Surface Area of Cylinders

Open a cylinder out into its net, find its curved and total surface area, work with open cylinders, rollers, labels and rolled sheets, find a height or a radius, and find the surface of a half-cylinder.

Before you start

A cylinder has two identical, parallel circular ends and a curved surface joining them. Its radius $r$ is the radius of an end, and its height $h$ is the distance between the ends. For a cylinder lying down (a roller, a pipe), its length plays the part of the height.
A circle of radius $r$ has circumference $2\pi r$ (or $\pi d$ , where the diameter $d = 2r$ ) and area $\pi r^{2}$ . For example, a circle of radius 5 cm has circumference $10\pi \approx 31.4$ cm and area $25\pi \approx 78.5\text{ cm}^{2}$ , correct to 3 significant figures.
Each question says which $\pi$ to use: the $\pi$ button on your calculator, $\pi = 3.142$ , $\pi = \frac{22}{7}$ , or "in terms of $\pi$ ". Give answers that are not exact correct to 3 significant figures, keep full values until the last line, give money to the nearest cent, and remember that $1\text{ m}^{2} = 10\,000\text{ cm}^{2}$ .

The Net of a Cylinder

Peel the paper label off a can of soup and flatten it: it is a rectangle. The label went once round the can, so the length of the rectangle is the distance round an end of the can, its circumference, $2\pi r$ . The width of the rectangle is the height of the can, $h$ .

Cut any cylinder along a straight line from one end to the other, open its curved surface out flat, and lay its two ends beside it. This flat shape is the net of the cylinder: one rectangle and two identical circles. When the net is folded up again, each long side of the rectangle wraps once round the edge of a circle.

The net of a cylinder

The net of a closed cylinder of radius $r$ and height $h$ is a rectangle and two circles of radius $r$ .

The rectangle is the curved surface opened out. Its length is the circumference of an end, $2\pi r$ , and its width is the height, $h$ .

Example 1: The net of a can

A closed can has radius 7 cm and height 15 cm. Take $\pi = \frac{22}{7}$ .

(a) Find the length and the width of the rectangle in its net. (b) A paper label covers the whole curved surface exactly, with no overlap. Find the area of the label. (c) Find the total area of the net.

Solution

(a) The length of the rectangle is the circumference of an end.

Length $= 2\pi r$

$= 2 \times \frac{22}{7} \times 7$

$= 44$ cm

The width of the rectangle is the height of the can, 15 cm.

(b) The label is the rectangle of the net.

Area of label $= 44 \times 15 = 660\text{ cm}^{2}$

Total area of the net $= 660 + 2 \times 154 = 968\text{ cm}^{2}$

(a) The rectangle is 44 cm long and 15 cm wide. (b) The label has area $660\text{ cm}^{2}$ . (c) The net has total area $968\text{ cm}^{2}$ .

Curved Surface Area and Total Surface Area

The curved surface area of a cylinder is the area of its curved surface only. Opened out, the curved surface is the rectangle of the net, so its area is length $\times$ width $= 2\pi r \times h$ .

The total surface area of a closed cylinder is the area of all its surfaces: the curved surface and the two circular ends. It is the area of the whole net.

Surface area of a cylinder

Curved surface area $= 2\pi rh$

Total surface area of a closed cylinder $= 2\pi r^{2} + 2\pi rh$ (the two ends and the curved surface)

The two terms have the common factor $2\pi r$ , so the total surface area may also be written $2\pi r(r + h)$ .

Here $r$ is the radius of an end and $h$ is the height. When the diameter is given, halve it first.

Example 2: Leaving the answer in terms of $\pi$

A closed cylinder has radius 5 cm and height 12 cm. Find (a) its curved surface area, (b) its total surface area. Leave each answer in terms of $\pi$ , and then give it correct to 3 significant figures using the $\pi$ button on your calculator.

Solution

(a) Curved surface area $= 2\pi rh$

$= 2 \times \pi \times 5 \times 12$

$= 120\pi\text{ cm}^{2} \approx 377\text{ cm}^{2}$

(b) The two ends: $2\pi r^{2} = 2 \times \pi \times 5^{2} = 50\pi\text{ cm}^{2}$

Total surface area $= 2\pi r^{2} + 2\pi rh$

$= 50\pi + 120\pi$

$= 170\pi\text{ cm}^{2} \approx 534\text{ cm}^{2}$

Check with the factorised form: $2\pi r(r + h) = 2 \times \pi \times 5 \times (5 + 12) = 170\pi$ .

The curved surface area is $120\pi \approx 377\text{ cm}^{2}$ and the total surface area is $170\pi \approx 534\text{ cm}^{2}$ .

Example 3: The diameter is given

A closed tin has diameter 9 cm and height 15 cm. Taking $\pi = 3.142$ , find the area of metal needed to make it, correct to 3 significant figures. (Ignore any overlap at the joins.)

Solution

The diameter is 9 cm, so the radius is $9 \div 2 = 4.5$ cm.

The tin is closed, so the metal covers its total surface area: two ends and the curved surface.

Two ends: $2\pi r^{2} = 2 \times 3.142 \times 4.5^{2} = 127.251\text{ cm}^{2}$

Curved surface: $2\pi rh = 2 \times 3.142 \times 4.5 \times 15 = 424.17\text{ cm}^{2}$

Total surface area $= 127.251 + 424.17 = 551.421\text{ cm}^{2} \approx 551\text{ cm}^{2}$

About $551\text{ cm}^{2}$ of metal is needed, correct to 3 significant figures.

Open Cylinders: Count Only the Surfaces That Are There

Many cylinders in use are not closed. A cup or a pencil holder has a bottom but no top. A pipe or a cardboard tube has no ends at all. The formula $2\pi r^{2} + 2\pi rh$ counts two ends, so for an open cylinder do not use it: list the surfaces that are there and add their areas.

Open cylinders

Open at one end (a cup, a tin without a lid): one end and the curved surface, $\pi r^{2} + 2\pi rh$ .

Open at both ends (a pipe, a tube): the curved surface only, $2\pi rh$ .

Read the question for the surfaces it asks about. A tank standing on the ground and painted on the outside has no paint on its bottom.

Example 4: A paper cup

A paper cup in the shape of a cylinder, open at the top, has radius 4 cm and height 10 cm. Using the $\pi$ button on your calculator, find the area of paper needed to make it, correct to 3 significant figures. (Ignore any overlap at the joins.)

Solution

The cup has a bottom and a curved surface, but no top.

Bottom: $\pi r^{2} = \pi \times 4^{2} = 16\pi\text{ cm}^{2}$

Curved surface: $2\pi rh = 2 \times \pi \times 4 \times 10 = 80\pi\text{ cm}^{2}$

Area of paper $= 16\pi + 80\pi = 96\pi\text{ cm}^{2} \approx 302\text{ cm}^{2}$

About $302\text{ cm}^{2}$ of paper is needed, correct to 3 significant figures.

Example 5: Painting a water tank

A cylindrical water tank on the roof of an HDB block has diameter 2.1 m and height 2.5 m. Its curved surface and its top are to be painted; its bottom stands on the roof and is not painted. Take $\pi = 3.142$ and give the area correct to 3 significant figures.

(a) Find the area to be painted. (b) Painting costs $8 per $\text{m}^{2}$ . Find the cost of painting the tank.

Solution

The radius is $2.1 \div 2 = 1.05$ m. The surfaces painted are the top and the curved surface.

Top: $\pi r^{2} = 3.142 \times 1.05^{2} = 3.464055\text{ m}^{2}$

Curved surface: $2\pi rh = 2 \times 3.142 \times 1.05 \times 2.5 = 16.4955\text{ m}^{2}$

Area painted $= 3.464055 + 16.4955 = 19.959555\text{ m}^{2} \approx 20.0\text{ m}^{2}$

(b) Use the full area, not the rounded 20.0: $19.959555 \times 8 = 159.67644$ , so the cost is $159.68 to the nearest cent.

(a) The area to be painted is about $20.0\text{ m}^{2}$ . (b) The cost of painting the tank is $159.68.

Rollers, Labels and Rolled Sheets

A roller turns about its axis as it rolls. In one complete turn every part of its curved surface touches the ground once, so it covers an area equal to its curved surface area. Its ends never touch the ground. In $n$ complete turns it covers $n \times 2\pi rh$ , where $h$ is the length of the roller.

Example 6: A garden roller

A garden roller is a cylinder of diameter 60 cm and length 1.2 m. Find the area of lawn, in square metres, that it rolls over in 25 complete turns. Use the $\pi$ button on your calculator and give your answer correct to 3 significant figures.

Solution

The answer is wanted in $\text{m}^{2}$ , so work in metres: the diameter is $60\text{ cm} = 0.6$ m, so the radius is 0.3 m.

Area covered in one turn $=$ curved surface area

$= 2\pi rh = 2 \times \pi \times 0.3 \times 1.2 = 0.72\pi\text{ m}^{2}$

Area covered in 25 turns $= 25 \times 0.72\pi = 18\pi\text{ m}^{2} \approx 56.5\text{ m}^{2}$

The roller rolls over about $56.5\text{ m}^{2}$ of lawn in 25 turns.

A label wrapped once round a can is a rectangle. If its ends overlap so that they can be glued, the overlap adds to the label's length, the side that goes round: length $=$ circumference $+$ overlap. The label's width is its own height, which may be less than the height of the can.

Example 7: A label that overlaps

A can of diameter 7 cm and height 12 cm has a paper label 9 cm tall wrapped once round it. The ends of the label overlap by 1 cm. Take $\pi = \frac{22}{7}$ and find the area of the label.

Solution

The radius is $7 \div 2 = 3.5$ cm.

Circumference $= 2\pi r = 2 \times \frac{22}{7} \times 3.5 = 22$ cm

Length of label $= 22 + 1 = 23$ cm (the overlap adds to the length that goes round)

Width of label $= 9$ cm (the label's own height, not the can's 12 cm)

Area of label $= 23 \times 9 = 207\text{ cm}^{2}$

The area of the label is $207\text{ cm}^{2}$ .

The other way round, a rectangular sheet rolled into a tube becomes the curved surface of a cylinder open at both ends. The edge that is rolled round becomes the circumference of the tube, and the other edge becomes its height. With no overlap, the curved surface area of the tube is the area of the sheet.

Example 8: A sheet rolled into a tube

A rectangular sheet of card 22 cm by 30 cm is rolled so that its two 30 cm edges just meet, making an open tube 30 cm tall. Take $\pi = \frac{22}{7}$ .

(a) Find the radius of the tube. (b) Find its curved surface area. (c) Two circles of card are cut to close the ends of the tube. Find their total area.

Solution

(a) The 30 cm edges meet, so the 22 cm edge goes round: the circumference of the tube is 22 cm.

$2\pi r = 22$

$2 \times \frac{22}{7} \times r = 22$

$\frac{44}{7}r = 22$

$r = 22 \times \frac{7}{44} = 3.5$ (multiply both sides by $\frac{7}{44}$ )

(b) Curved surface area $=$ area of the sheet $= 22 \times 30 = 660\text{ cm}^{2}$

Check: $2\pi rh = 2 \times \frac{22}{7} \times 3.5 \times 30 = 660$ .

(a) The radius is 3.5 cm. (b) The curved surface area is $660\text{ cm}^{2}$ . (c) The two circles have total area $77\text{ cm}^{2}$ .

Finding the Height or the Radius

When the curved surface area is given with the radius, or with the height, write $2\pi rh$ equal to it with the numbers you know put in. This gives an equation in the unknown length. Solve it as you solve any linear equation, by doing the same to both sides, and check by putting the answer back.

Example 9: The height from the curved surface area

A cylinder of radius 7 cm has curved surface area $440\text{ cm}^{2}$ . Take $\pi = \frac{22}{7}$ and find its height.

(not drawn to scale)

Solution

$2\pi rh = 440$

$2 \times \frac{22}{7} \times 7 \times h = 440$

$44h = 440$

$h = 10$ (divide both sides by 44)

Check: $2 \times \frac{22}{7} \times 7 \times 10 = 440$ .

The height of the cylinder is 10 cm.

Example 10: The radius from the curved surface area

A cylinder of height 12 cm has curved surface area $300\text{ cm}^{2}$ . Using the $\pi$ button on your calculator, find its radius correct to 3 significant figures.

(not drawn to scale)

Solution

$2\pi rh = 300$

$2 \times \pi \times r \times 12 = 300$

$24\pi r = 300$

$r = \frac{300}{24\pi}$ (divide both sides by $24\pi$ )

$= 3.9788\ldots \approx 3.98$

Check: $2 \times \pi \times 3.9788\ldots \times 12 = 300.0$ .

The radius of the cylinder is about 3.98 cm.

Half-Cylinders

Cut a cylinder in two along its length, through the centres of its ends, and each piece is a half-cylinder: the shape of a doorstop, a trough or a tunnel. Its surfaces are half of the curved surface, two semicircular ends, and a flat rectangle where the cut was. The rectangle is as wide as the diameter, $2r$ , and as long as the half-cylinder, $h$ .

Surface area of a half-cylinder of radius $r$ and length $h$

Curved surface $= \frac{1}{2} \times 2\pi rh = \pi rh$

Two semicircular ends $= 2 \times \frac{1}{2}\pi r^{2} = \pi r^{2}$

Flat rectangle $= 2r \times h = 2rh$

Total surface area of a closed half-cylinder $= \pi rh + \pi r^{2} + 2rh$

Leave out what is not there

An open trough (curved side down) has no lid: leave out the flat rectangle, which is its open top. Its surface is $\pi rh + \pi r^{2}$ .

A tunnel stands on the ground and is open at both ends: leave out the flat rectangle (the ground) and the two ends. Only its curved surface, $\pi rh$ , is left.

Example 11: A doorstop

A solid doorstop is a half-cylinder of radius 7 cm and length 20 cm. Take $\pi = \frac{22}{7}$ and find its total surface area.

Solution

The doorstop is closed: it has a curved surface, two semicircular ends and a flat rectangle.

Curved surface: $\pi rh = \frac{22}{7} \times 7 \times 20 = 440\text{ cm}^{2}$

Two semicircular ends: $\pi r^{2} = \frac{22}{7} \times 7^{2} = 154\text{ cm}^{2}$

Flat rectangle: $2r \times h = 14 \times 20 = 280\text{ cm}^{2}$

Total surface area $= 440 + 154 + 280 = 874\text{ cm}^{2}$

The total surface area of the doorstop is $874\text{ cm}^{2}$ .

Example 12: An open trough

A water trough for a farm is a half-cylinder of radius 30 cm and length 1.5 m, with its curved side down and its top open. It is made of metal sheet: the curved surface and the two semicircular ends. Using the $\pi$ button on your calculator, find the area of metal sheet in $\text{m}^{2}$ , correct to 3 significant figures.

Solution

Work in metres: the radius is $30\text{ cm} = 0.3$ m.

The top is open, so the flat rectangle is left out.

Curved surface: $\pi rh = \pi \times 0.3 \times 1.5 = 0.45\pi\text{ m}^{2}$

Two semicircular ends: $\pi r^{2} = \pi \times 0.3^{2} = 0.09\pi\text{ m}^{2}$

Area of metal $= 0.45\pi + 0.09\pi = 0.54\pi\text{ m}^{2} \approx 1.70\text{ m}^{2}$

About $1.70\text{ m}^{2}$ of metal sheet is needed, correct to 3 significant figures.

Common Mistakes

Using the diameter in place of the radius. Halve a diameter first.
Using $2\pi r^{2} + 2\pi rh$ for a cup, a tank standing on the ground or a pipe. That formula counts two ends; add only the surfaces that are there.
Reading $2\pi r^{2}$ as $(2\pi r)^{2}$ . Only $r$ is squared: $2\pi r^{2} = 2 \times \pi \times r \times r$ .
Rounding the area of one part before adding the parts. Keep full values and round once, at the end.
Mixing units. With a radius of 30 cm and a length of 1.5 m, change one of them first. To change $\text{cm}^{2}$ to $\text{m}^{2}$ , divide by $10\,000$ , not by 100.
Using the total surface area for a roller. Only its curved surface touches the ground.
Adding a label's overlap to its height. The overlap adds to the length that goes round.
For a half-cylinder, using $2\pi rh$ for the curved surface (it is half of that, $\pi rh$ ), or $r \times h$ for the flat face (it is $2r \times h$ ).
Rounding a number of tins of paint down. Enough paint is needed for the whole surface, so tins are rounded up.

Practice Questions

Try each question before you open its solution. Each question says how to use $\pi$ ; give answers that are not exact correct to 3 significant figures, and money to the nearest cent.

Practice 1: Curved and total surface area

A closed cylinder has radius 3 cm and height 8 cm. Find (a) its curved surface area, (b) its total surface area. Leave each answer in terms of $\pi$ , and then give it correct to 3 significant figures using the $\pi$ button on your calculator.

ShowHide solution

(a) Curved surface area $= 2\pi rh = 2 \times \pi \times 3 \times 8 = 48\pi\text{ cm}^{2} \approx 151\text{ cm}^{2}$

(b) Two ends: $2\pi r^{2} = 2 \times \pi \times 3^{2} = 18\pi\text{ cm}^{2}$

Total surface area $= 18\pi + 48\pi = 66\pi\text{ cm}^{2} \approx 207\text{ cm}^{2}$

Check: $2\pi r(r + h) = 2 \times \pi \times 3 \times 11 = 66\pi$ .

(a) $48\pi \approx 151\text{ cm}^{2}$ (b) $66\pi \approx 207\text{ cm}^{2}$

Practice 2: From the diameter

A closed cylinder has diameter 14 cm and height 20 cm. Take $\pi = \frac{22}{7}$ and find its total surface area.

ShowHide solution

The radius is $14 \div 2 = 7$ cm.

Two ends: $2\pi r^{2} = 2 \times \frac{22}{7} \times 7^{2} = 308\text{ cm}^{2}$

Curved surface: $2\pi rh = 2 \times \frac{22}{7} \times 7 \times 20 = 880\text{ cm}^{2}$

Total surface area $= 308 + 880 = 1188\text{ cm}^{2}$

The total surface area is $1188\text{ cm}^{2}$ .

Practice 3: From a net

The figure shows the net of a closed cylinder. Each circle has radius 4 cm and the rectangle is 10 cm wide. Using the $\pi$ button on your calculator, find (a) the length of the rectangle, (b) the total area of the net. Give each answer in terms of $\pi$ and correct to 3 significant figures.

ShowHide solution

(a) The length of the rectangle is the circumference of a circle: $2\pi r = 2 \times \pi \times 4 = 8\pi$ cm $\approx 25.1$ cm.

(b) Rectangle: $8\pi \times 10 = 80\pi\text{ cm}^{2}$ (using $8\pi$ , not the rounded 25.1)

Two circles: $2 \times \pi \times 4^{2} = 32\pi\text{ cm}^{2}$

Total area of the net $= 80\pi + 32\pi = 112\pi\text{ cm}^{2} \approx 352\text{ cm}^{2}$

(a) The rectangle is $8\pi \approx 25.1$ cm long. (b) The net has area $112\pi \approx 352\text{ cm}^{2}$ .

Practice 4: A pencil holder

Hui Min makes a pencil holder from card. It is a cylinder, open at the top, of diameter 7 cm and height 10 cm. Take $\pi = \frac{22}{7}$ and find the area of card she needs. (Ignore any overlap.)

ShowHide solution

The radius is $7 \div 2 = 3.5$ cm. The holder has a bottom and a curved surface, but no top.

Bottom: $\pi r^{2} = \frac{22}{7} \times 3.5^{2} = 38.5\text{ cm}^{2}$

Curved surface: $2\pi rh = 2 \times \frac{22}{7} \times 3.5 \times 10 = 220\text{ cm}^{2}$

Area of card $= 38.5 + 220 = 258.5\text{ cm}^{2}$

Hui Min needs $258.5\text{ cm}^{2}$ of card.

Practice 5: Which can uses more metal?

Two closed cans are made of thin metal. Can A has radius 3 cm and height 12 cm. Can B has radius 4 cm and height 7 cm.

(a) Find the total surface area of each can, in terms of $\pi$ . (b) Which can needs more metal, and how much more? Give the difference correct to 3 significant figures using the $\pi$ button on your calculator.

ShowHide solution

(a) Can A: $2\pi r^{2} + 2\pi rh = 2 \times \pi \times 3^{2} + 2 \times \pi \times 3 \times 12 = 18\pi + 72\pi = 90\pi\text{ cm}^{2}$

Can B: $2 \times \pi \times 4^{2} + 2 \times \pi \times 4 \times 7 = 32\pi + 56\pi = 88\pi\text{ cm}^{2}$

(b) $90\pi > 88\pi$ , so can A needs more metal. The difference is $90\pi - 88\pi = 2\pi\text{ cm}^{2} \approx 6.28\text{ cm}^{2}$ .

Writing both in terms of $\pi$ makes the comparison easy. (Can B still holds more: its volume is $\pi \times 4^{2} \times 7 = 112\pi\text{ cm}^{3}$ , against $\pi \times 3^{2} \times 12 = 108\pi\text{ cm}^{3}$ for can A.)

(a) Can A: $90\pi\text{ cm}^{2}$ ; can B: $88\pi\text{ cm}^{2}$ . (b) Can A needs more metal, by $2\pi \approx 6.28\text{ cm}^{2}$ .

Practice 6: A paint roller

Ravi paints a wall with a paint roller. The roller is a cylinder of diameter 5 cm and length 20 cm. Use the $\pi$ button on your calculator.

(a) Find the area, in $\text{m}^{2}$ , that the roller covers in 40 complete turns. (b) How many complete turns does it need to cover $3\text{ m}^{2}$ of wall?

ShowHide solution

(a) The radius is $5 \div 2 = 2.5$ cm.

One turn covers the curved surface area: $2\pi rh = 2 \times \pi \times 2.5 \times 20 = 100\pi\text{ cm}^{2}$

40 turns cover $40 \times 100\pi = 4000\pi\text{ cm}^{2} = 12\,566.37\ldots\text{ cm}^{2}$

In square metres: $12\,566.37\ldots \div 10\,000 = 1.2566\ldots \approx 1.26\text{ m}^{2}$

(b) $3\text{ m}^{2} = 3 \times 10\,000 = 30\,000\text{ cm}^{2}$

Number of turns $= \frac{30\,000}{100\pi} = 95.49\ldots$

95 turns cover only $95 \times 100\pi \approx 29\,845\text{ cm}^{2}$ , less than $30\,000\text{ cm}^{2}$ , so the number of turns is rounded up.

(a) About $1.26\text{ m}^{2}$ . (b) 96 complete turns.

Practice 7: A drink can label

A drink can has diameter 6.6 cm and height 11.5 cm. A label covers its whole curved surface, and the ends of the label overlap by 1 cm. Taking $\pi = 3.142$ , find the area of the label.

ShowHide solution

The radius is $6.6 \div 2 = 3.3$ cm.

Circumference $= 2\pi r = 2 \times 3.142 \times 3.3 = 20.7372$ cm

Length of label $= 20.7372 + 1 = 21.7372$ cm (the overlap adds to the length that goes round)

Width of label $= 11.5$ cm (it covers the whole height)

Area of label $= 21.7372 \times 11.5 = 249.9778\text{ cm}^{2} \approx 250\text{ cm}^{2}$

The area of the label is about $250\text{ cm}^{2}$ , correct to 3 significant figures.

Practice 8: The height from the curved surface area

A closed cylinder has radius 3.5 cm and curved surface area $264\text{ cm}^{2}$ . Take $\pi = \frac{22}{7}$ . (a) Find its height. (b) Find its total surface area.

(not drawn to scale)

ShowHide solution

(a) $2\pi rh = 264$

$2 \times \frac{22}{7} \times 3.5 \times h = 264$

$22h = 264$

$h = 12$ (divide both sides by 22)

Check: $2 \times \frac{22}{7} \times 3.5 \times 12 = 264$ .

(b) Two ends: $2\pi r^{2} = 2 \times \frac{22}{7} \times 3.5^{2} = 77\text{ cm}^{2}$

Total surface area $= 77 + 264 = 341\text{ cm}^{2}$

(a) The height is 12 cm. (b) The total surface area is $341\text{ cm}^{2}$ .

Practice 9: The radius from the area of an end

A closed cylinder is 10 cm high, and each of its circular ends has an area of $154\text{ cm}^{2}$ . Take $\pi = \frac{22}{7}$ . Find (a) its radius, (b) its total surface area.

(not drawn to scale)

ShowHide solution

(a) Let the radius be $r$ cm. The area of an end is $\pi r^{2}$ :

$\frac{22}{7} \times r^{2} = 154$

$r^{2} = 154 \times \frac{7}{22} = 49$ (multiply both sides by $\frac{7}{22}$ )

$r = \sqrt{49} = 7$ (the positive root, since $r$ is a length)

(b) Curved surface: $2\pi rh = 2 \times \frac{22}{7} \times 7 \times 10 = 440\text{ cm}^{2}$

Two ends: $2 \times 154 = 308\text{ cm}^{2}$ (the area of an end is given, so use it)

Total surface area $= 440 + 308 = 748\text{ cm}^{2}$

Check: $2\pi r(r + h) = 2 \times \frac{22}{7} \times 7 \times 17 = 44 \times 17 = 748$ .

(a) The radius is 7 cm. (b) The total surface area is $748\text{ cm}^{2}$ .

Practice 10: The radius from the curved surface area

A closed cylinder has height 15 cm and curved surface area $500\text{ cm}^{2}$ . Using the $\pi$ button on your calculator, find (a) its radius, (b) its total surface area, each correct to 3 significant figures.

(not drawn to scale)

ShowHide solution

(a) $2\pi rh = 500$

$2 \times \pi \times r \times 15 = 500$

$30\pi r = 500$

$r = \frac{500}{30\pi} = 5.3051\ldots \approx 5.31$ (divide both sides by $30\pi$ )

(b) Keep the full value of $r$ on the calculator.

Two ends: $2\pi r^{2} = 2 \times \pi \times 5.3051\ldots^{2} = 176.83\ldots\text{ cm}^{2}$

Total surface area $= 176.83\ldots + 500 = 676.83\ldots\text{ cm}^{2} \approx 677\text{ cm}^{2}$

(a) The radius is about 5.31 cm. (b) The total surface area is about $677\text{ cm}^{2}$ .

Practice 11: One sheet, two tubes

Mei Ling has a rectangular sheet of paper 30 cm by 20 cm. She can roll it into an open tube, with no overlap, in two ways. Tube A: the 30 cm edge goes round, and the tube is 20 cm tall. Tube B: the 20 cm edge goes round, and the tube is 30 cm tall. Use the $\pi$ button on your calculator.

(a) Find the radius of each tube. (b) Find the curved surface area of each tube. (c) Find the area of card needed to close both ends of tube A.

ShowHide solution

(a) Tube A: the circumference is 30 cm.

$2 \times \pi \times r = 30$ , so $r = \frac{30}{2\pi} = 4.7746\ldots \approx 4.77$ cm

Tube B: the circumference is 20 cm.

$2 \times \pi \times r = 20$ , so $r = \frac{20}{2\pi} = 3.1830\ldots \approx 3.18$ cm

(b) Each tube is the whole sheet rolled up, so each curved surface area is $30 \times 20 = 600\text{ cm}^{2}$ .

Check for tube A: $2\pi rh = 2 \times \pi \times 4.7746\ldots \times 20 = 600$ .

(c) Two ends of tube A, with the full value of $r$ : $2\pi r^{2} = 2 \times \pi \times 4.7746\ldots^{2} = 143.23\ldots \approx 143\text{ cm}^{2}$

(a) Tube A has radius about 4.77 cm and tube B about 3.18 cm. (b) Both have curved surface area $600\text{ cm}^{2}$ . (c) About $143\text{ cm}^{2}$ of card.

Practice 12: Tins of paint for a tank

Mr Tan paints the outside of a cylindrical tank standing on the ground: its top and its curved surface, but not its bottom. The tank has diameter 3 m and height 2 m. One tin of paint covers $4\text{ m}^{2}$ and costs $23.90. Use the $\pi$ button on your calculator.

Find (a) the area to be painted, (b) the number of tins Mr Tan must buy, (c) the cost of the paint.

ShowHide solution

(a) The radius is $3 \div 2 = 1.5$ m.

Top: $\pi r^{2} = \pi \times 1.5^{2} = 2.25\pi\text{ m}^{2}$

Curved surface: $2\pi rh = 2 \times \pi \times 1.5 \times 2 = 6\pi\text{ m}^{2}$

Area painted $= 2.25\pi + 6\pi = 8.25\pi\text{ m}^{2} = 25.918\ldots\text{ m}^{2} \approx 25.9\text{ m}^{2}$

(b) $25.918\ldots \div 4 = 6.479\ldots$ Six tins cover only $24\text{ m}^{2}$ , which is not enough, so the number is rounded up: 7 tins.

(a) About $25.9\text{ m}^{2}$ . (b) 7 tins. (c) $167.30.

Practice 13: A play tunnel

A play tunnel in a playground is a half-cylinder of radius 0.7 m and length 3 m. It stands on the ground and is open at both ends, so it has only its curved surface. Take $\pi = \frac{22}{7}$ .

(a) Find the area of its curved surface. (b) The curved surface is painted inside and outside. One tin of paint covers $5\text{ m}^{2}$ . How many tins are needed?

ShowHide solution

(a) The ground is its floor and both ends are open, so only the curved surface is there.

Curved surface: $\pi rh = \frac{22}{7} \times 0.7 \times 3 = 2.2 \times 3 = 6.6\text{ m}^{2}$

(b) Inside and outside: $2 \times 6.6 = 13.2\text{ m}^{2}$

$13.2 \div 5 = 2.64$ . Two tins cover only $10\text{ m}^{2}$ , which is not enough, so the number is rounded up: 3 tins.

(a) The curved surface is $6.6\text{ m}^{2}$ . (b) 3 tins are needed.

Practice 14: A garden shelter

A shelter in a community garden is a half-cylinder of radius 1.4 m and length 6 m, lying on its flat face. Its curved roof and its two semicircular ends are covered with plastic sheet; its floor is the ground. Use the $\pi$ button on your calculator.

(a) Find the area of plastic sheet used. (b) The sheet costs $4.50 per $\text{m}^{2}$ . Find the cost.

ShowHide solution

(a) The floor is the ground, so the flat rectangle is left out; the curved roof and the two ends are covered.

Curved roof: $\pi rh = \pi \times 1.4 \times 6 = 8.4\pi\text{ m}^{2}$

Two semicircular ends: $\pi r^{2} = \pi \times 1.4^{2} = 1.96\pi\text{ m}^{2}$

Area of sheet $= 8.4\pi + 1.96\pi = 10.36\pi\text{ m}^{2} = 32.546\ldots\text{ m}^{2} \approx 32.5\text{ m}^{2}$

(b) Cost, with the full area: $10.36\pi \times 4.50 = 146.46\ldots$ , so the cost is $146.46.

(a) About $32.5\text{ m}^{2}$ of plastic sheet. (b) The cost is $146.46.

Key Points

The net of a closed cylinder is a rectangle and two circles. The rectangle is $2\pi r$ long (the circumference) and $h$ wide (the height).
Curved surface area of a cylinder $= 2\pi rh$ .
Total surface area of a closed cylinder $= 2\pi r^{2} + 2\pi rh = 2\pi r(r + h)$ .
For an open cylinder, add only the surfaces that are there: a cup, $\pi r^{2} + 2\pi rh$ ; a pipe, $2\pi rh$ .
A roller covers its curved surface area in each complete turn: $n$ turns cover $n \times 2\pi rh$ .
A label's length is the circumference plus any overlap; its width is the label's own height.
A sheet rolled into a tube: the edge that goes round becomes the circumference, and the curved surface area is the area of the sheet.
To find $h$ or $r$ from the curved surface area, put the known numbers into $2\pi rh$ and solve the equation.
A half-cylinder has a curved surface $\pi rh$ , two semicircular ends $\pi r^{2}$ in all, and a flat rectangle $2rh$ ; leave out what is open or on the ground.
Change lengths to one unit first ( $1\text{ m}^{2} = 10\,000\text{ cm}^{2}$ ), round only at the end, and round a number of tins of paint up.

Before you start

The Net of a Cylinder

The net of a cylinder

Example 1: The net of a can

Curved Surface Area and Total Surface Area

Surface area of a cylinder

Example 2: Leaving the answer in terms of π\piπ

Example 3: The diameter is given

Open Cylinders: Count Only the Surfaces That Are There

Open cylinders

Example 4: A paper cup

Example 5: Painting a water tank

Rollers, Labels and Rolled Sheets

Example 6: A garden roller

Example 7: A label that overlaps

Example 8: A sheet rolled into a tube

Finding the Height or the Radius

Example 9: The height from the curved surface area

Example 10: The radius from the curved surface area

Half-Cylinders

Surface area of a half-cylinder of radius rrr and length hhh

Leave out what is not there

Example 11: A doorstop

Example 12: An open trough

Common Mistakes

Practice Questions

Practice 1: Curved and total surface area

Practice 2: From the diameter

Practice 3: From a net

Practice 4: A pencil holder

Practice 5: Which can uses more metal?

Practice 6: A paint roller

Practice 7: A drink can label

Practice 8: The height from the curved surface area

Practice 9: The radius from the area of an end

Practice 10: The radius from the curved surface area

Practice 11: One sheet, two tubes

Practice 12: Tins of paint for a tank

Practice 13: A play tunnel

Practice 14: A garden shelter

Key Points

Example 2: Leaving the answer in terms of $\pi$

Surface area of a half-cylinder of radius $r$ and length $h$