Answer
$\frac{1}{3}+\frac{2}{3(6r+1)}$
Work Step by Step
The volume of a sphere, $V_\text{sphere}$, is given by
$$
V_\text{sphere}=\frac{4}{3}\pi r^3
,$$where $r$ is the radius of the circle. Therefore, the volume of the $3$ tennis balls with radius $r$ is
$$\begin{aligned}
3V_\text{sphere}&=3\left(\frac{4}{3}\pi r^3\right)
\\&=
4\pi r^3
.\end{aligned}$$
The volume of a cylinder, $V_\text{cylinder}$ is given by
$$
V_\text{cylinder}=\pi r^2h
,$$ where $r$ is the radius and $h$ is the height of the cylinder. Therefore, the volume of the cylinder with radius, $r$, and height, $h=6r+1$, is
$$\begin{aligned}
V_\text{cylinder}&=\pi r^2(6r+1)
\\&=
6\pi r^3+\pi r^2
.\end{aligned}$$
When the $3$ tennis balls are placed inside the cylinder, the empty space, $V_\text{empty}$ is computed as
$$\begin{aligned}
V_\text{empty}&=V_\text{cylinder}-3V_\text{sphere}
\\&=
6\pi r^3+\pi r^2-4\pi r^3
\\&=
2\pi r^3+\pi r^2
.\end{aligned}
$$
Therefore, the part of the can that is empty is
$$\begin{aligned}
\frac{V_\text{empty}}{V_\text{cylinder}}&=\frac{2\pi r^3+\pi r^2}{6\pi r^3+\pi r^2}
\\&=
(2\pi r^3+\pi r^2)\div(6\pi r^3+\pi r^2)
.\end{aligned}$$
Using the long division method below, then
$$\begin{aligned}
&(2\pi r^3+\pi r^2)\div(6\pi r^3+\pi r^2)
\\&=
\frac{1}{3}+\frac{\color{red}{2\pi r^2}}{\color{red}{3}\color{blue}{(6\pi r^3+\pi r^2)}}
\\&=
\frac{1}{3}+\frac{2\pi r^2}{3\pi r^2(6r+1)}
\\&=
\frac{1}{3}+\frac{2}{3(6r+1)}
.\end{aligned}$$Hence, the fraction of the can that is empty is $\frac{1}{3}+\frac{2}{3(6r+1)}$.
