Volumes of Prisms and Cylinders
The volume of a prism or cylinder is the area of its cross-section times its length: \(V = Ah\), which becomes \(V = \pi r^2h\) for a cylinder. This NSW Year 9 Advanced maths subtopic finds volumes and capacities of prisms, cylinders and composite solids (NESA outcome MA5‑VOL‑C‑01).
You will find volumes with trapezium, triangle and quarter-circle cross-sections, change volume to capacity, and find a missing height. You will find the volume of a pipe, a block with a hole drilled through it and a shed with a curved roof, how long a rainwater tank takes to fill, and how far water rises when a cube is dropped into a glass. It is part of Year 9 Advanced Surface Area and Volume.
Theory
The volume of any right prism is the area of its cross-section times its length: \(V = Ah\). A cylinder is a prism with a circular cross-section, so \(V = \pi r^2 h\). This Year 9 Advanced maths page covers composite and hollow solids, capacity, displaced water and working backwards to a missing length.
Volume is the amount of space a solid takes up, measured in cubic units such as \(\text{cm}^3\) or \(\text{m}^3\). Capacity is how much a container holds, measured in mL, L or kL.
A right prism has the same cross-section all the way along, at right angles to its length. Stacking that cross-section along the length gives \(V = Ah\), where \(A\) is the cross-section area and \(h\) is the length (or height) of the prism.
The cross-section can be any shape: a triangle, trapezium, semicircle, quadrant, or a composite shape. Find its area with the right formula first.
A cylinder has a circular cross-section, so \(V = \pi r^2 h\). A composite solid is found by adding the volumes of its parts, or by subtracting a hole or cut-out from a larger solid.
Volume and capacity link through \(1~\text{cm}^3 = 1~\text{mL}\) and \(1~\text{m}^3 = 1000~\text{L}\).
How to find the volume of a prism or cylinder
- Identify the cross-section: the face that repeats along the solid.
- Find its area, using Pythagoras' theorem for a missing height if needed, and halving any diameter.
- Multiply by the length of the prism, in the same units.
- Convert to capacity if asked: \(\text{cm}^3 \to\) mL, \(\text{m}^3 \to\) kL (\(\times 1000\) for L).
Harder problems
- Composite solids: add the parts (a box with a half-cylinder roof) or subtract the hole (a block with a cylinder drilled through). The hole's height is the thickness it passes through.
- Working backwards: substitute the volume and solve. If \(V\) is in terms of \(\pi\), divide \(\pi\) out. If \(r\) appears three times, take a cube root.
- Displacement: a sunken object pushes up a layer of water with the same volume: base area of the container \(\times\) rise.
- Filling time: capacity \(\div\) flow rate, then convert minutes to hours and minutes.
- Comparisons: write both volumes with the same letters and divide, so the letters cancel and leave a number or a percentage.
The height of the triangle splits it into two right-angled triangles with base \(4\) m and hypotenuse \(5\) m.
| \(h\) | \(=\) | \(\sqrt{5^2 - 4^2} = 3~\text{m}\) |
| \(A\) | \(=\) | \(\dfrac{1}{2} \times 8 \times 3 = 12~\text{m}^2\) |
| \(V\) | \(=\) | \(Ah = 12 \times 12 = \mathbf{144}~\text{m}^3\) |
Two different heights appear: the triangle's height (\(3\) m) is used for its area; the prism's length (\(12\) m) multiplies that area.
(i) Find the volume of concrete, exactly and correct to \(3\) decimal places.
(ii) Concrete has a mass of \(2400\) kg per cubic metre. Find the mass of the pipe.
(i) Work in metres: \(R = 0.25\), \(r = 0.2\). Volume \(=\) outer cylinder \(-\) hole.
| \(V\) | \(=\) | \(\pi(0.25)^2(3) - \pi(0.2)^2(3)\) |
| \(\) | \(=\) | \(0.1875\pi - 0.12\pi\) |
| \(\) | \(=\) | \(\boldsymbol{0.0675\pi} \approx \mathbf{0.212}~\text{m}^3\) |
(ii)
| \(m\) | \(=\) | \(0.0675\pi \times 2400 = 508.938\ldots \approx \mathbf{508.94}~\text{kg}\) |
Subtract the two volumes (or the two squares), never the two radii first.
(i) A stone placed in the tank sinks and the water level rises by \(1.2\) cm. Find the volume of the stone.
(ii) The stone is replaced by a solid metal cylinder of radius \(3\) cm and height \(10\) cm, which is completely covered. By how much does the water rise, exactly and correct to \(2\) decimal places?
The water pushed up is a thin rectangular prism with the same volume as the object.
| \(V_{\text{stone}}\) | \(=\) | \(50 \times 30 \times 1.2 = \mathbf{1800}~\text{cm}^3\) |
(ii) Let the rise be \(d\) cm. The water layer has volume \(1500d\).
| \(1500d\) | \(=\) | \(\pi \times 3^2 \times 10 = 90\pi\) |
| \(d\) | \(=\) | \(\dfrac{90\pi}{1500} = \boldsymbol{0.06\pi} \approx \mathbf{0.19}~\text{cm}\) |
Write \(h = 2r\) and convert the capacity: \(2000~\text{L} = 2~\text{m}^3\).
| \(V\) | \(=\) | \(\pi r^2 (2r) = 2\pi r^3\) |
| \(2\pi r^3\) | \(=\) | \(2\) |
| \(r^3\) | \(=\) | \(\dfrac{1}{\pi} = 0.3183\ldots\) |
| \(r\) | \(=\) | \(\sqrt[3]{0.3183\ldots} = 0.6827\ldots~\text{m}\) |
| \(\) | \(\approx\) | \(\mathbf{68.3}~\text{cm}\) |
\(r\) appears three times (twice in \(r^2\), once in the height), so a cube root undoes it. Check: \(2\pi(0.6827\ldots)^3 = 2.000\) \(\checkmark\)
Common pitfalls
Frequently asked questions
What is the formula for the volume of a prism?
Volume \(=\) area of the cross-section \(\times\) length: \(V = Ah\). It works for any right prism.
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 convert volume to litres?
One cubic centimetre holds one millilitre, and one cubic metre holds \(1000\) litres. Find the volume, then convert.
How do you find the height of a cylinder from its volume?
Divide the volume by the area of the base: \(h = \dfrac{V}{\pi r^2}\).
How do you find the volume of a hollow pipe?
Find the volume of the outer cylinder and subtract the volume of the hole: \(\pi(R^2 - r^2)h\).
How do you find the volume of an object by putting it in water?
Measure how far the water level rises. The object's volume equals the base area of the container times the rise.