Volume of a hemisphere, sphere, cone and frustum: the formulas (⅔πr³, 4⁄3πr³, ⅓πr²h, ⅓πh(R² + Rr + r²)), where each comes from, the surface areas (including the hemisphere's 3πr² and the frustum's slant height), the composite solids exams love (a cone on a hemisphere, an ice-cream cone), working backwards to a radius, eight worked examples in terms of π and to 3 significant figures, the mistakes, and a printable formula sheet with practice

"Volume of a hemisphere" is searched about 1,200 times a month in the UK and "frustum" 1,700, by GCSE Higher students who have the sphere and cone formulas on the sheet and must adapt them, and the same four solids appear in the Australian Year 10 course, CBSE Class 9 and 10 (Surface Areas and Volumes) and US Geometry. This page gives the four volume formulas and where each comes from, the surface areas with their traps, the composite solids that exams build from them, the reverse problems, eight worked examples, the mistakes and a printable formula sheet with practice.

Download the volumes formula sheet with practice (PDF) (A4, the four solids with volume and surface area formulas and derivations on page 1, composite solids, reverse problems and the practice set with answers on page 2, free).

The four solids

The cylinder (πr²h, curved area 2πrh) is the reference the cone is a third of, and it turns up in the same composite questions.

Where each formula comes from

Sphere, 4⁄3πr³. Archimedes showed that a sphere has two thirds of the volume of the cylinder that just contains it (height 2r): ⅔ × πr² × 2r = 4⁄3πr³. He also showed the sphere's surface area equals the curved area of that cylinder, 2πr × 2r = 4πr². He was proud enough to have the sphere and cylinder carved on his tomb.

Hemisphere, ⅔πr³. Half a sphere. The flat face is a circle of radius r, area πr², which is why a solid hemisphere's surface is 2πr² + πr² = 3πr².

Cone, ⅓πr²h. One third of the cylinder with the same base and height (pour three cones of water into the cylinder). The slant height l is the hypotenuse of the right triangle with legs r and h: l² = r² + h². Curved area πrl comes from unrolling the curved surface into a sector of radius l and arc 2πr: area ½ × l × 2πr = πrl.

Frustum, ⅓πh(R² + Rr + r²). The full cone minus the small cone cut off. If the small cone has height x, similar triangles give x ÷ r = (x + h) ÷ R, so x = hr ÷ (R − r). Then V = ⅓πR²(x + h) − ⅓πr²x, which simplifies to the formula. In an exam you may use either route; show the similar-triangles step if you subtract cones.

Worked examples

1. Sphere, radius 6 cm. V = 4⁄3 × π × 216 = 288π = 905 cm³. Surface 4π × 36 = 144π = 452 cm².
2. Hemisphere, radius 6 cm, solid. V = ⅔ × π × 216 = 144π = 452 cm³. Total surface 3π × 36 = 108π = 339 cm² (curved 72π, flat 36π).
3. Cone, radius 5 cm, height 12 cm. V = ⅓ × π × 25 × 12 = 100π = 314 cm³. Slant l = √(25 + 144) = 13. Curved area π × 5 × 13 = 65π = 204 cm²; total 65π + 25π = 90π = 283 cm².
4. Frustum (a bucket), radii 15 cm and 10 cm, height 24 cm. V = ⅓ × π × 24 × (225 + 150 + 100) = 8π × 475 = 3,800π = 11,900 cm³ = 11.9 litres. Slant l = √(24² + 5²) = √601 = 24.5; curved area π × 25 × 24.5 = 1,925 cm².
5. Frustum by subtraction, same bucket. Small cone height x: x ÷ 10 = (x + 24) ÷ 15 → 15x = 10x + 240 → x = 48. Big cone: ⅓π × 225 × 72 = 5,400π. Small cone: ⅓π × 100 × 48 = 1,600π. Difference 3,800π. Same answer.
6. Hemisphere on a cylinder (a silo), radius 3 m, cylinder height 8 m. V = π × 9 × 8 + ⅔ × π × 27 = 72π + 18π = 90π = 283 m³.
7. Cone on a hemisphere (an ice cream), radius 3 cm, cone height 10 cm. V = ⅓ × π × 9 × 10 + ⅔ × π × 27 = 30π + 18π = 48π = 151 cm³.
8. Reverse: a sphere has volume 36π cm³. r³ = 3 × 36π ÷ (4π) = 27, r = 3 cm. A hemisphere has volume 18π: r³ = 3 × 18π ÷ (2π) = 27, r = 3 as well.

Give exact answers in terms of π where asked, and otherwise to 3 significant figures, keeping π in until the last step.

Composite solids: the exam pattern

Most Higher-tier questions glue two of these together and ask for the total volume or surface area, or give the total and ask for a dimension.

• Volume: add the parts (or subtract a hollow).
• Surface area: add only the exposed faces. A hemisphere sitting on a cylinder hides both the hemisphere's flat face and the cylinder's top: total = 2πr² (dome) + 2πrh (side) + πr² (base). A cone on a hemisphere hides both circles: total = πrl + 2πr².
• Similar solids: if lengths scale by k, areas scale by k² and volumes by k³. A frustum question that gives the two radii is a similar-triangles question in disguise.
• Units: 1 litre = 1,000 cm³; 1 m³ = 1,000 litres; a bucket of 11,900 cm³ holds 11.9 litres.

Which formulas you are given

Recent GCSE formula sheets (England) give the sphere's volume and surface area and the cone's volume and curved surface area; they do not give the hemisphere, the frustum, or the cone's slant height, which you derive. Check your exam year's arrangements. CBSE Class 10 expects every formula from memory. US Geometry tests usually provide a reference sheet with the sphere and cone.

The mistakes that cost marks

• Using the diameter as r (a 12 cm ball has r = 6).
• Using the hemisphere as if it were a sphere, or forgetting to add the flat face for a solid hemisphere.
• Using the slant height as the height in ⅓πr²h, or the height as the slant in πrl.
• Cubing πr instead of r (πr³, not (πr)³).
• Frustum: using the small cone's height for h, or the full cone's. The frustum formula wants the frustum's own height.
• Adding the hidden faces in a composite surface area.
• Taking a square root instead of a cube root when finding r from a volume.
• Rounding π early.

Where it sits in each curriculum

Practice (answers in the PDF)

1. Find the volume and total surface area of a solid hemisphere of radius 9 cm, in terms of π.
2. A cone has radius 8 cm and height 15 cm. Find its volume and its total surface area to 3 s.f.
3. A frustum has radii 12 cm and 6 cm and height 8 cm. Find its volume. Then find the height of the small cone that was removed.
4. A toy is a cone of height 7 cm on a hemisphere of radius 3 cm. Find its total volume to 3 s.f.
5. A sphere has volume 972π cm³. Find its radius.
6. A hemispherical bowl of inner radius 10 cm is full of water, which is poured into a cylinder of radius 5 cm. How high does the water reach?

Download the formula sheet and the practice set (PDF)

Where to get help

A 1-on-1 online maths class in UK and Australian after-school hours with a tutor who holds a master's in mathematics and teaches GCSE Higher and A-level, the Australian Years 9 to 10A and CBSE Class 9 and 10 every week: solids taught with the derivations drawn and every composite question started by naming the exposed faces. The GCSE maths page explains how the classes run; the free 30-minute first class is a real lesson on this week's homework. Related pages: area of a sector and arc length and the trapezium.

Questions parents ask

See how we teach this, 1-on-1 online →