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Cone Volume

V=13πr2hV = \frac{1}{3} \pi r^{2} h

Worked example: Cone r = 3 m, h = 4 m → V = 12 pi m3 — press Try an example to run it live, then adjust anything.

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π=3.141592653589793\pi = 3.141592653589793Pi · exact
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Cone Volume explained

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A cone holds exactly one-third the volume of the cylinder that encloses it — the same one-third rule that governs pyramids and every solid that tapers linearly to a point. Democritus guessed the ratio around 400 BC, and Eudoxus proved it rigorously with the method of exhaustion, a forerunner of integral calculus. The h in the formula is the perpendicular height from base to apex, not the slant height along the side; confusing the two overstates the volume.

A worked example: a conical stockpile of sand 6 m across (r = 3 m) and 2 m tall contains V = (1/3)π(3)²(2) ≈ 18.85 m³ — roughly two truckloads, which is exactly how aggregate yards estimate inventory from a tape measure and a clinometer. Solving for r takes the principal positive square root, and solving for h is a plain division, so both inversions are single-valued for physical inputs.

Cone Volume formula

V=13πr2hV = \frac{1}{3} \pi r^{2} h
Where
  • VV= Volume (L)
  • rr= Base radius (m)
  • hh= Height (m)

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