3D Compound Shapes (AI SL)
A compound solid is built from two or more simple 3D shapes joined together, or one shape cut out of another - a cone mounted on a cylinder, a hemisphere sitting on top of a cone, or a cone drilled out of a block. The skill isn't a new formula, it's spotting which simple solids make up the compound shape, then deciding whether to add or subtract their volumes. It's part of the broader Volume & Surface Area topic.
11 questions on this sub-topic.
The building-block formulas
Covered under IB syllabus reference SL3.1: volume of three-dimensional solids including right-pyramid, right cone, sphere, hemisphere and combinations of these solids. The pyramid, cone and cylinder volumes below are each printed in the formula booklet; the hemisphere isn't listed separately - halve the booklet's sphere volume \(\tfrac43\pi r^3\) instead.
Cone and cylinder
\(V_{\text{cone}} = \dfrac{1}{3}\pi r^2 h \qquad V_{\text{cylinder}} = \pi r^2 h\)
A cone has exactly a third of the volume of a cylinder with the same base radius and height - useful for a quick sanity check.
Pyramid and hemisphere
\(V_{\text{pyramid}} = \dfrac{1}{3} \times \text{base area} \times h \qquad V_{\text{hemisphere}} = \dfrac{2}{3}\pi r^3\)
A hemisphere is exactly half a sphere, so halve the booklet's sphere volume \(\tfrac43\pi r^3\) rather than re-deriving it from scratch.
Need the full syllabus wording and formula-booklet reference table? See Volume & Surface Area, or the calculator steps at using your GDC.
Worked examples
A solid cylinder of radius 5 cm and height 12 cm has a cone removed from it. The cone has the same base radius and height as the cylinder.
Find the remaining volume (3 significant figures).
Worked solution
\(V_C = \pi(5)^2(12) = 300\pi\). M1
\(V_{\text{cone}} = \tfrac13\pi(5)^2(12) = 100\pi\). M1
\(V = 300\pi - 100\pi = 200\pi = 628.3\ldots\) A1 \(\approx 628 \text{ cm}^3\). A1
A tent shape consists of a cuboid base of 6 m by 4 m by 2 m, with a square-based pyramid on top of height 1.5 m and the same 6 m by 4 m base.
Find the total volume.
Worked solution
\(V_1 = 6 \times 4 \times 2 = 48 \text{ m}^3\). M1
\(V_2 = \tfrac13 \times 6 \times 4 \times 1.5 = \tfrac13 \times 36 = 12 \text{ m}^3\). M1
\(V = 48 + 12\) A1 \(= 60 \text{ m}^3\). A1
Common mistakes
- Adding when you should subtract. "A cone is removed/drilled/cut from a cylinder" means the cone's volume comes out, not in - read the wording carefully before combining anything.
- Mixing up which radius or height belongs to which shape. When a cone sits on a cylinder, or is removed from one, each solid can have its own dimensions unless the question says they match - don't assume the smaller solid automatically shares the larger one's measurements.
- Forgetting the \(\tfrac13\) or \(\tfrac23\) factor. A cone is a third of the equivalent cylinder, a hemisphere is two-thirds of the equivalent sphere - dropping these factors is the single most common way marks are lost on this topic.
Ready to practise properly?
11 compound-shape questions, marked instantly like the real exam.
Quick answers
How do you find the volume of a compound 3D shape?
Split the solid into the simple shapes it is built from (cones, cylinders, pyramids, spheres or hemispheres), find each volume separately using the correct formula, then add the volumes together if the shapes are joined, or subtract if one has been removed from another.
How do you know whether to add or subtract volumes in a compound shape question?
Read the description carefully. If a shape is mounted on, attached to, or sits alongside another, add the volumes. If a shape has been removed, drilled out, or cut away from another, subtract the smaller volume from the larger one.