Volume of a Frustum

Statement

The volume of a frustum of a cone is given by:

\[ V = \tfrac{1}{3}\pi h \left(R^2 + r^2 + Rr\right) \]

where \(R\) is the radius of the larger base, \(r\) is the radius of the smaller base, and \(h\) is the vertical height of the frustum.

Why it’s true

• A frustum is a cone with its top cut off by a plane parallel to the base.
• The formula comes from subtracting the volume of the smaller cone (removed from the top) from the larger cone.
• After algebraic simplification, the result is \(\tfrac{1}{3}\pi h (R^2 + r^2 + Rr)\).

Recipe (how to use it)

1. Write down the radii of the two circular bases: \(R\) and \(r\).
2. Square each radius: \(R^2\) and \(r^2\).
3. Multiply the two radii: \(Rr\).
4. Add them together: \(R^2 + r^2 + Rr\).
5. Multiply by height \(h\) and \(\pi\).
6. Finally, divide by 3.

Spotting it

Look for cone-like shapes with the top sliced off parallel to the base — typical in GCSE “real-life” volume problems (buckets, vases, lampshades).

Common pairings

• Often appears with similar triangles when height of the missing cone is involved.
• Sometimes combined with surface area questions.

Mini examples

1. Given: \(R=6\), \(r=4\), \(h=10\). Find: Volume. Answer: \(\tfrac{1}{3}\pi \cdot 10 \cdot (36+16+24) = \tfrac{760}{3}\pi\).
2. Given: \(R=5\), \(r=2\), \(h=12\). Find: Volume. Answer: \(\tfrac{1}{3}\pi \cdot 12 \cdot (25+4+10) = 468\pi\).

Pitfalls

• Forgetting to use perpendicular height (not slant height).
• Forgetting the cross-term \(Rr\).
• Using diameter instead of radius.

Exam strategy

• Write the formula clearly before substitution.
• Always check which radius is larger and which is smaller.
• Leave answers in terms of \(\pi\) unless decimals are asked.

Summary

The volume of a frustum is one third of π times the height times the sum of the squares of the two radii plus their product. It’s derived from subtracting one cone’s volume from another.