Question

A cylindrical container holds three tennis balls, each with a diameter of 2.5 inches. The balls are stacked so they touch the container on the sides, top, and bottom. What is the volume of the empty space inside the container? Round your answer to the nearest hundredth, if necessary.

Answer 1

Answer:

the volume of the empty space is $12.22 in^{3}$

Step-by-step explanation:

This problem bothers on the mensuration of solid shapes, cylinder and sphere.

1. If we take a careful look at the problem, we will understand that the diameters of the four balls all sum up to give the height of the cylinder

2. Also to solve for the empty space, we have to subtract the volume of the 3 balls from the volume of the cylinder.

Given data

Diameter of sphere and cylinder d=  2.5 in

Radius r =  $\frac{diameter}{2} = \frac{2.5}{2} = 1.25 in$

Height h=  $3(2.5)= 7.5 in$

let us solve for the volume of the cylinder

$volume= \pi r^{2} h\\volume= 3.14*1.25^{2}* 7.5\\volume= 3.142*1.56*7.5\\volume= 36.76 in^{3}$

we can now solve for the volume of all three balls

we know that the volume of sphere is given as

$volume= \frac{4}{3} \pi r^{3}$

since we are solving for 3 balls we multiply by 3

$volume= 3*\frac{4}{3} \pi r^{3} \\volume= 4\pi r^{3} \\volume= 4*3.142*1.25^{3} \\volume= 4* 3.142*1.95\\volume= 24.54in^{3}$

hence the volume of the empty space is

= volume of cylinder- volume of all three balls

$=36.76- 24.5\\= 12.22 in^{3}$