Volume of Solids (AI SL)
Every volume question on this sub-topic reduces to picking the right formula and substituting carefully - the challenge is usually spotting which solid (or which combination of solids) you're actually looking at. This page gathers the cone, sphere, pyramid and cylinder volume formulas in one place, with worked examples showing a single solid and a composite one. It's part of the broader Volume & Surface Area topic.
54 questions on this sub-topic.
The four 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 cone, sphere and pyramid formulas are in the formula booklet; the cylinder formula is prior learning you're expected to already know.
Cylinder
\(V=\pi r^2 h\)
Prior learning - not in the formula booklet, but assumed knowledge.
Right cone
\(V=\dfrac13\pi r^2 h\)
One third of the cylinder with the same base radius and height.
Sphere
\(V=\dfrac43\pi r^3\)
A hemisphere is half this: \(\tfrac23\pi r^3\).
Right pyramid
\(V=\dfrac13\times\text{base area}\times\text{height}\)
The height here is the perpendicular height, not a slant length.
Need the full syllabus wording and formula-booklet reference table? See Volume & Surface Area.
Worked examples
Find the volume of a cone with radius 5 cm and height 12 cm (3 significant figures).
Worked solution
The volume of a cone is \(V = \tfrac13 \pi r^2 h\): one third of the cylinder with the same base and height. M1
\(r=5,\ h=12\): \(V = \tfrac13 \pi (5)^2 (12) = \tfrac13 \pi (25)(12) = 100\pi.\) A1
\(100\pi = 314.159\ldots\), so \(V \approx 314 \text{ cm}^3\) (3 significant figures). A1
A frustum is formed by removing a small cone (radius 3 cm, height 4 cm) from a large cone (radius 9 cm, height 12 cm).
Find the volume of the frustum (3 significant figures).
Worked solution
\(V_L = \tfrac13\pi(9)^2(12) = \tfrac13\pi(81)(12) = 324\pi\). M1
\(V_S = \tfrac13\pi(3)^2(4) = \tfrac13\pi(9)(4) = 12\pi\). M1
\(V = 324\pi - 12\pi = 312\pi\). A1
\(312\pi = 980.2\ldots \approx 980 \text{ cm}^3\) (3 significant figures). A1
Common mistakes
- Using the full sphere formula for a hemisphere. A hemisphere's volume is \(\tfrac23\pi r^3\), not \(\tfrac43\pi r^3\) - forgetting to halve it is one of the most common errors on composite solids.
- Mixing up radius and diameter. Every formula on this page uses the radius \(r\) - if a question gives a diameter, halve it before you substitute anything.
- Subtracting the wrong way round on a frustum. A frustum's volume is the large solid minus the small solid removed from its tip - swap the order and you'll get a negative answer that should immediately flag the mistake.
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55 volume-of-solids questions, marked instantly like the real exam.
Quick answers
What is the formula for the volume of a cone?
\(V = \tfrac13 \pi r^2 h\), one third of the volume of a cylinder with the same base radius \(r\) and height \(h\).
How do you find the volume of a frustum?
Find the volume of the full (large) cone or pyramid, find the volume of the smaller cone or pyramid removed from its tip, then subtract the second from the first.
GDC steps for evaluating expressions like these live on the parent page's using your GDC section.