A frustum is what remains when the top of a cone is cut off parallel to the base, like a bucket or a drinking glass. Enter both radii and the height to find its volume and surface areas.
Enter the larger and smaller radii.
Enter the vertical height.
Read the volume and surface areas.
Volume: V = ⅓ π h (R² + r² + Rr)
CSA: π l (R + r)
A frustum of a cone is the solid left when you slice the top off a cone with a cut parallel to its base. Buckets, tumblers, flower pots, lampshades, funnels and many chimney tops have this shape. This calculator takes the two radii and the vertical height and returns the volume, the slant height, the curved surface area and the total surface area.
The frustum is a Class 10 favourite, often in problems about how much milk a container holds or how much metal sheet is needed for a bucket. It is also the most formula-heavy of the school solids, so a calculator that shows each step helps you check where a hand calculation went wrong.
1. Measure the radius of the larger circle R, the radius of the smaller circle r, and the vertical height h between them.
2. Volume: V = ⅓ π h (R² + r² + Rr).
3. Slant height: l = √(h² + (R − r)²).
4. Curved surface area: CSA = π l (R + r).
5. Total surface area of a closed frustum: TSA = CSA + πR² + πr².
6. For an open container such as a bucket, add only the area of the circle that forms its base.
7. Convert cm³ to litres by dividing by 1000.
Imagine completing the frustum into the full cone. Let its full height be H. The small cone that was removed has height H − h. By similar triangles, the radii shrink in proportion to height: r ÷ R = (H − h) ÷ H, which gives H = hR ÷ (R − r). The frustum's volume is ⅓πR²H − ⅓πr²(H − h). Substituting H and simplifying, using R³ − r³ = (R − r)(R² + Rr + r²), leaves ⅓πh(R² + r² + Rr). The Rr term is what you would miss if you simply averaged the two circle areas.
Slice the frustum vertically through its axis. The side edge, the height and the difference in radii form a right triangle, so the slant height is √(h² + (R − r)²). The curved surface, cut and flattened, becomes a piece of a ring. Its area equals the slant height times the average of the two circumferences, l × (2πR + 2πr) ÷ 2 = πl(R + r). This is the same way a trapezium's area is its height times the average of its parallel sides.
The formulas are symmetric in R and r, so it does not matter which circle is on top; a bucket, which is wider at the mouth, works the same way as a lampshade. Put r = 0 and the volume becomes ⅓πR²h, a cone. Put r = R and it becomes πR²h, a cylinder, while the slant height becomes h. These checks are a quick way to catch a mistyped formula in an exam.
A dairy in Anand is ordering steel milk buckets that are 20 cm in radius at the mouth, 12 cm in radius at the base and 15 cm deep. Enter the larger radius as R and the smaller as r.
Volume = ⅓ π h (R² + r² + Rr): = ⅓ × π × 15 × (400 + 144 + 240) = 12,315.0432
Slant height = √(h² + (R − r)²): = 17
Curved surface area = π l (R + r): = 1,709.0264
Total surface area = CSA + πR² + πr²: = 3,418.0528
Answer: Volume 12,315.0432; Slant height 17; Curved surface area 1,709.0264
Using the vertical height where the slant height is needed in the curved surface area.
Averaging the two circle areas and multiplying by height, which misses the Rr term and underestimates the volume.
Using the total surface area for an open bucket, which has no lid.
Adding the wrong circle for an open container: the base of a bucket is its smaller circle, even though the calculator labels R as the bottom.
Entering diameters instead of radii.
Finding the capacity of buckets, tumblers, flower pots and milk cans in litres.
Estimating sheet metal for fabricating buckets, funnels and hoppers.
Calculating fabric or paper for lampshades.
Class 10 surface area and volume problems involving frustums.
Estimating the volume of truncated cone-shaped silos, chimneys and embankment sections.
How many litres does a bucket hold?
Measure in cm, compute the volume in cm³ and divide by 1000.
Does an open bucket use the total surface area?
No. A bucket is open at its wide top, so use the curved surface area plus only the base circle, πr² with the smaller radius.