OCR GCSE Mathematics: Surface area and volume of prisms, cylinders, cones, spheres and pyramids (J560)

This guide is written for the OCR GCSE (9-1) Mathematics J560 specification, Version 2.1 (August 2026), and covers section 10.04, volume and surface area calculations (10.04a to 10.04c). All of it is assessed on both the Foundation tier and the Higher tier. There is no Higher tier only content in this unit.

For condensed recall use the revision notes, and for exam-style questions use the practice set. The course hub is at OCR GCSE Mathematics, the printable list is the OCR GCSE Mathematics checklist, and you can test yourself with the free 10-minute diagnostics.

What this unit covers

Ref	What you must be able to do	Tier
10.04a	Calculate the surface area and volume of cuboids and other right prisms, including cylinders	Both tiers
10.04b	Calculate the surface area and volume of spheres, cones and simple composite solids (formulae will be given)	Both tiers
10.04c	Calculate the surface area and volume of a pyramid (the formula 1/3 × area of base × height will be given)	Both tiers

The specification says there is no formula sheet. The formulae it says you must know are listed under 6.02d, and they include the circumference and area of a circle. For spheres, cones, composite solids and pyramids it says the formulae will be given in the assessment. Learn the cuboid, prism and cylinder results below as well: they come straight from the circle area formula you must know.

Papers 2 and 5 do not permit a calculator, so you should be able to leave an answer in terms of π and to work out a slant height by hand.

Volume, surface area and units

Volume is the space inside a solid. It is measured in cubic units: mm³, cm³, m³. Surface area is the total area of all the faces, including curved ones. It is measured in square units.

Useful conversions:

From	To
1 cm³	1 millilitre
1000 cm³	1 litre
1 m³	1 000 000 cm³ (not 100 cm³)
1 m²	10 000 cm²

Always work in one unit before you calculate. Convert lengths first, then find the volume.

10.04a Cuboids and right prisms

A prism keeps one cross-section along its whole length. In a right prism, that length is at right angles to the cross-section.

volume of a prism = area of cross-section × length
surface area = 2 × cross-section area + (perimeter of cross-section × length)

For a cuboid, l × w × h, the volume is lwh and the surface area is 2(lw + lh + wh).

Worked example 1: a cuboid. A cuboid measures 11 cm by 6 cm by 4.5 cm. Find its volume and surface area.

Volume = 11 × 6 × 4.5 = 297 cm³
Faces: 11 × 6 = 66, 11 × 4.5 = 49.5, 6 × 4.5 = 27
Surface area = 2 × (66 + 49.5 + 27) = 2 × 142.5 = 285 cm²

Worked example 2: a triangular prism. The cross-section is a right-angled triangle with shorter sides 9 cm and 12 cm. The prism is 20 cm long. Find its volume and total surface area.

Hypotenuse = √(9² + 12²) = √225 = 15 cm
Cross-section area = ½ × 9 × 12 = 54 cm²
Volume = 54 × 20 = 1080 cm³
Perimeter of cross-section = 9 + 12 + 15 = 36 cm
Rectangular faces: 36 × 20 = 720 cm²
Surface area = 2 × 54 + 720 = 828 cm²

Notice that the hypotenuse was needed for the third rectangle. Missing it is the usual way to lose marks on a triangular prism.

10.04a Cylinders

A cylinder is a prism with a circular cross-section, so the same rules apply.

V = πr²h
Curved surface area = 2πrh        (the rolled-out rectangle: width 2πr, height h)
Total surface area = 2πr² + 2πrh

Worked example 3. A solid cylinder has radius 4 cm and height 11 cm. Give its volume and total surface area in terms of π, then to 1 decimal place.

V = π × 4² × 11 = 176π = 552.9 cm³ (1 d.p.)
Ends: 2 × π × 4² = 32π
Curved: 2 × π × 4 × 11 = 88π
Total surface area = 120π = 377.0 cm² (1 d.p.)

If the cylinder is open at the top, like a bucket, there is only one circular end: π × 16 + 88π = 104π = 326.7 cm².

Worked example 4: a hollow tube. A pipe has outer radius 5 cm, inner radius 4 cm and length 30 cm. The material is the difference of two cylinders.

V = π × 5² × 30 − π × 4² × 30 = π × (25 − 16) × 30 = 270π ≈ 848.2 cm³

Working backwards. If you know the volume, rearrange. For a cylinder of volume 250π cm³ and radius 5 cm: 250π = π × 25 × h, so h = 10 cm.

10.04b Spheres and hemispheres

These are the formulae you will normally be given.

Sphere: V = (4/3)πr³      surface area = 4πr²

A hemisphere is half a sphere. Its curved surface is 2πr². A solid hemisphere also has a flat circular face of area πr², so its total surface area is 3πr².

Worked example 5. A sphere has radius 7.5 cm.

V = (4/3) × π × 7.5³ = (4/3) × 421.875π = 562.5π ≈ 1767.1 cm³
Surface area = 4 × π × 56.25 = 225π ≈ 706.9 cm²

Worked example 6: a solid hemisphere of radius 9 cm.

V = (2/3) × π × 9³ = (2/3) × 729π = 486π ≈ 1526.8 cm³
Total surface area = 3 × π × 81 = 243π ≈ 763.4 cm²

Working backwards. A sphere has volume 2304π cm³. Find r.

(4/3)πr³ = 2304π
r³ = 2304 × 3/4 = 1728
r = 12 cm

10.04b Cones

Cone: V = (1/3)πr²h
Curved surface area = πrl, where l is the slant height
Total surface area = πr² + πrl
Slant height: l = √(r² + h²)

The height h is the perpendicular height. The slant height l runs along the sloping surface, so the two are not interchangeable.

Worked example 7. A cone has base radius 12 cm and perpendicular height 16 cm.

l = √(12² + 16²) = √400 = 20 cm
V = (1/3) × π × 144 × 16 = 768π ≈ 2412.7 cm³
Curved area = π × 12 × 20 = 240π
Base = π × 144 = 144π
Total surface area = 384π ≈ 1206.4 cm²

Working backwards. A cone of base radius 9 cm has curved surface area 135π cm². Find its volume.

π × 9 × l = 135π, so l = 15 cm
h = √(15² − 9²) = √144 = 12 cm
V = (1/3) × π × 81 × 12 = 324π ≈ 1017.9 cm³

10.04b Simple composite solids

A composite solid is two or more simple solids joined, or one removed from another. Split it, find each part, then add or subtract.

For surface area, include only the faces that are on the outside. When two solids are joined, the surface where they meet disappears from both. Draw each part and list its exposed faces before you calculate.

Worked example 8: a silo. A cylinder of radius 3 m and height 8 m has a solid hemisphere fixed on top with the same radius. The base is not painted. Find the volume and the area to be painted.

Volume of cylinder = π × 9 × 8 = 72π
Volume of hemisphere = (2/3) × π × 27 = 18π
Total volume = 90π ≈ 282.7 m³

Painted: curved wall 2π × 3 × 8 = 48π
         curved hemisphere 2π × 9 = 18π
Area = 66π ≈ 207.3 m²

The circle where the hemisphere meets the cylinder is hidden, so it is not counted. The base is excluded by the question.

Worked example 9: a cone removed from a cylinder. A cylinder has radius 3 cm and height 4 cm. A cone of the same radius and height is hollowed out of one end.

V = π × 9 × 4 − (1/3) × π × 9 × 4 = 36π − 12π = 24π ≈ 75.4 cm³
Slant height of cone = √(3² + 4²) = 5 cm
Surface area = base π × 9 + curved wall 2π × 3 × 4 + cone surface π × 3 × 5
             = 9π + 24π + 15π = 48π ≈ 150.8 cm²

The open end of the cylinder has been replaced by the cone’s curved surface, so there is no top circle.

10.04c Pyramids

V = (1/3) × area of base × perpendicular height

The base can be any polygon. The surface area is the base plus every sloping face. For a square-based pyramid, the four sloping faces are congruent triangles, and the height of each triangle (not the pyramid) is the slant height of the face.

Worked example 10. A square-based pyramid has base side 14 cm and perpendicular height 24 cm.

V = (1/3) × 14² × 24 = 1568 cm³
Distance from centre of base to the middle of a side = 7 cm
Slant height of a face = √(24² + 7²) = 25 cm
Each triangle = ½ × 14 × 25 = 175 cm²
Surface area = 196 + 4 × 175 = 896 cm²

Pyramid on a cuboid. A cuboid 8 cm by 8 cm by 5 cm has a square-based pyramid of height 3 cm on its top face, with the same base. The volume is 8 × 8 × 5 + (1/3) × 64 × 3 = 320 + 64 = 384 cm³.

Common errors

• Putting the slant height into a volume calculation. Volumes use perpendicular height.
• Forgetting the circular ends when the question says “total surface area”, or including one that the question has removed.
• Counting the joining face in a composite solid.
• Using a diameter as a radius. Halve it first.
• Mixing units, for example metres and centimetres in one product.
• Rounding too soon. Keep π in the working, or keep unrounded values, until the last line.
• Writing a volume in cm² or a surface area in cm³.

Next, try the practice set and tick off the skills on the checklist.

Official syllabus

OCR, GCSE (9-1) Mathematics J560, Version 2.1 (August 2026), for first assessment in 2017, section 10.04 Volume and surface area calculations.