Surface area and volume of a frustum

Big cone minus small cone

We will find the curved surface and the volume of a frustum by taking the small cone away from the complete cone. This works for every frustum, and it is the method to use if you forget a formula. Let the complete cone have height $H$ and slant height $L$. Let the small cone have height $H - h$ and slant height $L - l$.

The curved surface area

The curved surface of the frustum is the curved surface of the complete cone minus the curved surface of the small cone:

From lesson 1, $L = \dfrac{lR}{R - r}$ and $L - l = \dfrac{lr}{R - r}$. Put these in:

Now $R^2 - r^2 = (R - r)(R + r)$, and the factor $R - r$ cancels. So

The total surface area adds the two circular ends:

Which surfaces count depends on the object. A closed box of this shape has all three parts. A lampshade, open at both ends, has only the curved surface. A bucket is open at the top and closed at the bottom, so it has the curved surface and one circle, the small one if the wide end is the open top.

The volume

The volume of the frustum is the volume of the complete cone minus the volume of the small cone:

From lesson 1, $H = \dfrac{hR}{R - r}$ and $H - h = \dfrac{hr}{R - r}$. Put these in:

Now $R^3 - r^3 = (R - r)(R^2 + Rr + r^2)$, and the factor $R - r$ cancels. So

Two checks. If $r = R$, the bracket is $3R^2$ and $V = \pi R^2 h$, the volume of a cylinder. If $r = 0$, the bracket is $R^2$ and $V = \frac{1}{3}\pi R^2 h$, the volume of a cone. The formula behaves correctly at both ends.

Worked examples

Example 1. A frustum has bottom radius $8\ \text{cm}$, top radius $4\ \text{cm}$ and height $3\ \text{cm}$. Find its curved surface area, total surface area and volume.

1. The slant height. $l = \sqrt{3^2 + 4^2}\ \text{cm} = 5\ \text{cm}$.
2. The curved surface. $\pi(8 + 4)\ \text{cm} \times 5\ \text{cm} = 60\pi\ \text{cm}^2$.
3. The total surface. $60\pi + \pi \times 64 + \pi \times 16 = 140\pi\ \text{cm}^2$, about $439.8\ \text{cm}^2$.
4. The volume. $\dfrac{\pi \times 3\ \text{cm}}{3}(64 + 32 + 16)\ \text{cm}^2 = 112\pi\ \text{cm}^3$, about $351.9\ \text{cm}^3$.
5. A check with cones. The complete cone has height $6\ \text{cm}$ and the small cone $3\ \text{cm}$. The volumes are $\frac{1}{3}\pi \times 64 \times 6 = 128\pi$ and $\frac{1}{3}\pi \times 16 \times 3 = 16\pi$. The difference is $112\pi\ \text{cm}^3$.

Example 2. A bucket. A bucket is open at the top, with top radius $14\ \text{cm}$ and bottom radius $7\ \text{cm}$. It is $24\ \text{cm}$ high. Find the metal sheet that makes it, and its capacity in litres. Take $\pi = \frac{22}{7}$.

1. The slant height. $l = \sqrt{24^2 + 7^2}\ \text{cm} = 25\ \text{cm}$.
2. The curved surface. $\dfrac{22}{7} \times (14 + 7)\ \text{cm} \times 25\ \text{cm} = 1650\ \text{cm}^2$.
3. The base. $\dfrac{22}{7} \times (7\ \text{cm})^2 = 154\ \text{cm}^2$.
4. The sheet. $1650\ \text{cm}^2 + 154\ \text{cm}^2 = 1804\ \text{cm}^2$.
5. The capacity. $\dfrac{22}{7} \times \dfrac{24\ \text{cm}}{3} \times (196 + 98 + 49)\ \text{cm}^2 = 8624\ \text{cm}^3$, which is $8.624$ litres.

Practise with new numbers

Try it in Python

This program finds the curved surface, the total surface and the volume of a frustum by the formulas. Then it finds them again as the difference of two cones.

Common mistakes

• Using $R - r$ in the curved surface formula. The curved surface needs the sum, $R + r$.
• Using $h$ in place of $l$. The curved surface needs the slant height.
• Writing $(R + r)^2$ in the volume. The bracket is $R^2 + Rr + r^2$, not $R^2 + 2Rr + r^2$.
• Counting both circles when the vessel is open. Look at which ends are closed.

Remember these

• $\text{CSA} = \pi(R + r)\,l$, with $l = \sqrt{h^2 + (R - r)^2}$.
• $\text{TSA} = \pi(R + r)\,l + \pi R^2 + \pi r^2$ for a closed frustum.
• $V = \dfrac{\pi h}{3}(R^2 + Rr + r^2)$.
• Each formula comes from the big cone minus the small cone, and gives the cylinder when $r = R$.

Check yourself

Try each one on your own first. If you are stuck, open a hint. Then check with Show the answer.

1. A frustum has $R = 7\ \text{cm}$, $r = 3\ \text{cm}$ and slant height $5\ \text{cm}$. Find its curved surface area in terms of $\pi$.

2. A frustum has $R = 6\ \text{cm}$, $r = 3\ \text{cm}$ and $h = 4\ \text{cm}$. Find its volume in terms of $\pi$.

3. A lampshade open at both ends has $R = 12\ \text{cm}$, $r = 5\ \text{cm}$ and slant height $10\ \text{cm}$. How much cloth covers it?

4. Put $r = 0$ in the volume formula. Which solid do you get, and what is its volume?

5. Why is the volume of a frustum less than the volume of the cylinder with radius $R$ and the same height?

References

• NCERT, Mathematics for Class 10, the chapter on Surface Areas and Volumes. The frustum of a cone is in the older editions of this chapter.
• Cambridge IGCSE Mathematics (0580), syllabus topic Mensuration, for surface areas and volumes of cones, cylinders, spheres and frustums.