How this instrument works
V = 2π²Rr² is the second theorem of Pappus at work, the counterpart to the theorem that produces a torus's surface area from a curve instead of a region. Here the object being carried around the axis is not a line but a disk: the tube's solid cross-section, area πr². One full lap around the central axis, and that disk sweeps out the whole donut. Pappus's rule says a swept region's volume equals its own area multiplied by however far its middle point travels while sweeping — πr² for the disk, 2πR for the loop that point traces, giving (πr²)(2πR), which reduces to 2π²Rr².
Notice which radius gets squared and which does not, because that split is not arbitrary. The r² comes from the disk's own area, and a circle's area always carries the square of its radius. R stays unsquared because it measures how far the disk's middle point rides in one lap, a plain length rather than a region. An area multiplied by a length is exactly what a sweep produces, so the mismatched powers on R and r are baked into the geometry rather than the formula.
The companion sheet for surface area answers how much skin coats a torus; this one answers how much the solid inside actually holds — the rubber in an O-ring, the dough in a doughnut, the metal cast into a ring magnet. Shrink the minor radius down to nothing and the disk being carried around the axis has no area left to sweep, so the solid it traces vanishes with it — a clean check that this formula tracks real material rather than an outer skin.
- Find how far the ring's centre line sits from the torus's central axis and type that figure into Major radius (center to tube center).
- Type the tube's own cross-section radius — half its full thickness — into Minor radius (tube radius).
- Check that your minor-radius figure doesn't exceed the major radius; past that boundary the tube crosses through itself and stops describing a plain ring.
- Read off Volume, delivered in your input unit raised to the third power: feed it metres and it hands back cubic metres.
Worked example — R = 5, r = 3
Picture a ring shape whose axis-to-tube-centre span runs 5 units and whose tube itself is 3 units thick, then feed 5 into Major radius and 3 into Minor radius. The formula gives V = 2π² × 5 × 3² = 90π², which this sheet reports as 888.264396 cubic units for those two inputs, matching the golden figure of 888.2643960980422 to full precision.
A second route reaches the same number without touching the shortcut formula at all. Picture the tube's disk-shaped cross-section, area π × 3² = 28.274334 square units, riding around the axis once; its own centre traces a loop of length 2π × 5 = 31.415927 units along the way. Multiplying that area by that loop length lands on 888.264396 cubic units again — one identity, computed two ways.
Questions
How is the torus volume formula derived?
Pappus's second theorem gets you there: carry a flat region once around an axis outside it, and the solid it traces has a volume equal to that region's own area multiplied by how far its middle point rides during the trip. Here the flat region is the tube's circular cross-section, πr², and its middle point rides a loop of length 2πR, so V = (πr²)(2πR) = 2π²Rr².
Why is the minor radius squared but the major radius isn't?
The two factors describe different kinds of measurement. The cross-section's area scales with r² because a circle's area is πr², while the sweep path's length scales linearly with R because a circumference is 2πR. An area times a length gives a volume, so r ends up squared and R does not — a structural fact about the sweep, not a rounding convention.
How does torus volume relate to torus surface area?
Both start from the same major-and-minor-radius pair, but the minor radius carries different weight in each: this page's V = 2π²Rr² grows with r squared, while the surface figure grows with r alone. Think of pumping filling into a torus-shaped mold — the amount it holds scales with the square of the tube's thickness, while the mold's own wall area only scales in step with that thickness.
What is the most common mistake when computing torus volume?
Squaring the wrong radius. A frequent slip is writing 2π²R²r, squaring the major radius instead of the minor one, or leaving r unsquared entirely as 2π²Rr. The correct form squares only the tube's own radius: 2π²Rr². A quick sanity check is trying r = R and comparing against a source you trust.
What happens to the volume once the minor radius gets very small?
It heads to zero right along with r. With almost no tube thickness there is barely any cross-section left to sweep, so a torus with R = 5 and r = 0 has V = 2π² × 5 × 0² = 0 exactly — the formula collapses smoothly to nothing rather than jumping or misbehaving at that limit.
Can the minor radius be larger than the major radius?
Not for a real, simple torus. Once the minor radius exceeds the major radius, the tube overlaps its own far side and the shape becomes a self-crossing spindle torus rather than a ring with a clean hole. The formula still returns a number past that point, but it no longer measures one non-overlapping solid.