Cylinder surface area | Year 9

01 / LEARN

Understand the idea.

Unroll the curved face

Imagine cutting the cylinder wall vertically and unrolling it. The rectangle has height $h$ and length equal to the full circumference, $2 π r$ . Its area is therefore $2 π r h$ , rather than diameter multiplied by height.

Each circular end contributes $π r^{2}$ . A closed cylinder contains two such ends, so its total surface area combines the curved wall with $2 π r^{2}$ . All dimensions must use the same unit.

\begin{aligned} A_{curved} & = 2 π r h \\ A_{closed} & = 2 π r^{2} + 2 π r h \end{aligned}

Match the formula to the object

Choose surfaces from the object before choosing a formula. A label covers the wall alone; a tank without a lid includes its wall and one base; a closed can includes its wall and both ends.

If the question gives diameter $d$ , use radius $r = \frac{d}{2}$ . Keep $π$ throughout the calculation for an exact answer, then use the calculator value only when a rounded decimal is requested.

02 / WORKED EXAMPLES

See the method in action.

EXAMPLE 01

A closed cylinder

Find the total surface area of a closed cylinder with radius 3 cm and height 8 cm.

Worked solution

1
Calculate both circular ends.
$A_{ends} = 2 π (3^{2}) = 18 π {cm}^{2}$
The cylinder is closed, so both equal circles are included.
2
Calculate the curved wall.
$A_{wall} = 2 π (3) (8) = 48 π {cm}^{2}$
The unrolled wall has circumference $2 π r$ and height $h$ .
3
Add the surfaces exactly.
$A = 18 π + 48 π = 66 π {cm}^{2}$
The ends and wall together cover the entire outside.
4
Round the decimal value.
$66 π \approx 207.3 {cm}^{2}$
Rounding once at the end gives one decimal place.

Answer

$66 π$ cm², or 207.3 cm² to one decimal place.

EXAMPLE 02

One end is missing

A cylindrical tank has radius 2 m and height 5 m. It has a base but no lid. Find the area of material needed.

Worked solution

1
Identify the included surfaces.
$A = π r^{2} + 2 π r h$
The tank has one base and a curved wall, with no circular lid.
2
Calculate the base.
$A_{base} = π (2^{2}) = 4 π m^{2}$
One circular base has area $π r^{2}$ .
3
Calculate the wall.
$A_{wall} = 2 π (2) (5) = 20 π m^{2}$
The wall unrolls to a rectangle of length $4 π$ metres and height $5$ metres.
4
Add the material areas.
$A = 4 π + 20 π = 24 π m^{2}$
Only the surfaces that physically exist contribute material.

Answer

$24 π$ m²

03 / YOUR TURN

Try it yourself.

Check your understanding

What is the unrolled length of the curved face when the radius is 4 cm?

Show answer & reasoning

$8 π$ cm.

The unrolled rectangle reaches once around the circular rim. Its length is $2 π r = 2 π (4) = 8 π cm$ .

Keep practising

6 worksheet questions · 14 marks

Open worksheet Worked solutions

Read the questions online

Find the total surface area of a closed cylinder with radius 4 cm and height 6 cm. Give an exact answer.
2 marks
A closed cylinder has diameter 10 cm and height 7 cm. Find its surface area.
2 marks
Find the curved surface area of a cylinder with radius 3 cm and height 12 cm.
2 marks
An open cylindrical container has radius 5 cm and height 10 cm. It has one base. Find its material area.
3 marks
A label covers only the wall of a can with radius 2 cm and height 6 cm. Find the label area.
2 marks
A closed cylinder has radius 3 cm and total surface area $90 π$ cm². Find its height.
3 marks

Read the worked solutions

Equivalent correct methods earn the corresponding marks. Include units where asked.

Find the total surface area of a closed cylinder with radius 4 cm and height 6 cm. Give an exact answer.
2 marks
$2 π (4^{2}) + 2 π (4) (6) = 32 π + 48 π = 80 π$ cm².
$80 π$ cm²
A closed cylinder has diameter 10 cm and height 7 cm. Find its surface area.
2 marks
The radius is 5 cm. Area $= 2 π (25) + 2 π (5) (7) = 50 π + 70 π = 120 π$ cm².
$120 π$ cm²
Find the curved surface area of a cylinder with radius 3 cm and height 12 cm.
2 marks
$2 π r h = 2 π (3) (12) = 72 π$ cm².
$72 π$ cm²
An open cylindrical container has radius 5 cm and height 10 cm. It has one base. Find its material area.
3 marks
One base $25 π$ plus wall $2 π (5) (10) = 100 π$ gives $125 π$ cm².
$125 π$ cm²
A label covers only the wall of a can with radius 2 cm and height 6 cm. Find the label area.
2 marks
Only curved area is needed: $2 π (2) (6) = 24 π$ cm².
$24 π$ cm²
A closed cylinder has radius 3 cm and total surface area $90 π$ cm². Find its height.
3 marks
$18 π + 6 π h = 90 π$ . Divide by $π$ : $18 + 6 h = 90$ , so $h = 12$ cm.
12 cm

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Curriculum reference & lesson scope

Australian Curriculum v9 / South Australian Curriculum

Year 9 surface area and volume of right prisms and cylinders, capacity, absolute/relative/percentage measurement error, precision and practical measurement problems.

Official curriculum source ↗

In this lesson

Understand the idea.

Unroll the curved face

Match the formula to the object

See the method in action.

A closed cylinder

Calculate both circular ends.

Calculate the curved wall.

Add the surfaces exactly.

Round the decimal value.

One end is missing

Identify the included surfaces.

Calculate the base.

Calculate the wall.

Add the material areas.

Try it yourself.

Check your understanding

Keep practising