Substitute 221 for V in the formula
[tex]\begin{gathered} V=\frac{4}{3}\pi r^3 \\ \ast\ast221=\frac{4}{3}\pi r^3\ast\ast \\ 3\cdot221=4\pi r^3 \\ \frac{3\cdot221}{4\pi}=r^3 \\ 52.8=r^3 \end{gathered}[/tex]
So the radius of the ball is
[tex]\begin{gathered} 52.8cm^3=r^3 \\ \sqrt[3]{52.8cm^3\text{ }}=r \\ \ast\ast3.75cm=r\text{ }\ast\ast \end{gathered}[/tex]
The base of the cylindrical package will a radius equal to that of the tennis ball, or
[tex]\ast\ast3.75\text{ cm }\ast\ast[/tex]
The height of the package will equal the diameter of three tennis balls, or
[tex]\begin{gathered} \ast\ast3\lbrack2(3.75cm)\rbrack\ast\ast=\ast\ast3\lbrack7.5cm\rbrack\ast\ast=\ast\ast22.5\operatorname{cm}\ast\ast \\ \end{gathered}[/tex]
This is because the diameter is twice the radius and we have 3 tennis balls in the cylindrical package.
So, the volume of the package is
[tex]\begin{gathered} V=\pi r^2h \\ V=\ast\ast\pi(3.75cm)^2\ast\ast\cdot\ast\ast22.5\operatorname{cm}\ast\ast \\ V=\ast\ast316.41\ast\ast\operatorname{cm}^3 \end{gathered}[/tex]
And since we are asked to round to the next cubic centimeter, then the volume of the cylindrical package is
[tex]V=316\operatorname{cm}^3[/tex]
