Volume of Cylinders

Find the volume of a cylinder with $V = \pi r^{2}h$ , the capacity of cans and tanks in litres, a missing height or radius, and the volume of a half-cylinder.

Before you start

A circle of radius $r$ has area $\pi r^{2}$ and circumference $2\pi r$ . The diameter is $d = 2r$ , so the radius is half the diameter.
The volume of any solid with a uniform cross-section is area of cross-section × length (Volume of Prisms).
$1 \text{ cm}^{3} = 1$ mL, $1000 \text{ cm}^{3} = 1$ L, $1 \text{ m}^{3} = 1\,000\,000 \text{ cm}^{3}$ and $1 \text{ m}^{3} = 1000$ L (Converting Units of Area and Volume).
An equation such as $r^{2} = 49$ has $r = \sqrt{49} = 7$ as its answer when $r$ is a length (a length is positive).

The Cylinder

A cylinder is a solid with two identical, parallel circular ends joined by a curved surface: a drink can, a coin, a log, a water pipe. Cut it straight across, parallel to its ends, anywhere along it, and the cut face is a circle of the same radius. So a cylinder has a uniform cross-section, and that cross-section is a circle.

Its measurements are the radius $r$ of its ends (or the diameter $d = 2r$ ) and its height $h$ , the distance between the two ends. When a cylinder lies on its side, like a log or a rolling pin, its height is called its length: it is still the distance from one circular end to the other.

Cylinder

A cylinder has two identical, parallel circular ends and a curved surface. Its cross-section is a circle of the same radius all along it.

Radius $r$ , diameter $d = 2r$ , and height $h$ (called its length when the cylinder lies down).

Every cylinder here is a right cylinder: its curved surface meets its ends at right angles. A cylinder is not a prism, because its ends are circles, not polygons; but it is a solid with a uniform cross-section, so its volume is found in the same way.

The Volume of a Cylinder

Think of a cylinder 5 cm high as a stack of 5 layers, each 1 cm thick, like a stack of coins. Each layer is a disc whose face is a circle of area $\pi r^{2}$ , so each layer has volume $\pi r^{2} \times 1$ , and the 5 layers together have volume $\pi r^{2} \times 5$ . In the same way, a cylinder of height $h$ has volume $\pi r^{2} \times h$ : the area of its cross-section times its height, exactly as for a prism.

Volume of a cylinder

Volume of a cylinder = area of cross-section × height

$V = \pi r^{2}h$ , where $r$ is the radius and $h$ the height (the length, when the cylinder lies down).

The area of the circular end, $\pi r^{2}$ , is also called the base area, so the volume is also base area × height.

Use the radius, not the diameter, and have $r$ and $h$ in the same unit. The volume is then in cubic units ( $\text{cm}^{3}$ or $\text{m}^{3}$ ).

Which value of π?

Every question says which to use: the $\pi$ button on your calculator; $\pi = 3.142$ ; $\pi = \frac{22}{7}$ (useful when the radius is a multiple of 7 or of 3.5, since the 7 cancels); or "leave your answer in terms of $\pi$ ", writing, say, $300\pi \text{ cm}^{3}$ .

Keep the full values in your working and round only at the end, to 3 significant figures unless the question says otherwise. Write "≈" before a rounded value.

Example 1: From the radius and the height

A cylinder has a radius of 5 cm and a height of 12 cm.

Find its volume (a) in terms of $\pi$ , (b) correct to 3 significant figures, using the $\pi$ button on your calculator.

Solution

Volume of a cylinder $= \pi r^{2}h$

$= \pi \times 5^{2} \times 12$

$= \pi \times 25 \times 12$

(a) $= 300\pi \text{ cm}^{3}$

(b) $= 942.477\ldots \text{ cm}^{3} \approx 942 \text{ cm}^{3}$

The volume is $300\pi \text{ cm}^{3} \approx 942 \text{ cm}^{3}$ .

Example 2: From the diameter

A cylinder has a diameter of 14 cm and a height of 20 cm. Taking $\pi = \frac{22}{7}$ , find its volume.

Solution

The formula needs the radius: $r = 14 \div 2 = 7$ cm.

Volume $= \pi r^{2}h$

$= \frac{22}{7} \times 7^{2} \times 20$

$= \frac{22}{7} \times 49 \times 20$

$= 22 \times 7 \times 20$ (7 goes into 49 seven times)

$= 3080 \text{ cm}^{3}$

The volume is $3080 \text{ cm}^{3}$ .

Example 3: A cylinder lying down, in mixed units

A log lying on the ground is a cylinder of radius 15 cm and length 2.4 m.

Find its volume in $\text{m}^{3}$ . Use the $\pi$ button on your calculator and give your answer correct to 3 significant figures.

Solution

The log lies on its side, so its "height" is its length, 2.4 m.

Change the lengths to one unit first, here metres: $r = 15$ cm $= 0.15$ m.

Volume $= \pi r^{2}h$

$= \pi \times 0.15^{2} \times 2.4$

$= \pi \times 0.0225 \times 2.4$

$= 0.054\pi \text{ m}^{3} = 0.169\,646\ldots \text{ m}^{3} \approx 0.170 \text{ m}^{3}$

Check in centimetres: $\pi \times 15^{2} \times 240 = 54\,000\pi \text{ cm}^{3} = 169\,646.0\ldots \text{ cm}^{3}$ , and $169\,646.0\ldots \div 1\,000\,000 = 0.169\,646\ldots \text{ m}^{3}$ , the same.

The volume of the log is about $0.170 \text{ m}^{3}$ .

Capacity: How Much a Container Holds

The capacity of a container is the volume of liquid it holds when it is full. It is found from the internal (inside) measurements, the internal radius and the internal height, because the walls and the base take up room. Then $1 \text{ cm}^{3} = 1$ mL, $1000 \text{ cm}^{3} = 1$ L and $1 \text{ m}^{3} = 1000$ L.

Water in a cylindrical container that is not full is itself a cylinder: it has the same radius, and its height is the depth of the water.

Example 4: A water tank, full and partly full

A cylindrical water tank on the roof of an HDB block has an internal diameter of 2.4 m and an internal height of 1.5 m. The water in it is 0.9 m deep.

(a) Find the capacity of the tank in litres. (b) How many litres of water are in the tank? (c) How many more litres are needed to fill it? Use the $\pi$ button on your calculator and give each answer correct to 3 significant figures.

Solution

The internal radius is $r = 2.4 \div 2 = 1.2$ m.

(a) Capacity $= \pi r^{2}h = \pi \times 1.2^{2} \times 1.5 = 2.16\pi \text{ m}^{3} = 6.785\,84\ldots \text{ m}^{3}$

Since $1 \text{ m}^{3} = 1000$ L, the capacity is $6.785\,84\ldots \times 1000 = 6785.84\ldots$ L $\approx 6790$ L.

(b) The water is a cylinder of radius 1.2 m and height 0.9 m.

Volume of water $= \pi \times 1.2^{2} \times 0.9 = 1.296\pi \text{ m}^{3} = 4.071\,50\ldots \text{ m}^{3} = 4071.50\ldots$ L $\approx 4070$ L.

Its volume $= \pi \times 1.2^{2} \times 0.6 = 0.864\pi \text{ m}^{3} = 2.714\,33\ldots \text{ m}^{3} = 2714.33\ldots$ L $\approx 2710$ L.

Check with the full values: $6785.84\ldots - 4071.50\ldots = 2714.33\ldots$ L, the same.

(a) The capacity is about 6790 L. (b) There are about 4070 L of water. (c) About 2710 L more are needed.

Example 5: Which can holds more?

Can A has an internal diameter of 6.6 cm and an internal height of 11.5 cm. Can B is taller and narrower: its internal diameter is 5.8 cm and its internal height is 14.6 cm.

Which can holds more, and by how much? Give the difference in mL correct to 3 significant figures, using the $\pi$ button.

Solution

Can A: $r = 6.6 \div 2 = 3.3$ cm. Capacity $= \pi \times 3.3^{2} \times 11.5 = \pi \times 10.89 \times 11.5 = 125.235\pi \text{ cm}^{3}$

Can B: $r = 5.8 \div 2 = 2.9$ cm. Capacity $= \pi \times 2.9^{2} \times 14.6 = \pi \times 8.41 \times 14.6 = 122.786\pi \text{ cm}^{3}$

Both are a number times $\pi$ , so compare the numbers: $125.235 > 122.786$ . Can A holds more.

Difference $= 125.235\pi - 122.786\pi = 2.449\pi \text{ cm}^{3} = 7.693\,76\ldots \text{ cm}^{3} \approx 7.69$ mL

Here the taller can holds less. You cannot tell which of two cylinders holds more from the height alone or the width alone: work out both volumes and compare them (in Question 9 of the practice, the taller, narrower tin holds more).

Can A holds more, by about 7.69 mL.

Example 6: How many glasses can be filled?

A cylindrical jug has an internal radius of 7 cm and an internal height of 25 cm. Water is poured in to a depth of 21 cm.

All the water is then poured into glasses that each hold 250 mL. Taking $\pi = \frac{22}{7}$ , find how many glasses can be filled completely and how much water is left over.

Solution

The water is a cylinder of radius 7 cm and height 21 cm (its depth, not the height of the jug).

Volume of water $= \frac{22}{7} \times 7^{2} \times 21 = 22 \times 7 \times 21 = 3234 \text{ cm}^{3} = 3234$ mL

Number of glasses $= 3234 \div 250 = 12.936$

Only 12 glasses can be filled completely: the 13th would need 250 mL but only part of that is left, so round down.

Water left $= 3234 - 12 \times 250 = 3234 - 3000 = 234$ mL

12 glasses can be filled completely, and 234 mL of water is left over.

Finding the Height or the Radius

When the volume is known and the height or the radius is not, write the formula, put in the numbers you know, and solve the equation for the unknown, as you solve any equation. For the radius you reach $r^{2}$ = a number; $r$ is then the positive square root of that number, since a length is positive.

Finding an unknown height or radius

1. Write $V = \pi r^{2}h$ .

2. Substitute every number you know, including the volume (in a unit that matches the lengths: 1 L = $1000 \text{ cm}^{3}$ ).

3. Simplify, then solve: divide both sides by the number in front of the unknown. For the radius, take the positive square root at the end.

Example 7: The height from the volume

A cylindrical jug has an internal radius of 5 cm and holds exactly 1 litre when it is full.

Find its internal height. Use the $\pi$ button on your calculator and give your answer correct to 3 significant figures.

Solution

Let the internal height be $h$ cm. The capacity is 1 L $= 1000 \text{ cm}^{3}$ .

$\pi r^{2}h = V$

$\pi \times 5^{2} \times h = 1000$

$25\pi h = 1000$

$h = \frac{1000}{25\pi}$ (dividing both sides by $25\pi$ )

$= 12.732\ldots \approx 12.7$

The internal height is about 12.7 cm.

Example 8: The radius from the volume

A cylinder has a volume of $1540 \text{ cm}^{3}$ and a height of 10 cm.

(not drawn to scale)

Taking $\pi = \frac{22}{7}$ , find its radius and its diameter.

Solution

Let the radius be $r$ cm.

$\pi r^{2}h = V$

$\frac{22}{7} \times r^{2} \times 10 = 1540$

$\frac{220}{7}r^{2} = 1540$

$r^{2} = 1540 \times \frac{7}{220}$ (multiplying both sides by $\frac{7}{220}$ )

$r^{2} = 49$

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

Diameter $= 2 \times 7 = 14$ cm. Check: $\frac{22}{7} \times 49 \times 10 = 1540$ .

The radius is 7 cm and the diameter is 14 cm.

Starting from the Circumference or the Base Area

Sometimes a question gives the circumference of the circular end ( $2\pi r$ ) or its area ( $\pi r^{2}$ ) in place of the radius. From a circumference, find $r$ first by solving $2\pi r$ = the circumference. From a base area you need not find $r$ at all, since volume = base area × height.

Example 9: From the circumference

The circumference of the circular end of a cylinder is 44 cm, and the cylinder is 15 cm high.

Taking $\pi = \frac{22}{7}$ , find its volume.

Solution

Let the radius be $r$ cm.

$2\pi r = 44$

$2 \times \frac{22}{7} \times r = 44$

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

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

Volume $= \pi r^{2}h = \frac{22}{7} \times 7^{2} \times 15 = 22 \times 7 \times 15 = 2310 \text{ cm}^{3}$

The volume is $2310 \text{ cm}^{3}$ .

Example 10: From the base area

A cylindrical concrete pillar is 3.2 m tall, and the area of its circular cross-section is $0.5 \text{ m}^{2}$ .

(a) Find the volume of concrete in the pillar. (b) Find the radius of the pillar in centimetres, correct to 3 significant figures, using the $\pi$ button.

Solution

(a) Volume = base area × height $= 0.5 \times 3.2 = 1.6 \text{ m}^{3}$ (no $\pi$ is needed: the area of the circle is given).

(b) Let the radius be $r$ m. The area of the circle is $\pi r^{2}$ , so

$\pi r^{2} = 0.5$

$r^{2} = \frac{0.5}{\pi} = 0.159\,154\ldots$

$r = \sqrt{0.159\,154\ldots} = 0.398\,942\ldots$

In centimetres: $0.398\,942\ldots \times 100 = 39.894\ldots \approx 39.9$ cm.

(a) The volume is $1.6 \text{ m}^{3}$ . (b) The radius is about 39.9 cm.

Mass and Cost

When you know the mass of $1 \text{ cm}^{3}$ of a material (a rate, in grams per $\text{cm}^{3}$ ), the mass of a solid made of it is its volume × the mass of $1 \text{ cm}^{3}$ . In the same way, a cost is the volume × the cost of $1 \text{ m}^{3}$ . Find the volume first, in the unit the rate uses.

Example 11: The mass of a metal rod

A solid metal rod is a cylinder of diameter 4 cm and length 0.6 m. $1 \text{ cm}^{3}$ of the metal has a mass of 7.8 g.

Taking $\pi = 3.142$ , find the mass of the rod in kilograms, correct to 3 significant figures.

Solution

The rate is per $\text{cm}^{3}$ , so work in centimetres: $r = 4 \div 2 = 2$ cm, length $= 0.6$ m $= 60$ cm.

Volume $= \pi r^{2}h = 3.142 \times 2^{2} \times 60 = 3.142 \times 240 = 754.08 \text{ cm}^{3}$

Mass $= 754.08 \times 7.8 = 5881.824$ g

$= 5881.824 \div 1000 = 5.881\,824$ kg $\approx 5.88$ kg

The mass of the rod is about 5.88 kg.

Half-Cylinders

Cut a cylinder in two along its length, through the centres of its two ends, and each piece is a half-cylinder. Its cross-section is a semicircle of area $\frac{1}{2}\pi r^{2}$ , the same all along it, so a half-cylinder is also a solid with a uniform cross-section (and, like the cylinder, not a prism). A trough whose cross-section is a semicircle, or a tunnel whose cross-section is a semicircle, is a half-cylinder. (A tunnel with upright walls under a curved roof is not: it is a cuboid and a half-cylinder put together, as a later lesson shows.)

Volume of a half-cylinder

Volume of a half-cylinder = area of semicircle × length $= \frac{1}{2}\pi r^{2}h$

It is half the volume of the whole cylinder with the same radius and length. Halve the volume, not the radius.

Example 12: The capacity of a trough

A trough for watering plants has a semicircular cross-section of internal radius 21 cm and an internal length of 1.2 m.

Taking $\pi = \frac{22}{7}$ , find how many litres of water it holds when it is full.

Solution

Work in centimetres: length $= 1.2$ m $= 120$ cm.

Volume of a half-cylinder $= \frac{1}{2}\pi r^{2}h$

$= \frac{1}{2} \times \frac{22}{7} \times 21^{2} \times 120$

$= \frac{1}{2} \times \frac{22}{7} \times 441 \times 120$

$= \frac{1}{2} \times 1386 \times 120$ (since $\frac{22}{7} \times 441 = 22 \times 63 = 1386$ )

$= 83\,160 \text{ cm}^{3}$

Since $1000 \text{ cm}^{3} = 1$ L, this is $83\,160 \div 1000 = 83.16$ L.

The trough holds 83.16 L of water when full.

Common mistakes

Using the diameter in place of the radius. A cylinder of diameter 14 cm has $r = 7$ cm. Putting 14 into $\pi r^{2}h$ gives a volume four times too big.
Squaring the wrong thing. $\pi r^{2}h$ means $\pi \times r \times r \times h$ : only the radius is squared. And $2\pi r$ is the circumference, not the area of the circle.
Mixing units. A radius in cm and a height in m must be changed to one unit before you multiply.
Converting a volume with the wrong number. $1 \text{ m}^{3} = 1\,000\,000 \text{ cm}^{3}$ , not 100 or 1000; and $1 \text{ m}^{3} = 1000$ L.
Using outside measurements for a capacity. A container holds only what fits inside it: use the internal radius and height. For water that does not fill a container, use the depth of the water, not the height of the container.
Rounding too early. Keep the full value of a radius found from a circumference or a volume, and of $\pi$ , until the last line.
Rounding a count the wrong way. 12.936 glasses means 12 full glasses, not 13.
Halving the radius of a half-cylinder. Its volume is $\frac{1}{2}\pi r^{2}h$ , half the cylinder's volume; the radius stays the same.

Practice

Try each question before you open its solution. Each question says which value of π to use; give answers that are not exact correct to 3 significant figures, and money to the nearest cent.

Question 1

Find the volume of the cylinder. Leave your answer in terms of $\pi$ .

ShowHide solution

Volume $= \pi r^{2}h = \pi \times 4^{2} \times 9 = \pi \times 16 \times 9 = 144\pi \text{ cm}^{3}$

The volume is $144\pi \text{ cm}^{3}$ .

Question 2

A cylindrical pencil holder has an internal radius of 3.5 cm and an internal height of 10 cm. Taking $\pi = \frac{22}{7}$ , find how many millilitres of sand would fill it.

ShowHide solution

Capacity $= \pi r^{2}h = \frac{22}{7} \times 3.5^{2} \times 10 = \frac{22}{7} \times 12.25 \times 10$

$= 22 \times 1.75 \times 10$ (since $12.25 \div 7 = 1.75$ )

$= 385 \text{ cm}^{3} = 385$ mL

385 mL of sand would fill the pencil holder.

Question 3

A coin is a cylinder of diameter 24 mm and thickness 2 mm. The metal it is made of has a mass of 8.9 g per $\text{cm}^{3}$ .

(not drawn to scale)

Find (a) the volume of the coin in $\text{cm}^{3}$ , (b) its mass in grams. Use the $\pi$ button on your calculator and give each answer correct to 3 significant figures.

ShowHide solution

Change the lengths to cm first: diameter 24 mm $= 2.4$ cm, so $r = 1.2$ cm; the thickness (the coin's height) is 2 mm $= 0.2$ cm.

(a) Volume $= \pi \times 1.2^{2} \times 0.2 = \pi \times 1.44 \times 0.2 = 0.288\pi \text{ cm}^{3} = 0.904\,778\ldots \text{ cm}^{3} \approx 0.905 \text{ cm}^{3}$

(b) Mass $= 0.904\,778\ldots \times 8.9 = 8.052\,53\ldots$ g $\approx 8.05$ g (the full volume used, not 0.905)

(a) The volume is about $0.905 \text{ cm}^{3}$ . (b) The mass is about 8.05 g.

Question 4

A drainage pipe lying on the ground has an internal diameter of 30 cm and is 2.5 m long. Find how many litres of water it holds when it is full. Use the $\pi$ button on your calculator and give your answer correct to 3 significant figures.

ShowHide solution

The water in the full pipe is a cylinder lying on its side, so its "height" is its length.

Change the lengths to one unit first, here centimetres: $r = 30 \div 2 = 15$ cm, and the length is 2.5 m $= 250$ cm.

Capacity $= \pi r^{2}h = \pi \times 15^{2} \times 250 = \pi \times 225 \times 250$

$= 56\,250\pi \text{ cm}^{3} = 176\,714.5\ldots \text{ cm}^{3}$

In litres: $176\,714.5\ldots \div 1000 = 176.714\,5\ldots$ L $\approx 177$ L, since $1000 \text{ cm}^{3} = 1$ L.

Check in metres: $\pi \times 0.15^{2} \times 2.5 = 0.056\,25\pi \text{ m}^{3} = 0.176\,714\ldots \text{ m}^{3}$ , and $0.176\,714\ldots \times 1000 = 176.714\ldots$ L, the same.

The pipe holds about 177 L of water when it is full.

Question 5

Ali's drink can holds 330 mL when full, and its internal radius is 3.2 cm. Find its internal height. Use the $\pi$ button on your calculator and give your answer correct to 3 significant figures.

ShowHide solution

Let the internal height be $h$ cm. 330 mL $= 330 \text{ cm}^{3}$ .

$\pi \times 3.2^{2} \times h = 330$

$10.24\pi h = 330$

$h = \frac{330}{10.24\pi} = 10.258\ldots \approx 10.3$

The internal height is about 10.3 cm.

Question 6

A cylindrical water tank holds 2000 L when full. Its internal height is 1.6 m. Find its internal radius in metres. Use the $\pi$ button on your calculator and give your answer correct to 3 significant figures.

ShowHide solution

The lengths are in metres, so change the capacity to $\text{m}^{3}$ : 2000 L $= 2000 \div 1000 = 2 \text{ m}^{3}$ .

Let the radius be $r$ m.

$\pi \times r^{2} \times 1.6 = 2$

$1.6\pi r^{2} = 2$

$r^{2} = \frac{2}{1.6\pi} = 0.397\,887\ldots$

$r = \sqrt{0.397\,887\ldots} = 0.630\,783\ldots \approx 0.631$

The internal radius is about 0.631 m.

Question 7

Siti's cylindrical cake tin is 7 cm deep inside, and the inside of its circular base has an area of $380 \text{ cm}^{2}$ .

(not drawn to scale)

(a) Find the capacity of the tin in litres. (b) Find the radius of its base, correct to 3 significant figures, using the $\pi$ button.

ShowHide solution

(a) Capacity = base area × height $= 380 \times 7 = 2660 \text{ cm}^{3} = 2660 \div 1000 = 2.66$ L

(b) Let the radius be $r$ cm. $\pi r^{2} = 380$ , so $r^{2} = \frac{380}{\pi} = 120.957\ldots$

$r = \sqrt{120.957\ldots} = 10.998\ldots \approx 11.0$

(a) The capacity is 2.66 L. (b) The radius is about 11.0 cm.

Question 8

A cylindrical pillar at a void deck is 3 m high, and the circumference of its circular cross-section is 220 cm. Taking $\pi = \frac{22}{7}$ , find the volume of the pillar in $\text{m}^{3}$ .

ShowHide solution

Let the radius be $r$ cm. $2\pi r = 220$ , so $2 \times \frac{22}{7} \times r = 220$ , that is $\frac{44}{7}r = 220$ .

$r = 220 \times \frac{7}{44} = 35$ cm

Height $= 3$ m $= 300$ cm.

Volume $= \frac{22}{7} \times 35^{2} \times 300 = \frac{22}{7} \times 1225 \times 300 = 3850 \times 300 = 1\,155\,000 \text{ cm}^{3}$

In $\text{m}^{3}$ : $1\,155\,000 \div 1\,000\,000 = 1.155 \text{ m}^{3}$ .

(Or in metres from the start: $r = 0.35$ m, and $\frac{22}{7} \times 0.35^{2} \times 3 = 1.155 \text{ m}^{3}$ .)

The volume of the pillar is $1.155 \text{ m}^{3}$ .

Question 9

Tin A has a diameter of 10 cm and a height of 12 cm. Tin B has a diameter of 12 cm and a height of 8 cm.

Which tin has the greater volume? Find the difference in terms of $\pi$ , and then correct to 3 significant figures using the $\pi$ button.

ShowHide solution

Tin A: $r = 5$ cm. Volume $= \pi \times 5^{2} \times 12 = 300\pi \text{ cm}^{3}$

Tin B: $r = 6$ cm. Volume $= \pi \times 6^{2} \times 8 = 288\pi \text{ cm}^{3}$

$300 > 288$ , so tin A has the greater volume.

Difference $= 300\pi - 288\pi = 12\pi \text{ cm}^{3} = 37.699\ldots \approx 37.7 \text{ cm}^{3}$

Tin A has the greater volume, by $12\pi \text{ cm}^{3} \approx 37.7 \text{ cm}^{3}$ .

Question 10

A cylindrical tank has an internal diameter of 80 cm and an internal height of 1.2 m. It contains water to a depth of 45 cm.

(a) How many litres of water are in the tank? (b) How many more litres of water are needed to fill it? Use the $\pi$ button on your calculator and give each answer correct to 3 significant figures.

ShowHide solution

Work in centimetres: $r = 80 \div 2 = 40$ cm, internal height $= 1.2$ m $= 120$ cm.

(a) Water $= \pi \times 40^{2} \times 45 = 72\,000\pi \text{ cm}^{3} = 226\,194.6\ldots \text{ cm}^{3}$

$= 226\,194.6\ldots \div 1000 = 226.19\ldots$ L $\approx 226$ L

(b) The empty part is a cylinder of height $120 - 45 = 75$ cm.

Its volume $= \pi \times 40^{2} \times 75 = 120\,000\pi \text{ cm}^{3} = 376\,991.1\ldots \text{ cm}^{3} = 376.99\ldots$ L $\approx 377$ L

(a) About 226 L of water are in the tank. (b) About 377 L more are needed.

Question 11

Mei Ling's cylindrical flask has an internal radius of 6 cm and an internal height of 25 cm, and is full of barley water. She pours it into cups that each hold 180 mL.

How many cups can she fill completely, and how much barley water is left over? Use the $\pi$ button, and give the amount left over in mL correct to 3 significant figures.

ShowHide solution

Volume $= \pi \times 6^{2} \times 25 = 900\pi \text{ cm}^{3} = 2827.43\ldots$ mL

Number of cups $= 2827.43\ldots \div 180 = 15.70\ldots$

She can fill 15 cups completely (round down: there is not enough for a 16th full cup).

Left over $= 2827.43\ldots - 15 \times 180 = 2827.43\ldots - 2700 = 127.43\ldots$ mL $\approx 127$ mL

She can fill 15 cups, and about 127 mL is left over.

Question 12

Mr Tan is building 4 cylindrical concrete pillars, each of diameter 40 cm and height 3 m. Concrete costs $120 per $\text{m}^{3}$ .

Find the cost of the concrete for the 4 pillars, to the nearest cent. Use the $\pi$ button.

ShowHide solution

The cost is per $\text{m}^{3}$ , so work in metres: $r = 40 \div 2 = 20$ cm $= 0.2$ m, $h = 3$ m.

Volume of one pillar $= \pi \times 0.2^{2} \times 3 = 0.12\pi \text{ m}^{3}$

Volume of 4 pillars $= 4 \times 0.12\pi = 0.48\pi \text{ m}^{3} = 1.507\,96\ldots \text{ m}^{3}$

Cost $= 1.507\,96\ldots \times 120 = 180.955\ldots$ , so the cost is $180.96.

The concrete costs $180.96.

Question 13

Ravi's trough has a semicircular cross-section of internal radius 14 cm. It holds 46.2 L of water when full.

Taking $\pi = \frac{22}{7}$ , find the internal length of the trough in metres.

ShowHide solution

Let the length be $x$ cm. 46.2 L $= 46.2 \times 1000 = 46\,200 \text{ cm}^{3}$ .

Area of semicircle × length = volume:

$\frac{1}{2}\pi r^{2} \times x = 46\,200$

$\frac{1}{2} \times \frac{22}{7} \times 14^{2} \times x = 46\,200$

$\frac{1}{2} \times \frac{22}{7} \times 196 \times x = 46\,200$

$308x = 46\,200$ (since $\frac{22}{7} \times 196 = 616$ , and half of 616 is 308)

$x = 46\,200 \div 308 = 150$

150 cm $= 1.5$ m.

The trough is 1.5 m long inside.

Key points

A cylinder has two identical, parallel circular ends and a uniform circular cross-section. It is not a prism.
Volume of a cylinder = area of cross-section × height $= \pi r^{2}h$ (base area × height).
A cylinder lying down has its height along the ground: its length.
Use the radius ( $r = d \div 2$ ), and have every length in one unit before you multiply.
Capacity uses the internal measurements: $1 \text{ cm}^{3} = 1$ mL, $1000 \text{ cm}^{3} = 1$ L, $1 \text{ m}^{3} = 1000$ L.
Water in a cylinder is a cylinder of the same radius, with the depth of the water as its height.
To find a height or a radius, substitute the numbers into $V = \pi r^{2}h$ and solve; the radius is the positive square root.
From a circumference, find $r$ first; from a base area, volume = base area × height.
Use the value of $\pi$ the question gives; round only at the end, to 3 significant figures; count whole things by sense (full glasses: round down).
Volume of a half-cylinder $= \frac{1}{2}\pi r^{2}h$ .