Year 10 Maths Core Measurement

Surface Area of Cylinders

Q: What is the formula for the surface area of a cylinder?

For a closed cylinder, surface area equals 2 pi r squared plus 2 pi r h. The first part is the two circular ends and the second part is the curved surface.

Q: How do you find the surface area of an open cylinder?

If it is open at one end, use one circle plus the curved surface: pi r squared plus 2 pi r h. If it is open at both ends, like a pipe, use only the curved surface, 2 pi r h.

Q: Is the net of a cylinder a rectangle and two circles?

Yes. The net of a closed cylinder is two circles of radius r and a rectangle h high and 2 pi r long, the circumference of the circles.

20 practice questions 1 video lesson Theory + worked examples

Surface Area of Cylinders in NSW Year 10 Core maths unrolls a cylinder's curved surface into a rectangle with area $2\pi rh$, then adds the circular ends, giving $A = 2\pi r^2 + 2\pi rh$ for a closed cylinder (NESA outcome MA5‑ARE‑C‑01). Open cans and pipes simply leave out one or both ends.

You will find the surface areas of closed cylinders, cans open at one end and pipes open at both ends, giving answers in terms of $\pi$ or rounded, and work backwards to find a radius or a height. You will also solve practical problems: sheet metal for a rainwater tank, paint for a grain silo, a label wrapped around a can and the cost of metal for 500 tins. It is part of Year 10 Core Measurement.

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Theory

The surface area of a cylinder is the two circular ends plus the curved surface, which unrolls into a rectangle: $SA = 2\pi r^{2} + 2\pi rh$. This Year 10 maths page covers closed and open cylinders, pipes and labels, exact answers in $\pi$, and working backwards to a height or radius.

A cylinder has two identical circular ends joined by a curved surface. Its radius $r$ is the radius of the ends and its height $h$ is the distance between them.

If you cut the curved surface straight down and unroll it, it becomes a rectangle. The rectangle's width is the height $h$. Its length is the distance once around the end: the circumference $2\pi r$. So the curved surface area is $2\pi rh$.

A closed cylinder adds both circles: $2\pi r^{2}$. A cylinder open at one end, such as a cup, a tank with no lid or a pot, has only one circle. A tube or pipe open at both ends has no circles at all.

An exact answer is left in terms of $\pi$, such as $180\pi\text{ cm}^2$.

A closed cylinder has two circular ends and one curved surface.

Unroll the curved surface and it is a rectangle: length $2\pi r$ (the circumference), width $h$. So its area is $2\pi rh$.

\[SA = 2\pi r^{2} + 2\pi rh \qquad \text{curved surface} = 2\pi rh\]

S A = 2 π r^{2} + 2 π r h

Cylinder	Surfaces	Area
Closed (solid)	2 circles + curved surface	$2\pi r^{2} + 2\pi rh$
Open at one end	1 circle + curved surface	$\pi r^{2} + 2\pi rh$
Open at both ends (tube, pipe)	curved surface only	$2\pi rh$

Diameter given? Halve it first: \(r = \dfrac{d}{2}\). The curved surface can also be written \(\pi dh\).

Height equal to the diameter. Put \(h = 2r\) into the formula: \(SA = 2\pi r^{2} + 2\pi r(2r) = 6\pi r^{2}\). Then solve for \(r\).

How to find the surface area of a cylinder

Read which surfaces are there: two ends, one end, or none.
Find the radius (halve a diameter) and put $r$ and $h$ in the same unit.
Calculate each part: $\pi r^{2}$ for each end and $2\pi rh$ for the curved surface.
Add. For an exact answer, collect the $\pi$ terms; otherwise round at the end and give square units.

Problems

Painting: include only the surfaces painted. A tank sitting on the ground has no painted base; the inside of an open tank is its base plus the curved wall.
A pipe painted inside and out: add two curved surfaces, one with the outer radius and one with the inner radius.
Label with an overlap: length $=$ circumference $+$ overlap, then multiply by the height.
Working backwards: substitute the known values, divide by $\pi$ if every term has it, and solve for $h$ or $r$.
Many items or a cost: find one surface area, multiply by the number of items, then by the price per unit of area. Painting the pipe in Example 2 at $\$9.50$ per square metre costs $2.4\pi \times 9.50 = 71.628\ldots \approx \$71.63$.

Example 1 — Closed cylinder

Find the surface area of a closed cylinder with radius $6$ cm and height $9$ cm, as an exact value and correct to two decimal places.

Solution

$SA$	$=$	$2\pi r^{2} + 2\pi rh$
	$=$	$2\pi \times 6^{2} + 2\pi \times 6 \times 9$
	$=$	$72\pi + 108\pi$
	$=$	$\mathbf{180\pi}\text{ cm}^2$
	$=$	$565.4866\ldots$
	$\approx$	$\mathbf{565.49}\text{ cm}^2$

S A = 2 π \times 6^{2} + 2 π \times 6 \times 9 = 180 π

Collect the $\pi$ terms before you use the calculator.

Example 2 — Open end, no ends

(i) A vase is a cylinder with a closed base and an open top. Its diameter is $9$ cm and its height is $20$ cm. Find the area of its outside surface, correct to two decimal places.
(ii) A pipe with no ends has radius $0.4$ m and length $3$ m. Find its curved surface area, correct to two decimal places.

Solution

(i) One circle (the base) plus the curved surface. The radius is $9 \div 2 = 4.5$ cm.

$A$	$=$	$\pi \times 4.5^{2} + 2\pi \times 4.5 \times 20$
	$=$	$20.25\pi + 180\pi = 200.25\pi$
	$=$	$629.1039\ldots$
	$\approx$	$\mathbf{629.10}\text{ cm}^2$

(ii) No ends: only the curved surface.

$A$	$=$	$2\pi \times 0.4 \times 3$
	$=$	$2.4\pi$
	$\approx$	$\mathbf{7.54}\text{ m}^2$

A = π {4.5}^{2} + 2 π \times 4.5 \times 20 \approx 629.10

Example 3 — The curved surface is a rectangle

A label covers the curved surface of a can of radius $4$ cm and height $10$ cm, with no overlap. Find
(i) the length of the label
(ii) its area, each correct to two decimal places.

Solution

Unrolled, the label is a rectangle. Its length goes once around the can.

$\text{length}$	$=$	$2\pi \times 4 = 8\pi$
	$\approx$	$\mathbf{25.13}\text{ cm}$
$\text{area}$	$=$	$8\pi \times 10 = 80\pi$
	$\approx$	$\mathbf{251.33}\text{ cm}^2$

A = 2 π \times 4 \times 10 = 80 π

If the label overlaps itself, add the overlap to the length before you multiply.

Example 4 — Working backwards

A closed cylinder has a radius of $4$ cm and a total surface area of $120\pi\text{ cm}^2$. Find its height.

Solution

Substitute what you know, divide every term by $\pi$, then solve.

$2\pi \times 4^{2} + 2\pi \times 4 \times h$	$=$	$120\pi$
$32\pi + 8\pi h$	$=$	$120\pi$
$32 + 8h$	$=$	$120$
$8h$	$=$	$88$
$h$	$=$	$\mathbf{11}\text{ cm}$

32 π + 8 π h = 120 π, h = 11

If the area is a decimal (not a multiple of $\pi$), work out $2\pi r^{2}$ and $2\pi r$ on the calculator and solve the same way.

Common pitfalls

Radius, not diameter. Both parts of the formula use $r$. Halve a diameter before you substitute.

Check the ends. Closed, open at one end and open at both ends give three different answers. Read the question for a lid, a base or an open pipe.

$2\pi r^{2}$ is not $(2\pi r)^{2}$. Square only the radius, then multiply by $2\pi$.

Same units. A radius in centimetres and a height in metres must be matched before you multiply.

Frequently asked questions

What is the formula for the surface area of a cylinder?

For a closed cylinder, $SA = 2\pi r^{2} + 2\pi rh$. The first part is the two circular ends and the second part is the curved surface.

Why is the curved surface area 2 pi r h?

Unrolled, the curved surface is a rectangle. Its length is the circumference, $2\pi r$, and its width is the height $h$, so its area is $2\pi r \times h$.

How do you find the surface area of an open cylinder?

Open at one end: $\pi r^{2} + 2\pi rh$. Open at both ends, like a pipe: $2\pi rh$ only.

How do you find the height of a cylinder from its surface area?

Substitute the surface area and the radius into the formula, then solve for $h$. If every term contains $\pi$, divide through by $\pi$ first.

Is the net of a cylinder a rectangle and two circles?

Yes. The net of a closed cylinder is two circles of radius $r$ and a rectangle $h$ high and $2\pi r$ long, the circumference of the circles.

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Surface Area of Prisms

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Volumes of Prisms and Cylinders

\(SA\)	\(=\)	\(2\pi r^{2} + 2\pi rh\)
\(\)	\(=\)	\(2\pi \times 6^{2} + 2\pi \times 6 \times 9\)
\(\)	\(=\)	\(72\pi + 108\pi\)
\(\)	\(=\)	\(\mathbf{180\pi}\text{ cm}^2\)
\(\)	\(=\)	\(565.4866\ldots\)
\(\)	\(\approx\)	\(\mathbf{565.49}\text{ cm}^2\)

\(A\)	\(=\)	\(\pi \times 4.5^{2} + 2\pi \times 4.5 \times 20\)
\(\)	\(=\)	\(20.25\pi + 180\pi = 200.25\pi\)
\(\)	\(=\)	\(629.1039\ldots\)
\(\)	\(\approx\)	\(\mathbf{629.10}\text{ cm}^2\)

\(A\)	\(=\)	\(2\pi \times 0.4 \times 3\)
\(\)	\(=\)	\(2.4\pi\)
\(\)	\(\approx\)	\(\mathbf{7.54}\text{ m}^2\)

\(\text{length}\)	\(=\)	\(2\pi \times 4 = 8\pi\)
\(\)	\(\approx\)	\(\mathbf{25.13}\text{ cm}\)
\(\text{area}\)	\(=\)	\(8\pi \times 10 = 80\pi\)
\(\)	\(\approx\)	\(\mathbf{251.33}\text{ cm}^2\)

\(2\pi \times 4^{2} + 2\pi \times 4 \times h\)	\(=\)	\(120\pi\)
\(32\pi + 8\pi h\)	\(=\)	\(120\pi\)
\(32 + 8h\)	\(=\)	\(120\)
\(8h\)	\(=\)	\(88\)
\(h\)	\(=\)	\(\mathbf{11}\text{ cm}\)

Surface Area of Cylinders

📖 Theory

How to find the surface area of a cylinder

Problems

Common pitfalls

Frequently asked questions

Theory