Cylinder Volume Calculator
The cylinder volume formula, explained and computed: enter radius and height to get the volume, exact (in terms of π) and decimal, with substitution steps.
How to use this calculator
Enter the radius of the circular base and the height, then press Calculate. The calculator substitutes into V = πr²h, shows the exact form (a multiple of π) and the decimal volume, and walks through each substitution.
How it works
A cylinder is a prism with a circular base: volume = base area × height. The base area is πr², so V = πr²h. That is the whole derivation, one idea, not a formula to memorize blindly.
Worked example
Example: r = 3, h = 5
Step 1, Substitute: V = π × 3² × 5.
Step 2, Simplify: V = 45π.
Step 3, Decimal: V ≈ 141.372 cubic units.
Common mistakes
- Using diameter as r: the formula needs the radius, halve the diameter first. This is the most common cylinder error.
- Forgetting to square: πr²h is not πrh. Square the radius before multiplying.
- Unit mismatch: radius in cm and height in m gives nonsense, convert to one unit first.
When to use it
Cylinders are everywhere: cans, pipes, tanks, batteries, tree trunks, and silos. Volume answers “how much does it hold?”, capacity, concrete pours, and displacement all run through πr²h.
Frequently asked questions
V = πr²h: pi times the radius squared times the height. It is the area of the circular base (πr²) multiplied by the height.
Halve the diameter to get the radius first, then use V = πr²h. For diameter 10 (radius 5) and height 7: V = π × 25 × 7 = 175π ≈ 549.78.
The exact form keeps π symbolic (45π); the decimal approximates it (141.372). Exact form is preferred in school math; decimals suit real measurements.
Cubic units matching your inputs: cm inputs give cm³, inches give in³. The calculator reports “cubic units”, attach your unit.
More worked examples, with different starting points
The calculator takes a radius and a height, but real problems rarely hand you both numbers in the same unit. These examples start from what you actually have: a radius, a diameter, a known volume, or mixed units. Every example shows the exact form first, then the decimal, so you can match school answers and real measurements.
Example: radius given, r = 4 cm, h = 10 cm
Step 1, Substitute: V = π × 4² × 10.
Step 2, Square first: 4² = 16, so V = π × 16 × 10 = 160π.
Step 3, Decimal: V ≈ 502.655 cm³.
Example: diameter given, d = 14 cm, h = 9 cm
Step 1, Halve the diameter: r = 14 ÷ 2 = 7 cm.
Step 2, Substitute: V = π × 7² × 9.
Step 3, Simplify: 7² = 49 and 49 × 9 = 441, so V = 441π ≈ 1,385.442 cm³.
Example: working backwards, find the height from the volume
A cylinder holds 1,000 cm³ and has radius 5 cm. How tall is it?
Step 1, Rearrange the formula: h = V ÷ (πr²).
Step 2, Substitute: h = 1,000 ÷ (π × 25) = 40 ÷ π.
Step 3, Decimal: h ≈ 12.73 cm.
Check: putting h = 40 ÷ π back gives π × 25 × (40 ÷ π) = 1,000 exactly, because the π cancels. The height is about 12.73 cm.
Reading the answer in litres and gallons
The calculator reports cubic units. For liquids, litres or gallons are more useful, and the conversions are exact: 1 cm³ = 1 mL, 1 litre = 1,000 cm³, and 1 m³ = 1,000 litres. In US units, 1 in³ = 16.387064 cm³, 1 US gallon = 231 in³ = 3.78541 litres, and 1 ft³ = 28.3168 litres. Say which gallon you mean: a US gallon is 3.78541 litres, while an imperial gallon is 4.54609 litres.
Example: cubic centimetres to litres, r = 5 cm, h = 20 cm
Step 1, Volume: V = π × 25 × 20 = 500π ≈ 1,570.796 cm³.
Step 2, Millilitres: cm³ and mL are the same unit, so that is 1,570.8 mL.
Step 3, Litres: 1,570.796 ÷ 1,000 ≈ 1.571 litres.
Example: cubic inches to US gallons, r = 6 in, h = 12 in
Step 1, Volume: V = π × 36 × 12 = 432π ≈ 1,357.168 in³.
Step 2, Gallons: one US gallon is 231 in³, so 1,357.168 ÷ 231 ≈ 5.88 US gallons.
Step 3, Cross-check in litres: 1,357.168 × 16.387064 ≈ 22,240 cm³ ≈ 22.24 litres, and 22.24 ÷ 3.78541 ≈ 5.88 gallons. Both routes give the same answer.
Where cylinder volume shows up in real life
Cans, pipes, tanks and silos all ask the same question, how much does it hold, but each one asks it a different way. The worked cases below use the dimensions you would actually measure, and say which dimensions to measure.
Water tanks
A cylindrical water tank with inner radius 0.6 m and inner height 1.5 m holds V = π × 0.36 × 1.5 = 0.54π ≈ 1.696 m³, which is about 1,696 litres when full. Measure the inside dimensions for capacity: the outside dimensions include the wall thickness and overstate what the tank holds. For paint or insulation, use the outside dimensions instead, because that is the surface you are covering.
Pipes and hollow cylinders
A pipe is a hollow cylinder: it has an outer radius R and an inner radius r. The material in the wall is π(R² – r²)h, and the liquid the pipe carries fills πr²h. Take a pipe with outer radius 6 cm, inner radius 5 cm and length 200 cm. The wall material is π × (36 – 25) × 200 = 2,200π ≈ 6,911.5 cm³. The capacity is π × 25 × 200 = 5,000π ≈ 15,708 cm³, about 15.708 litres. Two different volumes from one pipe, so be clear which one you need before you calculate.
Cans and bottles
A drink can with radius 3.3 cm and height 12 cm has a full geometric volume of π × 3.3² × 12 = 130.68π ≈ 410.5 cm³, about 410 mL. A can labelled 330 mL holds less than that: the headspace at the top and the thickness of the walls take the rest. The cylinder formula gives the full volume of the shape, not the labelled fill.
Silos and grain stores
A grain silo with radius 2 m and height 8 m holds π × 4 × 8 = 32π ≈ 100.531 m³, about 100,531 litres. Silo capacity is usually quoted in cubic metres, and the conversion is one multiplication: cubic metres × 1,000 = litres. Note that a silo is rarely filled to the brim, so the working capacity is the volume up to the fill level, not the full height.
Cylinder vs cone vs sphere
With the same radius and the same height, a cone holds exactly one third of the cylinder, and a sphere holds exactly two thirds of the cylinder that encloses it. This is the comparison Archimedes is said to have been proudest of. The table below uses radius 3 and height 6 for all three shapes. The sphere has no height of its own, so it is compared with the cylinder of height 2r that just encloses it, which is the case here because 6 = 2 × 3.
Cone: V = (1/3)πr²h = 18π ≈ 56.549
Sphere: V = (4/3)πr³ = 36π ≈ 113.097
Check the ratios: 56.549 is exactly one third of 169.646, and 113.097 is exactly two thirds of 169.646. If you already know a cylinder volume, you can read off the matching cone and sphere volumes without a fresh calculation: divide by 3 for the cone, multiply by 2/3 for the sphere.
More questions
Use both radii. The wall material is V = π(R² – r²)h, where R is the outer radius and r is the inner radius. For a pipe with outer radius 6 cm, inner radius 5 cm and length 200 cm, that is π × (36 – 25) × 200 = 2,200π ≈ 6,911.5 cm³. The liquid inside the pipe is a separate volume, πr²h = 5,000π ≈ 15,708 cm³, about 15.708 litres.
Rearrange the formula to h = V ÷ (πr²). For a volume of 1,000 cm³ and a radius of 5 cm, h = 1,000 ÷ (25π) = 40 ÷ π ≈ 12.73 cm. Check it by substituting back: π × 25 × (40 ÷ π) = 1,000 exactly, because the π cancels.
Compute the volume in cm³, then divide by 1,000. A cylinder with radius 5 cm and height 20 cm has volume 500π ≈ 1,570.8 cm³, which is 1.571 litres. If your inputs are in metres, multiply the m³ answer by 1,000 instead. If your inputs are in inches, divide the in³ answer by 231 for US gallons, or multiply by 16.387064 and divide by 1,000 for litres.
No, it quadruples, because the radius is squared in V = πr²h. With height 10 cm, radius 2 cm gives 40π ≈ 125.664 cm³, while radius 4 cm gives 160π ≈ 502.655 cm³, exactly four times as much. Doubling the height, by contrast, exactly doubles the volume, because height appears only once in the formula.
Yes, with one condition: h must be the perpendicular distance between the two bases, not the length measured along the slant. A slanted cylinder with radius 3 cm and perpendicular height 8 cm has volume π × 9 × 8 = 72π ≈ 226.195 cm³, the same as an upright cylinder with the same radius and height. If you measure along the slant instead, you will overstate the volume.