Case 1
- Outer radius R
- 5
- Inner radius r
- 3
- Height
- 10
502.654825
Open with these valuesCylindrical Shell Volume Calculator
Change any value to update the result.
502.654825units³
Shell volume: 502.654825 units³Examples
A pipe wall is the full cylinder minus the empty bore, so V = π × h × (R² − r²). Outer radius 5, inner radius 3 and height 10 give 160π, about 502.65 cubic units. Enter the two radii, not a radius and a wall thickness — the wall is simply R − r.
The numbers at a glance
Held fixed: Outer radius R 5.000, Inner radius r 3.000.
| Height | Shell volume |
|---|---|
| 2.500 | 125.663706 |
| 5.000 | 251.327412 |
| 7.500 | 376.991118 |
| 10.000Your value | 502.654825 |
| 12.500 | 628.318531 |
| 15.000 | 753.982237 |
| 17.500 | 879.645943 |
| 20.000 | 1,005.309649 |
Worked examples
Case 2
- Outer radius R
- 2
- Inner radius r
- 1
- Height
- 1
9.424778
Open with these valuesCase 3
- Outer radius R
- 10
- Inner radius r
- 5
- Height
- 20
4,712.388980
Open with these valuesHow it's calculated
V = π × h × (R² − r²)
- StepMeasure the outer radius — half the outside diameter.Outer radius R=5
- StepMeasure the inner radius of the bore, in the same unit.Inner radius r=3
- StepMeasure the height along the axis, same unit again.Height=10
- ResultRead the volume of the wall material, not of the bore.502.654825units³
Reference table
| R, r, height | Exact | Volume |
|---|---|---|
| 3, 5, 10 | radii swapped: −160π | -502.654825 |
| 2, 1, 1 | 3π | 9.424778 |
| 3, 2.5, 8 | 22π | 69.115038 |
| 4, 2, 5 | 60π | 188.495559 |
| 5, 3, 10 | 160π | 502.654825 |
| 10, 5, 20 | 1500π | 4712.388980 |
Questions
How do I calculate the volume of a hollow cylinder?
Subtract the inner radius squared from the outer radius squared, multiply by π and then by the height: V = π × (R² − r²) × h. For R = 5, r = 3 and h = 10 that is π × 16 × 10 = 502.654825 cubic units.
Why does the formula use R² − r²?
The wall is the full outer cylinder, π × R² × h, minus the empty bore, π × r² × h; factoring out π and h leaves π × (R² − r²) × h. The ring-shaped end face is an annulus, and its area is the big circle minus the small one.
I only know the wall thickness — what do I enter?
The wall thickness t is the gap between the radii, so the inner radius is r = R − t. An outer radius of 5 with a 2 thick wall means r = 3, which is the second field.
What if I put the inner radius in the outer field?
You get the right size with a minus sign in front: 3, 5, 10 returns −502.654825 where 5, 3, 10 returns +502.654825. A negative answer therefore means the two radii are the wrong way round, and there is no hollow cylinder with a bore wider than its outside.
Does this give the volume the pipe can carry?
No, it gives the material of the wall itself. For the contents of the bore use a plain cylinder of radius r: π × r² × h.