Spherical Segment Volume Calculator
From two base radii and a height, get the volume of a spherical zone — the slice of a sphere lying between two parallel cutting planes.
Three inputs, one answer
Enter the bottom base radius, the top base radius, and the height between the planes and the calculator returns the segment's volume, (πh/6)(3a² + 3b²
- h²), at once.
Keep units consistent
The two radii and the height are unit-agnostic — your volume comes back in cubic units of whatever length unit you use, so don't mix centimetres with inches.
What is a spherical segment volume calculator?
Two radii and a height in, the zone's volume out
A spherical segment — also called a spherical zone — is the portion of a sphere caught between two parallel planes, like a horizontal belt sliced from a ball. It has two flat circular faces: a bottom one of radius a and a top one of radius b, separated by a height h. This is different from a spherical cap, which has only one flat face because its second cut passes through the very top of the sphere (so the top radius shrinks to zero). This calculator turns the two base radii and the height into the volume of that zone — handy for tank fill levels, dome and bowl sections, lens and ball-cut problems, and geometry homework. The single value is fixed once you know a, b, and h, so those three inputs are all you need.
Enter the two base radii and the height in any length unit to get the spherical segment's volume instantly.
One compact formula, built from the two base radii, the height, and the constant π (about 3.14159).
volume = (π × h / 6) × (3 × a² + 3 × b² + h²)The term in brackets, 3a² + 3b² + h², blends the two end-circle radii with the segment's own thickness. Multiplying by π × h / 6 turns that into the volume of the curved slice. If the top radius b shrinks to zero, the formula collapses to the spherical-cap volume (πh²/6)(3a² + ... ) — the segment becomes a cap with one flat face.
Suppose you have a spherical segment with a bottom base radius of 4, a top base radius of 3, and a height of 2.
Square the radii and height
a² = 16, b² = 9, h² = 4.
Build the bracket
3 × 16 + 3 × 9 + 4 = 48 + 27 + 4 = 79.
Multiply by π × h / 6
(π × 2 / 6) × 79 ≈ 82.728607 cubic units — the volume of the zone.
The result (about 82.728607 cubic units for a = 4, b = 3, h = 2) is the actual volume of material — or fluid — held in that horizontal band of the sphere. The most useful thing to notice is that the volume depends only on the two end-circle radii and the height, not on where the band sits on the sphere or even on the sphere's full radius: a thin belt near the equator and a thin belt near the pole have the same volume as long as their base radii and thickness match. That is why the formula is so handy for partly filled spherical tanks — you measure the liquid surface radius, the bottom-contact radius, and the depth, and you have the trapped volume directly. Because the bracket adds 3a² and 3b², a wider band always holds more than a narrow one of the same height, and the extra h² term gives a small additional contribution from the curvature across the thickness. Keep the geometry honest: a and b are the radii of the flat circular faces, not the sphere's radius, and h is the straight vertical gap between the two planes.
The formula is exact, but a couple of practical points are worth keeping in mind.
A true sphere, parallel cuts, and consistent units
This formula describes a segment of a perfect sphere cut by two parallel planes. It does not apply to an ellipsoid, an egg shape, or a tank whose ends are not truly spherical, and the two cuts must be parallel — a tilted cut gives a different solid. A spherical cap (one flat face, second cut through the top) is a special case where the top radius is zero, so use a cap calculator there rather than entering a tiny b. The two radii and the height are also unit-agnostic, so the answer is only meaningful if you keep one unit throughout: radii and a height in centimetres give a volume in cubic centimetres, never a mix.