Volume of a Cylinder
Geometry & MeasurementThe volume of a cylinder equals pi times the radius squared times the height.
Formula
V = \pi r^2 h
Definition
To find the volume of a cylinder, find the area of the circular base ($\pi r^2$) and then multiply by the height. $V = \pi r^2 h$.
Example
A can with radius $3$ cm and height $10$ cm: $V = \pi \times 9 \times 10 = 90\pi = 282.7$ cm$^3$. A water tank with radius $2$ m and height $5$ m: $V = \pi \times 4 \times 5 = 20\pi = 62.8$ m$^3$.
Key Insight
A cylinder is just a prism with a circle as its base. The base area is $\pi r^2$ (the circle area formula), and you multiply by height exactly like any prism.
Definition
A right circular cylinder with radius $r$ and height $h$ has volume $V = \pi r^2 h$. This equals the base area ($\pi r^2$) times the height, consistent with the general prism formula $V = Bh$. Doubling the radius quadruples the volume ($r$ is squared); doubling the height doubles it.
Example
A cylindrical pool $6$ m in diameter and $1.5$ m deep: $r = 3$ m. $V = \pi(9)(1.5) = 13.5\pi = 42.4$ m$^3$. Since $1$ m$^3 = 1000$ liters, the pool holds $42{,}412$ liters.
Key Insight
The cylinder volume is exactly $3$ times the cone volume with the same base and height (since $V_{cone} = (1/3)\pi r^2 h$). This ratio explains why three ice cream scoops from a cone-shaped server equal one cylindrical cup.
Definition
$V = \pi r^2 h$ is derived by integrating circular cross-sections: $$V = \int_0^h \pi r^2 \, dz = \pi r^2 h.$$ Alternatively, using the shell method: $$V = \int_0^r 2\pi t h \, dt = \pi r^2 h.$$ Both methods agree, illustrating that volume is independent of the integration approach.
Example
Optimizing a cylinder of fixed surface area $SA = 2\pi r^2 + 2\pi r h$ for maximum volume: substituting $h = (SA - 2\pi r^2)/(2\pi r)$ into $V$ and differentiating, the maximum occurs at $h = 2r$ (height equals diameter). This is the "ideal can" problem in calculus optimization.
Key Insight
The optimal can problem shows that minimizing material for a fixed volume and maximizing volume for fixed material both give $h = 2r$. Many real cans deviate from this because of manufacturing costs, stacking efficiency, and lid-to-body material ratios.