Case 1
- Bottom radius
- 5
- Top radius
- 3
- Vertical height
- 10
513.126800
Open with these valuesChange any value to update the result.
513.126800units³
Volume: 513.126800 units³A frustum is a cone with its tip sliced off parallel to the base: a bucket, a plant pot, a lampshade. It is not the average of the two circles — the middle term R·r is what makes the sides straight instead of curved. Equal radii give back the cylinder.
Held fixed: Bottom radius 5.000, Top radius 3.000.
| Vertical height | Volume |
|---|---|
| 2.500 | 128.281700 |
| 5.000 | 256.563400 |
| 7.500 | 384.845100 |
| 10.000Your value | 513.126800 |
| 12.500 | 641.408500 |
| 15.000 | 769.690200 |
| 17.500 | 897.971900 |
| 20.000 | 1,026.253600 |
513.126800
Open with these values3.141593
Open with these values3,665.191429
Open with these valuesV = ⅓ × π × h × (R² + R·r + r²)
Averaging them gives 534.07 for the bucket of radii 5 and 3 over a height of 10, instead of 513.13 — about four per cent too much. The side is a straight slope, not a step.
It is the only reason the bracket is not simply two circle areas added up. Set R = r and it turns into 3R², collapsing the whole formula back to the cylinder πR²h.
Each of those needs three measurements: the radius of the wide opening, the radius of the narrow one, and the height straight between the two circles.
MisconceptionI averaged the two circle areas and multiplied by the height.
CorrectionThat overstates the bucket of radii 5 and 3 by about four per cent, 534.07 instead of 513.13. Use R² + R·r + r² in the bracket.
MisconceptionI put the wider radius in the top field by mistake.
CorrectionNo harm done. The formula is symmetric in R and r, so a bucket upside down comes out the same.
MisconceptionBoth openings are the same size, so this is the wrong calculator.
CorrectionIt still works: equal radii turn the formula into πR²h, the cylinder. Entering 1, 1 and 1 returns π, which is a good check that the inputs landed where you meant.
| Bottom, top, height | Note | Volume |
|---|---|---|
| 0.5, 0.25, 3 | small pot | 1.374447 |
| 1, 1, 1 | equal radii → cylinder π | 3.141593 |
| 2, 1, 6 | halves at each end | 43.982297 |
| 5, 3, 10 | typical bucket | 513.126800 |
| 10, 6, 20 | bucket doubled → eight times over | 4105.014401 |
One third of π times the height times R² + R·r + r². A bucket 10 units tall with radii 5 and 3 holds about 513.13 cubic units.
Because the side is a straight slope, not a step. Averaging the end areas gives 534.07 for the bucket above instead of 513.13 — about four per cent too much.
No. The formula is symmetric in R and r, so swapping them returns the same volume. A bucket and the same bucket upside down hold the same amount.
Then it is a cylinder and the formula collapses to πR²h. That is a useful check: enter 1, 1 and 1 and you should read π.
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