Pick the unit your lengths are in. Areas are then labelled in square units and volumes in cubic units; nothing is converted.
R is the radius of the bottom circle, r that of the top, and h the height, the perpendicular distance between the two. A bucket, a lampshade and a flowerpot are frustums.
A worked example
R = 5, r = 3 and h = 6 cm: V = π · 6 · (25 + 15 + 9) / 3 = π · 6 · 49 / 3 = 98π ≈ 307.9 cm³. A bucket 30 cm across at the top, 20 at the bottom and 30 cm deep holds π · 30 · (225 + 150 + 100) / 3 = 4,750π ≈ 14,900 cm³, about 15 litres.
Two limits
With r = 0 the top shrinks to a point and the formula becomes πR²h / 3, the cone. With r = R the sides are vertical and it becomes πR²h, the cylinder. The frustum sits between the two, and the middle term Rr is what carries it from one to the other.
Where the Rr comes from
A frustum is a large cone with a small cone cut off the top. Writing the two cones’ volumes and subtracting, with the small cone’s height found by similar triangles, gives πh(R³ − r³) / (3(R − r)), and R³ − r³ factors as (R − r)(R² + Rr + r²). The (R − r) cancels and the three terms are what is left.
Common mistakes
- Dropping the middle term. (R² + r²) is too small: for the example it gives 68π instead of 98π.
- Using the slant height for h. The height is the perpendicular distance between the circles; the slant is longer.
- Averaging the radii. π · ((R + r) / 2)² · h is a cylinder of the mean radius, and smaller than the frustum.
Frequently asked questions
What is the surface area?
π(R² + r²) + π(R + r) · l with the slant height l, on the
surface area of a frustum page.
Does it work for a pyramid?
The same shape of formula with the areas of the two bases: h(A₁ + √(A₁A₂) + A₂) / 3. The circle is the
case where the areas are πR² and πr².
How do I find the height of a bucket?
Measure the depth straight down, not along the side. If only the slant l is known, h = √(l² − (R − r)²)
by Pythagoras.