Hollow cylinder volume:
formula and worked example

Material in a tube

A hollow cylinder is a solid tube: a steel pipe, a concrete culvert section, a bush or sleeve, a bearing race, a rolled-paper core. The volume of the wall tells you how much material it contains, and combined with density it gives the mass. That is the number for shipping weights, structural loads, cost per metre and cutting stock.

It should be kept separate from the capacity of the bore. The empty space inside a pipe is a different volume, usually used for fluid capacity, and is worked out with the inner radius alone.

Common cases include structural hollow sections, drilled bars, hydraulic cylinders and the cardboard core of a roll. In each, the solid material carries the load while the bore saves weight, so the ring volume is the figure that drives both strength and cost per metre.

The formula and its three inputs

Volume equals pi times the difference of the squares of the outer and inner radii, multiplied by the length. The bracket pi x (R^2 - r^2) is the area of the ring-shaped cross-section, and multiplying by length extrudes it along the pipe. R is the outer radius, r is the inner radius and L is the length.

The most important point is that the squares are subtracted, not the radii. R^2 - r^2 equals (R - r) x (R + r), so the wall thickness enters together with the average radius.

Written in terms of wall thickness t and outer diameter D, the same volume is pi x t x (D - t) x L. That form is convenient when the supplier lists a pipe by its outside diameter and wall, as most catalogues do.

A pipe of outer radius 0.07 m, inner 0.05 m, length 2 m

Take a tube whose outer radius is 0.07 m, inner radius 0.05 m (wall 20 mm) and length 2 m.

1. 1Square the outer radius: 0.07 x 0.07 = 0.0049.
2. 2Square the inner radius: 0.05 x 0.05 = 0.0025.
3. 3Subtract: 0.0049 - 0.0025 = 0.0024.
4. 4Multiply by pi and length: 0.0024 x 3.14159 x 2 = 0.0150796 cubic metres.

A cross-check is to find the outer solid cylinder, pi x 0.0049 x 2 = 0.030788, and subtract the bore, pi x 0.0025 x 2 = 0.015708. The difference is 0.015080, the same result.

From volume to mass and capacity

The wall volume of 0.01508 cubic metres, which is about 15.1 litres of material, becomes mass when multiplied by density. For illustration, a material with a density of 7,800 kg per cubic metre would give about 117.6 kg for this 2 m length, or roughly 59 kg per metre. Always take density from the material datasheet rather than a round figure.

The bore holds pi x 0.0025 x 2 = 0.0157 cubic metres, so 15.7 litres, which is the capacity if the pipe is filled. The ring volume and bore volume add up to the outer solid cylinder, a useful identity that checks both numbers at once.

Remember to apply the same unit throughout. A radius of 70 mm is 0.07 m; entering 70 with a length in metres would return a volume a million times too large, an error that is easy to make when mixing millimetre drawings and metre-based stock lengths.

Growing only the outer radius

Taking R from 0.07 to 0.077 raises the volume from 0.0150796 to 0.021545 cubic metres, an increase of nearly 43%, though the outer radius grew by only 10%. The reason is that the wall thickened from 20 mm to 27 mm, a 35% change, and the area added lies at a larger radius where each unit of thickness carries more material.

This sensitivity matters in tolerancing: a modest increase in outside diameter on a tube with a fixed bore can add a large fraction to weight and cost. Conversely, shaving a little from the outer radius saves proportionally far more material than the dimension suggests.

Specific traps with hollow cylinders

• Subtracting the radii first and squaring the difference, which gives (0.02)^2 = 0.0004 and a result six times too small.
• Entering diameters as radii, or the reverse, either quadrupling or quartering the volume.
• Unit mixing such as 70 mm and 0.05 m in the same line, producing a result off by factors of a thousand or a million.
• Forgetting length, so the answer is the cross-section area in square units rather than a volume.

Scenario: ordering cut lengths of tube

A fabricator needs 25 sleeves, each 150 mm long, cut from a tube with a 70 mm outer radius and 50 mm inner radius. One sleeve is pi x (0.0049 - 0.0025) x 0.15 = 0.001131 cubic metres, so 25 sleeves total about 0.0283 cubic metres. Stock tube is supplied in 6 m lengths, giving up to 40 sleeves per length, or 39 once saw kerf is allowed for, so one length is enough. Knowing the volume per sleeve lets the same number feed mass for shipping and cost per piece.

For tapered or non-uniform tubes the single-formula approach no longer holds. Split the length into short constant-radius segments, calculate each, and add them, or use the average of the radius squares weighted by length for a quick estimate.