How this instrument works
This sheet solves V = (4⁄3)πr³ for the radius: r = (3V ⁄ 4π)^(1⁄3). The rearrangement itself is three algebra moves — multiply by 3, divide by 4π, then take a cube root of what's left — and that last step is the one doing the real work, because a cube root is the only operation that undoes an exponent of three. A square root only ever undoes squaring; a variable raised to the third power needs a root of matching order to come back out cleanly.
Volume is a genuinely three-dimensional quantity, so it scales with the cube of the radius: double a sphere's radius and its volume multiplies by eight, not two. That cubic relationship is exactly why this formula needs a cube root rather than the square root that inverts a sphere's surface area — A = 4πr² solves backwards to r = √(A ⁄ 4π), because area only scales with the square of the radius, one dimension short of volume. Dividing the volume by 4⁄3 π and stopping there, reporting r³ as if it were r, is the single most common slip when working this identity by hand.
Archimedes proved the ratio behind this formula around 250 BCE: a sphere's volume is exactly two-thirds that of the tightest cylinder enclosing it, one sharing the same radius with a height equal to the diameter. That cylinder's volume works out to πr² × 2r = 2πr³, and two-thirds of that collapses to (4⁄3)πr³ — the sphere's own volume, arrived at independently of the formula this page rearranges. Unlike a cone or a cylinder, a sphere's volume depends on a single measurement, its radius, with no separate height to fix first — which makes this the one member of the family solvable from one input alone. At the low end, a volume of zero forces the radius all the way to zero too: no positive radius, however small, could ever enclose nothing.
- Enter the sphere's volume into the Volume field, in cubic units — cubic centimetres, cubic metres, whatever the source measurement uses.
- Read the result in the Radius field: the sheet computes r = (3V ⁄ 4π)^(1⁄3) instantly and updates live as Volume changes.
- Check the answer by cubing it and running the volume formula forward — 4⁄3 × π × Radius³ should reproduce your original Volume.
- Keep units consistent: a Volume entered in cubic feet returns a Radius in feet, never in any other length unit.
Worked example — a 523.6 cubic-unit sphere
A cast bronze sphere displaces 523.5987755982989 cubic units of water when submerged, a foundry's exact reading off a calibrated tank. Applying r³ = 3V ⁄ 4π: 3 × 523.5987755982989 = 1570.7963267948967, divided by 4π (12.566370614359172), gives r³ = 125 exactly. The cube root of 125 is 5, since 5 × 5 × 5 = 125, and that is the Radius this sheet reports the instant 523.5987755982989 is entered into Volume.
Running the volume formula forward confirms it: 4⁄3 × π × 5³ = 4⁄3 × π × 125 = 500π ⁄ 3, which comes to 523.5987755982989 at full double precision — the same figure the tank started with, to the last digit. A smaller check makes the scaling clear too: a sphere with volume exactly 4⁄3 π, about 4.1887902047863905 cubic units, has a radius of exactly 1, the unit sphere's own defining case.
Questions
What is the formula for the radius of a sphere from its volume?
r = (3V ⁄ 4π)^(1⁄3), the algebraic inverse of V = (4⁄3)πr³. Multiply the volume by 3, divide by 4π, then take the cube root of the result. For a volume of 523.5987755982989, that chain gives r³ = 125 and then r = 5 exactly, since 5 × 5 × 5 reproduces 125.
Why does this formula need a cube root instead of a square root?
Because volume is a three-dimensional quantity: it scales with r³, so undoing it means undoing an exponent of three, which only a cube root does. A sphere's surface area, A = 4πr², scales with r² instead and inverts with a square root — the exponent inside the original formula always dictates the order of root needed to solve it backwards.
How is this different from finding the radius of a cone or a cylinder?
A cone's volume is (1⁄3)πr²h and a cylinder's is πr²h — both depend on radius and height together, so recovering radius from volume alone is impossible without also knowing height. A sphere's volume, (4⁄3)πr³, depends on radius by itself, with no second measurement to fix first, which is why this page needs only one input field where its cone and cylinder cousins need two.
What did Archimedes prove about a sphere and its enclosing cylinder?
That a sphere's volume is exactly two-thirds the volume of the tightest cylinder around it — one sharing the sphere's radius, with a height equal to the diameter. That cylinder's volume is πr² × 2r = 2πr³, and two-thirds of that is (4⁄3)πr³, the sphere's own volume exactly. It is one of the earliest results tying a curved solid's volume to a plain ratio against a straight-edged one.
What is the radius of a sphere with zero volume?
Exactly zero. The formula r = (3V ⁄ 4π)^(1⁄3) sends V = 0 straight to r = 0, the single degenerate case where a sphere shrinks to a dimensionless point. There is no smaller sphere with positive volume, and no negative volume for a real sphere to have in the first place, so zero sits at the bottom of the domain, not the middle of it.
What mistake do people make solving this formula by hand?
Stopping one step early — dividing the volume by 4⁄3 π to get r³, then reporting that number as the radius instead of taking its cube root. r³ = 125 is not the radius; ∛125 = 5 is. The other common slip is reaching for a square root out of habit, borrowed from inverting the area formula, when a volume-to-radius conversion always needs a cube root instead.