This problem is one that Portland Community College has donated to the Open Problem Library.
A cylinder’s base’s radius is \({6\ {\rm m}}\text{,}\) and its height is \({10\ {\rm m}}\text{.}\)
 This cylinder’s volume, in terms of \(\pi\text{,}\) is .
 This cylinder’s volume, rounded to the hundredth place, is .
Answer 1.
\(360\pi\ {\rm m^{3}}\)
Answer 2.
\(1130.97\ {\rm m^{3}}\)
Solution.
We use \(r\) to represent the base’s radius, and \(h\) to represent the cylinder’s height.
A cylinder’s volume formula is \(V= (\text{base area}) \cdot \text{height}\text{.}\) A cylinder’s base is a circle, with its area formula \(A = \pi r^{2}\text{.}\)
Putting together these two formulas, we have a cylinder’s volume formula:
\(\displaystyle{ V= \pi r^{2} h }\)
Throughout these computations, all quantities have units attached, and we only show them in the final step.

Using the volume formula, we have:\(\displaystyle{\begin{aligned} V \amp = \pi r^{2} h \\ \amp = \pi \cdot 6^{2} \cdot 10 \\ \amp = \pi \cdot 360 \\ \amp = 360 \pi \textrm{ m}^3 \end{aligned}}\)Don’t forget the volume unit \(\textrm{m}^3\text{.}\)

To find the decimal version, we replace \(\pi\) with its decimal value, and we have:\(\displaystyle{\begin{aligned}[t] V\amp = 360 \pi \\ \amp \approx 360 \cdot 3.14\ldots \\ \amp \approx {1130.97\ {\rm m^{3}}} \end{aligned}}\)Don’t forget the volume unit \(\textrm{m}^3\text{.}\)