Volume of a Cylinder
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Theory
A cylinder is a prism with a circle for its cross-section, so its volume is \(V = \pi r^2 h\). This Year 9 maths page covers exact and rounded answers, capacity in mL, L and kL, half-cylinders and pipes, and working backwards to a height or radius.
A cylinder has two identical, parallel circular ends. Its cross-section is a circle of radius \(r\), the same all the way through, and its height (or length) \(h\) is the distance between the ends.
Like any prism, its volume is the area of the cross-section times the height: \(\pi r^2 \times h\). Volume is measured in cubic units, such as \(\text{cm}^3\) or \(\text{m}^3\).
The capacity of a container is how much it holds: \(1 \text{ cm}^3\) holds \(1\) mL, \(1000 \text{ cm}^3\) holds \(1\) L, and \(1 \text{ m}^3\) holds \(1000\) L, which is \(1\) kilolitre (kL).
A half-cylinder has a semicircle as its cross-section, so its volume is half of \(\pi r^2 h\). A pipe is a big cylinder with a smaller one removed from its middle.
An exact answer leaves \(\pi\) in, such as \(144\pi \text{ cm}^3\).
| Solid | Volume |
|---|---|
| Cylinder | \(\pi r^2 h\) |
| Half-cylinder | \(\dfrac{1}{2}\pi r^2 h\) |
| Pipe (outer radius \(R\), inner radius \(r\)) | \(\pi R^2 h - \pi r^2 h\) |
Working backwards:
| Volume | Capacity |
|---|---|
| \(1 \text{ cm}^3\) | \(1\) mL |
| \(1000 \text{ cm}^3\) | \(1\) L |
| \(1 \text{ m}^3\) | \(1000\) L \(= 1\) kL |
How to find the volume of a cylinder
- Find the radius. If you are given the diameter, halve it. Put all lengths in the same unit.
- Find the area of the circular cross-section, \(\pi r^2\). Use half of it for a half-cylinder.
- Multiply by the height or length.
- Finish: leave \(\pi\) in for an exact answer, or round once at the end. Convert to mL, L or kL if asked.
Working backwards
- Change a capacity to a volume first: \(500\) mL is \(500 \text{ cm}^3\).
- Substitute into \(V = \pi r^2 h\) and solve. For the height, divide by \(\pi r^2\). For the radius, divide by \(\pi h\), then take the square root.
(i) Find its exact volume.
(ii) Find its volume correct to two decimal places.
(iii) How many millilitres does it hold, to the nearest millilitre?
The radius is half the diameter: \(r = 8 \div 2 = 4\) cm.
(i) The cross-section is a circle, so \(A = \pi r^2\). Leave \(\pi\) in the answer:
| \(V\) | \(=\) | \(\pi r^2 h\) |
| \(\) | \(=\) | \(\pi \times 4^2 \times 9\) |
| \(\) | \(=\) | \(\pi \times 16 \times 9\) |
| \(\) | \(=\) | \(\mathbf{144\pi} \textbf{ cm}^\mathbf{3}\) |
(ii) Now use the \(\pi\) key:
| \(V\) | \(=\) | \(452.389\ldots\) |
| \(\) | \(\approx\) | \(\mathbf{452.39} \textbf{ cm}^\mathbf{3}\) |
(iii) \(1 \text{ cm}^3\) holds \(1\) mL, so the jar holds about 452 mL.
The cross-section is a semicircle, so take half of \(\pi r^2\), then multiply by the length.
| \(V\) | \(=\) | \(\dfrac{1}{2} \times \pi \times 2^2 \times 10\) |
| \(\) | \(=\) | \(20\pi = 62.831\ldots\) |
| \(\) | \(\approx\) | \(\mathbf{62.83} \textbf{ m}^\mathbf{3}\) |
\(1 \text{ m}^3 = 1\) kL, so the greenhouse holds about \(62.83\) kL of air.
The plastic is the big cylinder minus the hole. Both are \(50\) cm long.
| \(V\) | \(=\) | \(\pi \times 6^2 \times 50 - \pi \times 5^2 \times 50\) |
| \(\) | \(=\) | \(1800\pi - 1250\pi\) |
| \(\) | \(=\) | \(\mathbf{550\pi} \textbf{ cm}^\mathbf{3}\) |
| \(\) | \(=\) | \(1727.875\ldots\) |
| \(\) | \(\approx\) | \(\mathbf{1727.88} \textbf{ cm}^\mathbf{3}\) |
Subtract the areas first if you like: \(\pi(36 - 25) = 11\pi\), then \(11\pi \times 50 = 550\pi\).
(ii) A can holds \(500\) mL and has a radius of \(4\) cm. Find its height, correct to two decimal places.
(i) Substitute, then divide both sides by \(9\pi\). The \(\pi\) cancels.
| \(\pi \times 3^2 \times h\) | \(=\) | \(72\pi\) |
| \(9\pi h\) | \(=\) | \(72\pi\) |
| \(h\) | \(=\) | \(\mathbf{8} \textbf{ cm}\) |
(ii) \(500\) mL is \(500 \text{ cm}^3\). Substitute and divide by \(\pi r^2 = 16\pi\).
| \(16\pi h\) | \(=\) | \(500\) |
| \(h\) | \(=\) | \(\dfrac{500}{16\pi} = \dfrac{500}{50.265\ldots}\) |
| \(\) | \(=\) | \(9.947\ldots\) |
| \(\) | \(\approx\) | \(\mathbf{9.95} \textbf{ cm}\) |
Check (ii): \(\pi \times 4^2 \times 9.947\ldots = 500\) \(\checkmark\)
Common pitfalls
Frequently asked questions
What is the formula for the volume of a cylinder?
\(V = \pi r^2 h\): the area of the circular end times the height.
How do you find the volume of a cylinder from the diameter?
Halve the diameter to get the radius, then use \(V = \pi r^2 h\). A jar with diameter \(8\) cm and height \(9\) cm has \(V = 144\pi \approx 452.39 \text{ cm}^3\).
How do you convert the volume of a cylinder to litres?
\(1000 \text{ cm}^3 = 1\) L and \(1 \text{ m}^3 = 1000\) L. Find the volume, then divide cubic centimetres by \(1000\), or multiply cubic metres by \(1000\).
How do you find the height of a cylinder from its volume?
Divide the volume by the area of the end: \(h = \dfrac{V}{\pi r^2}\). A \(500\) mL can of radius \(4\) cm is about \(9.95\) cm tall.
What happens to the volume of a cylinder if you triple the radius?
The volume becomes nine times as big, because the radius is squared in \(\pi r^2 h\).