How this instrument works
A hollow cylinder — a pipe, a length of tubing, a bored-out shaft — is really two ordinary cylinders sharing one axis and one height: a solid outer cylinder with a smaller cylindrical cavity removed from its middle. V = π(R² − r²)h states that plainly: πR²h is the full outer cylinder, πr²h is the cavity taken back out of it, and what remains is the volume of actual material — steel, PVC, concrete — rather than the empty bore running through the centre.
The subtraction works because volume is additive: split the outer cylinder into the wall and the cavity, and the two pieces must add back up to πR²h, so the wall's volume is whatever is left once πr²h is removed. It collapses into one tidy product only because the outer and inner cylinders share the same height and axis — any solid whose cross-section keeps the same shape all the way along has volume equal to that cross-section's area times the length, and here the cross-section is a ring, unchanged at every point along h.
A useful limit sits at the thin end of the range. Write the formula as V = π(R − r)(R + r)h and let t = R − r stand for the wall thickness: when t is small next to R, the sum R + r sits close to 2R, so V comes out near 2πRth — thickness times the circumference through the wall's middle, times the length. That shortcut is how a machinist estimates a thin tube's material by hand, skipping the squared terms entirely and trading a small, predictable error for speed.
- Enter the pipe's outer radius into Outer radius, R — the distance from the axis out to the exterior wall.
- Enter the radius of the hollow bore into Inner radius, r — the distance from that same axis out to where the cavity begins.
- Enter the tube's length along its axis into Height.
- Read Volume for the material volume, π(R² − r²)h, updating as soon as any figure changes.
- Keep Inner radius, r below Outer radius, R — a bore that reaches the outer wall leaves no material to measure.
Worked example — a pipe with R = 5, r = 3, h = 10
A steel pipe has an outer radius of 5 units, an inner bore of radius 3 units, and runs 10 units long — the exact figures this sheet's own example uses. The outer cylinder alone would hold πR²h = π × 25 × 10 = 250π ≈ 785.398163 cubic units, and the bore removes πr²h = π × 9 × 10 = 90π ≈ 282.743339 cubic units of empty space from the middle of it.
Subtracting one from the other gives the steel's actual volume: 250π − 90π = 160π = 502.6548245743669 cubic units, the exact figure this sheet's Volume field returns for R = 5, r = 3, h = 10. The factored check agrees: (R − r)(R + r)h = (5 − 3)(5 + 3) × 10 = 2 × 8 × 10 = 160, and π × 160 lands on that same 502.6548245743669 — a foundry costing the pipe by weight would multiply that figure by the steel's density next.
Questions
What is the formula for the volume of a hollow cylinder?
V = π(R² − r²)h, where R is the outer radius, r is the inner radius, and h is the height or length along the shared axis. It is the outer cylinder's volume, πR²h, with the inner cylindrical cavity's volume, πr²h, subtracted out — the material left over once the bore is removed, not the space the bore occupies.
How is the hollow cylinder volume formula actually derived?
Treat the pipe as two ordinary cylinders sharing an axis and a height, then subtract the inner one's volume from the outer one's: πR²h − πr²h, which factors to π(R² − r²)h. The subtraction is valid because volume adds up — wall volume plus cavity volume equals the full outer cylinder's volume, so wall volume is whatever remains once the cavity is taken out.
How is this different from finding the ring's cross-sectional area?
A cross-sectional area describes one flat slice — a washer or a pipe seen end-on, measured in square units. This calculator carries that same ring shape through the pipe's full length by multiplying by h, turning a flat area into a solid, cubic-unit volume: the actual material running the tube's whole length, not just one slice of it.
What mistake do people most often make with this formula?
Two, usually. The first is squaring the difference of the radii instead of taking the difference of the squares — (R − r)² is not R² − r² except when r is zero. The second is entering wall thickness where the inner radius belongs: if a pipe's wall thickness t is known instead of its bore, the inner radius is r = R − t, not t itself.
Does the formula still work if I only know the pipe's wall thickness?
Yes — set r = R − t, where t is the wall thickness, and volume becomes π(R² − (R − t)²)h, which simplifies to πt(2R − t)h. For a thin wall, where t is small next to R, that is close to 2πRth: thickness times the circumference at the mean radius times the length, the shortcut engineers use to estimate a thin pipe's material by hand.
What happens at the edges — a zero inner radius or an inner radius equal to the outer?
At r = 0 there is no bore at all, and the formula collapses to πR²h, the volume of an ordinary solid cylinder. At r = R the wall has thinned to nothing, and V falls to exactly zero — every bit of the outer cylinder has been hollowed away, leaving no material left to measure.