Cylinder Volume Calculator
Calculate the volume of a cylinder from its radius and height.
About
Cylinder Volume Calculator
The volume of a cylinder is V = πr²h: the area of the circular end multiplied by the height. Enter a radius and a height above and the calculator returns the volume with the substitution shown, plus the same figure in litres and US gallons.
What the formula means
A cylinder is a circle dragged along a straight line. Every horizontal slice through it is the same circle, so the volume is simply that circle's area repeated for the full height. The circle's area is πr², and stacking it through a height h gives πr²h.
That reasoning holds for any prism. A rectangular tank is a rectangle dragged upward, so its volume is length × width × height. The pattern is always base area times height, and the only thing that changes between shapes is how you work out the base.
Slanted cylinders
The formula still works if the cylinder leans. Cavalieri's principle says two solids with equal cross-sections at every height have equal volume, so a leaning stack of coins holds exactly as much as a straight one. What matters is the perpendicular height, not the length of the slanted side. Measuring along the slope is a common way to overestimate.
A worked example
Take a radius of 2 units and a height of 4 units.
- Square the radius: 2² = 4
- Multiply by π for the base area: 4π, which is 12.566 square units
- Multiply by the height: 4π × 4 = 16π
- As a decimal: 16 × 3.14159265 = 50.2655 cubic units
Again the exact form, 16π, is the more useful answer when the question asks for one.
Working from the diameter
Pipes, cans and tanks are specified by diameter, not radius, so halving comes first. A tank 1.2 m across and 1.5 m tall has a radius of 0.6 m, giving π × 0.36 × 1.5 = 1.6965 m³, which is 1696.5 litres.
Skipping the halving step quadruples the answer, because the radius is squared. If a tank calculation comes out four times larger than the label claims, this is why.
Pipes and hollow cylinders
A pipe is a cylinder with a cylinder removed from the middle. Work out the volume using the outer radius, work it out again using the inner radius, and subtract. The result is πh(R² − r²), where R is outer and r is inner.
This matters for anyone sizing a heating system, because the water sitting inside the pipework is part of the system volume, and for anyone costing material, because the metal is the difference between the two figures rather than either one.
Capacity against contents
A cylinder's volume is the space inside it, which is rarely the same as what it holds in practice. A 355 ml drink can measures roughly 6.6 cm across and 12.3 cm tall, which gives a geometric volume near 423 ml. The gap is headspace, the deliberate air gap that lets the contents expand without bursting the can.
Fuel tanks, water butts and pressure vessels all carry similar allowances. Treat the geometric figure as the ceiling rather than the working capacity.
Units
| Measurements in | Volume in | Reads as |
|---|---|---|
| centimetres | cm³ | millilitres |
| metres | m³ | 1000 litres each |
| inches | in³ | 231 in³ per US gallon |
| feet | ft³ | 7.481 US gallons each |
The radius and the height must be in the same unit before you multiply. A radius in centimetres with a height in metres is the most frequent source of a hundredfold error.
Where it gets used
Tanks and vessels. Water storage, fuel, brewing and chemical processing all size vessels this way, then apply a fill factor for headspace.
Plumbing and heating. System volume determines how much inhibitor to add and how large an expansion vessel must be. It comes from the pipe bore, the length, and this formula.
Construction. Concrete for a cylindrical footing or pile is priced by the cubic metre, and the calculation is πr²h with an allowance for waste.
Engines. Swept volume per cylinder is πr²s, where s is the stroke. Multiply by the number of cylinders and you have the engine's displacement, which is why a 2.0 litre engine is described by a volume rather than a size.
Horizontal tanks and partial fills
A cylinder standing upright is straightforward: the contents are πr² multiplied by the depth. Lying on its side it is not, because the surface of the liquid is a chord across the circular end rather than a full circle, and depth stops scaling with volume.
For a horizontal cylinder filled to depth d, the cross-sectional area of the liquid is r² arccos((r − d)/r) − (r − d)√(2rd − d²), and the volume is that area multiplied by the length. Half full is the one easy case: exactly half the total, by symmetry.
This is why a dipstick on a horizontal fuel tank has uneven markings. A tank a quarter full by depth holds less than a quarter of its capacity, because the narrow bottom of the circle contributes little area.
Surface area, and what it costs
A closed cylinder has two circular ends and one curved side that unrolls into a rectangle. Its surface area is 2πr² + 2πrh, or 2πr(r + h). The curved part alone is 2πrh, which is the figure that matters when sizing insulation for a pipe or a label for a can.
For a fixed volume, material use is lowest when the height equals the diameter. A can shaped purely for efficiency would be as tall as it is wide. Real drink cans are taller than that because they are shaped for the hand and the shelf rather than for minimum aluminium.
Converting to mass
Multiply volume by density. A steel bar 50 mm across and 1 m long has a radius of 25 mm and a volume of π × 625 × 1000 = 1.963 × 10⁶ mm³, which is 1963 cm³. At 7.85 g/cm³ that is about 15.4 kg. Getting from cubic millimetres to cubic centimetres means dividing by 1000, which is the step most often missed.
For liquids the shortcut is easier: one litre of water weighs one kilogram, so cubic metres and tonnes map directly for anything close to the density of water.
Oval tanks and rounded rectangles
Plenty of tanks are not truly circular. An oval or obround cross-section, common in transport tanks where height is limited, is two semicircles joined by a straight section. Its area is πr² + 2rw, where w is the length of the straight part, and the volume is that multiplied by the length.
Treating such a tank as a plain cylinder using the widest measurement overstates its capacity, sometimes by a fifth. If the ends look like a running track rather than a circle, use the obround area.
Checking a result quickly
A cylinder holds a little under 80% of the box that would enclose it, because π/4 is 0.7854. So a tank 1 m across and 2 m tall sits inside a 1 × 1 × 2 m box holding 2000 litres, and holds about 1571 litres.
That ratio is worth memorising. It gives a sanity check on any cylinder calculation in a couple of seconds, and it catches the diameter-for-radius error immediately, since that mistake produces an answer larger than the enclosing box.
Common mistakes
Using diameter as radius. The most common error, and it multiplies the answer by four.
Measuring the slanted side. Only the perpendicular height belongs in the formula.
Forgetting the wall thickness. Outer dimensions overstate capacity. For thin cans the difference is small; for a thick-walled vessel it is not.
Assuming full capacity. Geometric volume is the maximum, not the usable amount.
Common questions
Frequently asked questions
V = πr²h. Square the radius, multiply by π to get the area of the circular end, then multiply by the height. A cylinder of radius 2 and height 4 has a volume of 16π, or 50.2655 cubic units.
Halve the diameter first. A tank 1.2 m across and 1.5 m tall has a radius of 0.6 m and holds π × 0.36 × 1.5 = 1.6965 m³, which is 1696.5 litres.
Work in metres to get cubic metres, then multiply by 1000. Alternatively work in centimetres to get cubic centimetres, which are millilitres directly.
Calculate with the outer radius, calculate again with the inner radius, and subtract. The formula is πh(R² − r²), where R is the outer radius and r the inner.
Yes. Cavalieri's principle means a leaning cylinder holds the same as an upright one of the same perpendicular height. Use the vertical height, not the slanted length.
Headspace. A 355 ml can has a geometric volume near 423 ml, and the difference is the deliberate air gap that allows the contents to expand.
The volume quadruples, because the radius is squared. Doubling the height only doubles the volume, since height appears to the first power.
Swept volume per cylinder is πr²s, where s is the stroke length. Multiplied by the number of cylinders it gives the displacement, which is why engines are described in litres.