divide\text{ each side by }2\pi ]\\ \Rightarrow r_1\approx 2.23inches\end{array}$

circumference of the balloon at final inflation =

  $\begin{array}{l}{C}_2=2\pi r_2=18inches\\ \Rightarrow r_2=\frac{18}{2\pi }\mathrm{................}[divide\text{ each side by }2\pi ]\\ \Rightarrow r_2\approx 2.9inches\end{array}$

The volume of the sphere at initial inflation = $V_1=\frac{4}{3}\pi r_1{}^3=\frac{4}{3}\pi {(2.23)}^3\approx 46.45i{n}^3$

The volume of the sphere at final inflation = $V_2=\frac{4}{3}\pi r_2{}^3=\frac{4}{3}\pi {(2.9)}^3\approx 102.16i{n}^3$

Volume of the sphere that was added to the balloon = $V_2-V_1\approx 102.16-46.45\approx 55.7i{n}^3$

Hence ,

Volume of the sphere that was added to the balloon is $55.7i{n}^3$