Volume of Composite Solids
The volume of a composite solid is found by splitting it into simpler solids, finding each volume and then adding them, or subtracting a part that has been cut out. In NSW Year 10 Advanced maths the parts include prisms, cylinders, pyramids, cones and hemispheres, as in a medicine capsule or an ice-cream cone topped with a scoop (NESA outcome MA5‑VOL‑P‑01).
You will find volumes of solids such as a shed with a gable roof, a capsule, a bollard and a rainwater tank with a cone-shaped base, and of solids with a cone or hemisphere carved out. You will find the volume of a frustum by completing the pyramid or cone, the empty space around three tennis balls in a can, and how many small spheres a melted ornament makes. Harder questions solve for an unknown height or radius. It is part of Year 10 Advanced Measurement.
Every question with a fully worked solution.
- Volume of Composite Solids Watch
Theory
The volume of a composite solid is found by adding the volumes of the parts joined together and subtracting any part cut out. This Year 10 Advanced maths page covers cylinders, cones, pyramids, spheres and hemispheres together, frustums, holes and hollows, melting and recasting, and finding an unknown length.
A composite solid is made from simple solids: prisms, cylinders, pyramids, cones, spheres and hemispheres.
If the parts are joined, add their volumes. If one shape is cut out of another (a hole, a bowl carved into the top, a hollow), subtract it.
A hemisphere is half a sphere, so its volume is \(\dfrac{2}{3}\pi r^3\). Two equal hemispheres on the ends of a cylinder make one whole sphere.
A frustum is a cone or pyramid with its top cut off parallel to the base. Its volume is the full solid minus the small one removed. The small solid is similar to the full one, so lengths are in the same ratio.
When a solid is melted and recast, or a liquid is poured into another container, the volume stays the same.
Volumes you will combine:
| Solid | Volume |
|---|---|
| Prism / cylinder | \(Ah\) / \(\pi r^2 h\) |
| Pyramid / cone | \(\dfrac{1}{3}Ah\) / \(\dfrac{1}{3}\pi r^2 h\) |
| Sphere | \(\dfrac{4}{3}\pi r^3\) |
| Hemisphere | \(\dfrac{2}{3}\pi r^3\) |
| Half-cylinder | \(\dfrac{1}{2}\pi r^2 h\) |
How to find the volume of a composite solid
- Split the solid into simple parts. Decide which are added and which are cut out.
- Find the measurements of each part. Shared radii and heights are often hidden in the diagram.
- Calculate each volume, keeping \(\pi\) exact.
- Combine and round once at the end; convert to litres if asked.
Other questions
- Objects in a container (balls in a can): the can's height and radius come from the objects; empty space = container \(-\) objects.
- Melt and recast: total volume \(\div\) volume of one new object, rounded down. For one new sphere, solve \(\dfrac{4}{3}\pi r^3 = V\) with a cube root.
- Unknown height or radius: write each part in terms of the unknown, set the total equal to the volume and solve.
- Two expressions equal: set the volumes equal and divide both sides by the common factor, such as \(\dfrac{4}{3}\pi r^2\).
- Buying material: round up to the amount you can buy (for example, multiples of \(0.2\text{ m}^3\)), then find the cost.
- Rain filling a tank: depth of rain \(\times\) roof area = volume of the tank.
A hemisphere is half a sphere, so its volume is \(\dfrac{1}{2} \times \dfrac{4}{3}\pi r^3 = \dfrac{2}{3}\pi r^3\).
| \(V_{\text{cyl}}\) | \(=\) | \(\pi \times 2.4^2 \times 5 = 28.8\pi\) |
| \(V_{\text{hemi}}\) | \(=\) | \(\dfrac{2}{3}\pi \times 2.4^3 = 9.216\pi\) |
| \(V\) | \(=\) | \(28.8\pi + 9.216\pi\) |
| \(\) | \(=\) | \(\mathbf{38.016\pi}\text{ m}^3\) |
| \(\) | \(=\) | \(119.4307\ldots\) |
| \(\) | \(\approx\) | \(\mathbf{119.43}\text{ m}^3\) |
The hole is a cylinder as tall as the cube. Subtract it:
| \(V\) | \(=\) | \(10^3 - \pi \times 2^2 \times 10\) |
| \(\) | \(=\) | \(1000 - 40\pi\) |
| \(\) | \(=\) | \(874.3362\ldots\) |
| \(\) | \(\approx\) | \(\mathbf{874.34}\text{ cm}^3\) |
Complete the cone below the base. Let the missing small cone have height \(x\). The full cone has height \(x + 16\), and the two cones are similar, so radius \(\div\) height is the same:
| \(\dfrac{9}{x}\) | \(=\) | \(\dfrac{15}{x + 16}\) |
| \(9x + 144\) | \(=\) | \(15x\) |
| \(x\) | \(=\) | \(24\text{ cm}\) |
Full cone (\(r = 15\), \(h = 40\)) minus small cone (\(r = 9\), \(h = 24\)):
| \(V\) | \(=\) | \(\dfrac{1}{3}\pi(15^2)(40) - \dfrac{1}{3}\pi(9^2)(24)\) |
| \(\) | \(=\) | \(3000\pi - 648\pi = 2352\pi\) |
| \(\) | \(=\) | \(7389.0259\ldots\text{ cm}^3\) |
| \(\) | \(\approx\) | \(\mathbf{7.39}\text{ L}\) |
\(1000\text{ cm}^3 = 1\) L.
The volume of metal stays the same.
| \(V_{\text{cyl}}\) | \(=\) | \(\pi \times 4^2 \times 9 = 144\pi\) |
| \(\dfrac{4}{3}\pi r^3\) | \(=\) | \(144\pi\) |
| \(r^3\) | \(=\) | \(144 \times \dfrac{3}{4} = 108\) |
| \(r\) | \(=\) | \(\sqrt[3]{108} = 4.7622\ldots\) |
| \(\) | \(\approx\) | \(\mathbf{4.76}\text{ cm}\) |
If many small objects are made, divide the total volume by the volume of one and round down to whole objects.
Common pitfalls
Frequently asked questions
How do you find the volume of a composite solid?
Split it into simple solids, find the volume of each, then add the parts that are joined together and subtract any part that is cut out.
What is the volume of a hemisphere?
Half the volume of a sphere: \(\dfrac{2}{3}\pi r^3\).
How do you find the volume of a frustum?
Complete the cone or pyramid. Use similar triangles to find the height of the small piece that was cut off, then subtract its volume from the volume of the full solid.
What happens to the volume when a solid is melted and recast?
It stays the same. Set the original volume equal to the volume of the new shape, or divide by the volume of one small object to find how many can be made.
How do you find the radius of a sphere from its volume?
Solve \(\dfrac{4}{3}\pi r^3 = V\): \(r = \sqrt[3]{\dfrac{3V}{4\pi}}\).