Surface Area of Cylinders
A closed cylinder has two circular ends and a curved surface that opens out into a rectangle, so its surface area is \(A=2\pi r^2+2\pi rh\). In NSW Year 10 Advanced maths you adapt the formula for open containers and composite shapes, give exact answers in terms of \(\pi\), and work backwards from a surface area to the radius or height (NESA outcome MA5‑ARE‑C‑01).
You will find surface areas of tanks, cans, bins, hay bales and a flower pot, and work out how many cans can be cut from a sheet of metal. You will find how many turns of a paint roller cover a wall, which of two cans of equal volume uses less metal, and the icing on a two-tier cake. Harder questions form and solve a quadratic equation for the radius and reject the negative root. It is part of Year 10 Advanced Measurement.
Every question with a fully worked solution.
- Surface Area of Cylinders - Video - Surface Area Of A Cylinder Watch
Theory
The surface area of a closed cylinder is two circles plus the curved surface: \(\text{SA} = 2\pi r^2 + 2\pi rh\). This Year 10 Advanced maths page covers open and hollow cylinders, exact answers in terms of \(\pi\), working backwards to the radius or height, and practical problems with paint, metal and cost.
A cylinder is a solid with two identical circular ends joined by a curved surface. A right cylinder has its height \(h\) at right angles to the ends.
The surface area is the total area of every surface on the outside. Unroll the curved surface and it becomes a rectangle: one side is the height \(h\) and the other is the circumference of the end, \(2\pi r\). So the curved surface has area \(2\pi rh\).
A closed cylinder has both ends. An open cylinder, such as a cup or a bin, is missing one end. A pipe or tube is open at both ends, so only the curved surface counts.
An answer in terms of \(\pi\) (an exact answer) leaves \(\pi\) as a symbol, such as \(90\pi\text{ cm}^2\). Collect the \(\pi\) terms, and only use the \(\pi\) key when a rounded answer is asked for.
If a question gives the diameter or the circumference, find the radius first: \(r = \dfrac{d}{2}\), or solve \(2\pi r = C\).
For a closed cylinder with radius \(r\) and height \(h\):
Choose the surfaces that are really there:
| Cylinder | Surfaces | Surface area |
|---|---|---|
| Closed (lid and base) | 2 circles + curved surface | \(2\pi r^2 + 2\pi rh\) |
| Open at one end (cup, bin, tank) | 1 circle + curved surface | \(\pi r^2 + 2\pi rh\) |
| Open at both ends (pipe, tube) | curved surface only | \(2\pi rh\) |
How to find the surface area of a cylinder
- Read which surfaces are there: two ends, one end, or none (a pipe). Leave out any surface that is not painted, wrapped or made.
- Find the radius from the diameter or circumference, and put every length in the same unit.
- Write each surface: circles \(\pi r^2\), curved surface \(2\pi rh\).
- Add the exact \(\pi\) terms, then round once at the end if a decimal is needed.
Harder problems
- Unknown radius or height: substitute the known surface area and solve. A quadratic in \(r\) usually factorises; a radius must be positive.
- One length given in terms of another (for example \(h = r + 3\) or \(h = 4r\)): substitute it into the formula first, then simplify.
- Volume given: use \(V = \pi r^2 h\) to find the missing length, then the surface area.
- Rim-to-rim diagonal given: use Pythagoras' theorem with the diameter and height.
- Rolling: one full turn of a roller covers its curved surface \(2\pi rh\).
- Stacked cylinders (a tiered cake, a step): the tops you can see always add up to the top of the largest cylinder, because the smaller top replaces the area it covers.
- Rounding counts: round up for tins, rolls or turns you must buy or make; round down for how many whole items a sheet can make.
(i) exactly, in terms of \(\pi\)
(ii) in square metres, correct to two decimal places.
The radius is half the diameter: \(r = 28\) cm.
(i)
| \(\text{SA}\) | \(=\) | \(2\pi r^2 + 2\pi rh\) |
| \(\) | \(=\) | \(2\pi \times 28^2 + 2\pi \times 28 \times 85\) |
| \(\) | \(=\) | \(1568\pi + 4760\pi\) |
| \(\) | \(=\) | \(\mathbf{6328\pi}\text{ cm}^2\) |
(ii) \(1\text{ m}^2 = 100 \times 100 = 10\,000\text{ cm}^2\), so divide by \(10\,000\):
| \(6328\pi\) | \(=\) | \(19\,879.99\ldots\text{ cm}^2\) |
| \(\) | \(=\) | \(1.9879\ldots\text{ m}^2\) |
| \(\) | \(\approx\) | \(\mathbf{1.99}\text{ m}^2\) |
Keep the exact form \(6328\pi\) until the last step, then round once.
Substitute into the formula. Every term has a factor of \(\pi\), so divide it out.
| \(2\pi r^2 + 2\pi r(7)\) | \(=\) | \(120\pi\) |
| \(2r^2 + 14r\) | \(=\) | \(120\) |
| \(r^2 + 7r - 60\) | \(=\) | \(0\) |
| \((r + 12)(r - 5)\) | \(=\) | \(0\) |
| \(r\) | \(=\) | \(-12 \text{ or } r = 5\) |
A radius cannot be negative, so reject \(r = -12\).
| \(r\) | \(=\) | \(\mathbf{5}\text{ cm}\) |
Check: \(2\pi(25) + 2\pi(5)(7) = 50\pi + 70\pi = 120\pi\) ✓
If the radius is given and the height is unknown, the equation is linear instead: one step of subtracting and one of dividing.
(i) the painted area, correct to two decimal places
(ii) the number of tins
(iii) the cost.
(i) One circle (the roof) plus the curved surface:
| \(A\) | \(=\) | \(\pi r^2 + 2\pi rh\) |
| \(\) | \(=\) | \(\pi(2.5)^2 + 2\pi(2.5)(9)\) |
| \(\) | \(=\) | \(51.25\pi\) |
| \(\) | \(=\) | \(161.0066\ldots\) |
| \(\) | \(\approx\) | \(\mathbf{161.01}\text{ m}^2\) |
(ii) Two coats, then litres, then tins:
| \(2 \times 161.006\ldots\) | \(=\) | \(322.013\ldots\text{ m}^2\) |
| \(322.013\ldots \div 10\) | \(=\) | \(32.20\ldots\text{ L}\) |
| \(32.20\ldots \div 10\) | \(=\) | \(3.22\ldots\) |
Three tins are not quite enough, so round up: \(\mathbf{4}\) tins.
(iii)
| \(\text{cost}\) | \(=\) | \(4 \times \$89\) |
| \(\) | \(=\) | \(\mathbf{\$356}\) |
(i) Show that its total surface area is \(10\pi r^2\), where \(r\) is the radius.
(ii) The surface area is \(250\pi\text{ cm}^2\). Find the radius and the height.
(i) Twice as tall as it is wide means \(h = 2 \times\) diameter \(= 2 \times 2r = 4r\).
| \(\text{SA}\) | \(=\) | \(2\pi r^2 + 2\pi r(4r)\) |
| \(\) | \(=\) | \(2\pi r^2 + 8\pi r^2\) |
| \(\) | \(=\) | \(10\pi r^2\) |
(ii)
| \(10\pi r^2\) | \(=\) | \(250\pi\) |
| \(r^2\) | \(=\) | \(25\) |
| \(r\) | \(=\) | \(\mathbf{5}\text{ cm} \quad (r \gt 0)\) |
| \(h\) | \(=\) | \(4r = \mathbf{20}\text{ cm}\) |
Check: \(2\pi(25) + 2\pi(5)(20) = 50\pi + 200\pi = 250\pi\) ✓
Common pitfalls
Frequently asked questions
What is the formula for the surface area of a cylinder?
For a closed cylinder, \(\text{SA} = 2\pi r^2 + 2\pi rh\): two circular ends plus the curved surface. Leave out a circle for each open end.
Why is the curved surface area 2 pi r h?
If you unroll the curved surface it becomes a rectangle. One side is the height \(h\) and the other is the circumference of the end, \(2\pi r\). Length times width gives \(2\pi rh\).
What does in terms of pi mean?
It means an exact answer that keeps \(\pi\) as a symbol, such as \(90\pi\text{ cm}^2\). Collect the \(\pi\) terms and do not press the \(\pi\) key.
How do you find the radius of a cylinder from its surface area?
Substitute the surface area and height into the formula, divide every term by pi, and rearrange into a quadratic equal to zero. Factorise and keep the positive solution.
How do you convert square centimetres to square metres?
Divide by \(10\,000\), because \(1\text{ m}^2\) is \(100\text{ cm} \times 100\text{ cm}\). For example, \(25\,000\text{ cm}^2 = 2.5\text{ m}^2\).