Volume of Prisms and Cylinders
The volume of any prism or cylinder is the area of its cross-section times its length, \(V=Ah\), so a cylinder has volume \(V=\pi r^2h\). In NSW Year 10 Advanced maths the cross-sections include rhombuses, trapeziums, sectors and semicircles, and answers are often changed to litres, kilolitres or millilitres (NESA outcome MA5‑VOL‑C‑01).
You will find volumes of prisms, half-cylinders and cylinders, and work backwards to a radius or a missing length. Practical problems include filling a pool and a cake tin, the water in a garden hose, the concrete in a wheelchair ramp, the steel in a pipe, rain running off a roof into a tank and comparing two farm tanks. You will also find the percentage change in volume when both the radius and the height of a cylinder change. It is part of Year 10 Advanced Measurement.
Theory
The volume of a prism or cylinder is the area of its cross-section times its length: \(V = Ah\), and for a cylinder \(V = \pi r^2 h\). This Year 10 Advanced maths page covers cross-sections of every shape, capacity in litres and kilolitres, finding a missing length, pouring between containers, rates and percentage changes.
A prism has the same cross-section all the way through. A cylinder is like a prism with a circular cross-section.
The volume is the space inside, measured in cubic units (mm\(^3\), cm\(^3\), m\(^3\)). For both solids, volume = area of the cross-section \(\times\) the distance through it.
Capacity is the amount a container holds, in mL, L or kL. \(1\text{ cm}^3\) holds \(1\) mL, and \(1\text{ m}^3\) holds \(1000\) L, which is \(1\) kL.
The "height" \(h\) in \(V = Ah\) is the length at right angles to the cross-section, so a prism lying on its side still uses its length. Find the cross-section first: it may be a triangle, trapezium, sector, a ring (for a pipe), or a shape with a piece cut out.
Common cross-sections:
| Cross-section | Area \(A\) |
|---|---|
| Triangle | \(\dfrac{1}{2}bh\) |
| Trapezium | \(\dfrac{1}{2}(a + b)h\) |
| Rhombus or kite (diagonals \(x\), \(y\)) | \(\dfrac{1}{2}xy\) |
| Circle / semicircle | \(\pi r^2\) / \(\dfrac{1}{2}\pi r^2\) |
| Sector with angle \(\theta\) | \(\dfrac{\theta}{360^\circ} \times \pi r^2\) |
| Ring (outer \(R\), inner \(r\)) | \(\pi(R^2 - r^2)\) |
How to find a volume
- Identify the cross-section and the length through it.
- Convert all lengths to one unit (often cm for litres, m for kilolitres).
- Find the area of the cross-section, using Pythagoras' theorem or a formula from the table.
- Multiply by the length, then convert to a capacity if asked.
Other questions
- Missing length: substitute the volume and solve. For a radius, \(r = \sqrt{\dfrac{V}{\pi h}}\). If the volume is in litres, change it to cm\(^3\) first.
- Radius from a circumference: \(r = \dfrac{C}{2\pi}\); keep the full value.
- Pipes and tubes: use the ring \(\pi(R^2 - r^2)\) as the cross-section. Mass = volume \(\times\) density.
- Pools with a sloping floor: the side view is a trapezium; it is the cross-section and the width is the length.
- Partly full: use the depth of the liquid, not the depth of the container. In a sloping-sided trough the width of the water changes evenly with depth.
- Rates: time = volume \(\div\) rate. If water flows in and out at once, use the net rate.
- Unknown in an expression: form the volume equation, rearrange to a quadratic equal to zero and keep the positive solution.
- Best buy: compare litres per dollar (or dollars per litre).
The height of the triangle splits the base in half, making a right-angled triangle with sides \(1.5\), \(h\) and \(2.5\):
| \(h^2\) | \(=\) | \(2.5^2 - 1.5^2 = 4\) |
| \(h\) | \(=\) | \(2\text{ m}\) |
| \(A\) | \(=\) | \(\dfrac{1}{2} \times 3 \times 2 = 3\text{ m}^2\) |
| \(V\) | \(=\) | \(Ah = 3 \times 4\) |
| \(\) | \(=\) | \(\mathbf{12}\text{ m}^3\) |
(i) Find its capacity, to the nearest litre.
(ii) How many litres are in it when it is \(65\%\) full?
(i) The radius is \(1.6\) m. Find the volume in cubic metres, then use \(1\text{ m}^3 = 1000\) L.
| \(V\) | \(=\) | \(\pi r^2 h = \pi \times 1.6^2 \times 2.1\) |
| \(\) | \(=\) | \(5.376\pi\) |
| \(\) | \(=\) | \(16.8892\ldots\text{ m}^3\) |
| \(\) | \(=\) | \(16\,889.2\ldots\text{ L}\) |
| \(\) | \(\approx\) | \(\mathbf{16\,889}\text{ L}\) |
(ii)
| \(0.65 \times 16\,889.2\ldots\) | \(=\) | \(10\,977.9\ldots\) |
| \(\) | \(\approx\) | \(\mathbf{10\,978}\text{ L}\) |
Use the full value of the capacity in part (ii), not the rounded one.
The volume of water does not change. Find it in the tank, then set it equal to \(\pi r^2 d\) for the drum.
| \(V\) | \(=\) | \(80 \times 50 \times 40 = 160\,000\text{ cm}^3\) |
| \(\pi \times 25^2 \times d\) | \(=\) | \(160\,000\) |
| \(d\) | \(=\) | \(\dfrac{160\,000}{625\pi}\) |
| \(\) | \(=\) | \(81.4873\ldots\) |
| \(\) | \(\approx\) | \(\mathbf{81.5}\text{ cm}\) |
The same idea finds the rise in a tank when rain falls on a roof: volume of rain on the roof = volume added to the tank.
Call the original volume \(V = \pi r^2 h\). The new radius is \(1.1r\) and the new height is \(0.75h\).
| \(V_{\text{new}}\) | \(=\) | \(\pi(1.1r)^2(0.75h)\) |
| \(\) | \(=\) | \(1.21 \times 0.75 \times \pi r^2 h\) |
| \(\) | \(=\) | \(0.9075\,V\) |
\(0.9075 \lt 1\), so the volume goes down, by \(1 - 0.9075 = 0.0925\):
| \(\text{decrease}\) | \(=\) | \(\mathbf{9.25\%}\) |
The radius is squared, so a \(10\%\) increase in \(r\) multiplies the volume by \(1.1^2 = 1.21\), not \(1.1\).
Common pitfalls
Frequently asked questions
What is the formula for the volume of a prism?
\(V = Ah\), where \(A\) is the area of the cross-section and \(h\) is the length of the prism measured at right angles to it. It works for any shape of cross-section.
How do you find the volume of a cylinder?
Multiply the area of the circle by the height: \(V = \pi r^2 h\). Use the radius, not the diameter.
How many litres are in a cubic metre?
One cubic metre holds 1000 litres, which is one kilolitre. One cubic centimetre holds one millilitre, so 1000 cubic centimetres is one litre.
How do you find the radius of a cylinder from its volume?
Divide the volume by \(\pi h\), then take the square root: \(r = \sqrt{\dfrac{V}{\pi h}}\). Make sure the volume and the height are in matching units first.
What happens to the volume if you double the radius of a cylinder?
The volume is multiplied by 4, because the radius is squared. Doubling the height only doubles the volume.