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

These notes condense OCR GCSE (9-1) Mathematics J560, Version 2.1 (August 2026), section 10.04 (10.04a to 10.04c). Everything here is for both the Foundation and the Higher tier; nothing is Higher tier only. For full explanations and longer worked examples, read the study guide, then try the practice set.

Links: OCR GCSE Mathematics hub, checklist, free 10-minute diagnostics.

What the specification says about formulae

• The specification says there is no formula sheet, and that the formulae you must know are the ones listed under 6.02d. They include the circumference and area of a circle.
• For spheres, cones and composite solids (10.04b) it says the formulae will be given. For pyramids (10.04c) it says the formula 1/3 × area of base × height will be given.
• Learn every formula below anyway. It saves time and lets you check what you are given.
• Papers 2 and 5 do not permit a calculator, so practise leaving answers in terms of π.

Formula table

Solid	Volume	Surface area
Cuboid l × w × h	lwh	2(lw + lh + wh)
Any right prism	cross-section area × length	2 × cross-section + perimeter × length
Cylinder (r, h)	πr²h	2πr² + 2πrh (curved part 2πrh)
Sphere (r)	(4/3)πr³	4πr²
Hemisphere (solid)	(2/3)πr³	3πr² (curved part 2πr²)
Cone (r, h, slant l)	(1/3)πr²h	πr² + πrl (curved part πrl)
Pyramid	(1/3) × base area × height	base + all sloping faces

Slant height of a cone: l = √(r² + h²). Slant height of a face of a square-based pyramid: √(h² + (half the base side)²).

Units

Quantity	Fact
Volume units	cm³, m³ and so on
1 cm³	1 millilitre
1000 cm³	1 litre
1 m³	1 000 000 cm³
1 m²	10 000 cm²

Convert all lengths to one unit before you multiply. Area conversions use the square of the length factor, and volume conversions use the cube: 1 m is 100 cm, so 1 m² is 100² = 10 000 cm² and 1 m³ is 100³ = 1 000 000 cm³.

Method boxes

Prism or cylinder volume

1. Identify the cross-section (it may be a trapezium, triangle, L-shape or circle).
2. Find its area, using Pythagoras if a length is missing.
3. Multiply by the length.

Total surface area of a prism

1. List the faces: two ends and the rectangles around.
2. The rectangles together make perimeter × length.
3. Add the two ends.

Cone or pyramid with a missing slant height

1. Draw the right-angled triangle: perpendicular height, radius (or half-side), slant height.
2. Use Pythagoras.
3. Use the slant height for area, the perpendicular height for volume.

Composite solid

1. Sketch and label each part.
2. For volume, add the parts, or subtract a removed part.
3. For surface area, list only the outside faces. A joined face is hidden for both solids.
4. Add up and give units.

Working backwards

1. Write the formula with known values substituted.
2. Rearrange for the unknown.
3. For a cube root or square root, check the answer back in the formula.

Small worked reminders

• A cylinder of radius 2.5 cm and height 8 cm: V = π × 6.25 × 8 = 50π ≈ 157.1 cm³.
• A sphere of diameter 10 cm has radius 5 cm: V = (4/3) × π × 125 = 500π/3 ≈ 523.6 cm³.
• A cone of radius 6 cm and height 8 cm: l = 10, curved area = π × 6 × 10 = 60π ≈ 188.5 cm².
• A solid hemisphere of radius 2 cm: total surface area = 3π × 4 = 12π ≈ 37.7 cm².
• A pyramid with a 6 cm by 5 cm rectangular base and height 9 cm: V = (1/3) × 30 × 9 = 90 cm³.

Three full reminders

A prism with a trapezium cross-section. Parallel sides 6 cm and 12 cm, distance between them 4 cm, slanted sides 5 cm each (the horizontal overhang on each side is 3 cm, and 3² + 4² = 5²), length 15 cm.

Cross-section = ½ × (6 + 12) × 4 = 36 cm²
Volume = 36 × 15 = 540 cm³
Perimeter = 6 + 12 + 5 + 5 = 28 cm
Surface area = 2 × 36 + 28 × 15 = 72 + 420 = 492 cm²

A capsule. A cylinder of radius 2 cm and length 7 cm with a hemisphere of radius 2 cm on each end.

Cylinder: π × 4 × 7 = 28π
Two hemispheres make one sphere: (4/3) × π × 8 = 32π/3
Volume = 28π + 32π/3 = 116π/3 ≈ 121.5 cm³
Surface: curved cylinder 2π × 2 × 7 = 28π, plus a sphere's surface 4π × 4 = 16π
Surface area = 44π ≈ 138.2 cm²

Litres and depth. A cylindrical tank has radius 50 cm and holds 200 litres. Convert: 200 litres = 200 000 cm³. Then 200 000 = π × 50² × h, so h = 200 000 ÷ (2500π) = 80/π ≈ 25.5 cm.

Writing answers in terms of π

• Multiply the numbers and write π once: 4² × 11 × π gives 176π.
• Add like terms: 28π + 16π = 44π.
• A fraction of π is fine: 500π/3.
• Convert to a decimal only on the last line, and only if the question asks for it. Give the accuracy the question states.
• Square roots can often be removed by hand before you start. A Pythagorean triple such as 3, 4, 5 or 5, 12, 13 often appears, and noticing it saves time on a paper without a calculator.

Spotting the right formula in a question

The question mentions	Think
“How much water / paint / metal”	volume
“How much material / wrapping / label”	surface area, possibly only part of it
“Open at the top”	leave out one end
“Same shape melted and recast”	volumes equal
“Litres”, “millilitres”	convert to cm³
“Slant height”, “sloping side”	Pythagoras with the perpendicular height
“Hollow”, “removed”, “cut out”	subtract volumes, check which surfaces are now exposed

Must-know distinctions

• Perpendicular height h and slant height l. Volumes use h. Curved cone area uses l.
• Curved surface area and total surface area. Total includes every end or base that is present.
• Diameter and radius. Formulae use the radius.
• Volume (cubic units, space inside) and surface area (square units, outer covering).
• Exact and rounded answers. Give π in the answer if the question asks for it, otherwise follow the stated accuracy.
• A hemisphere as a bowl (open, one curved surface 2πr²) and as a solid (add the flat face πr²).

Quick self-test

1. A cube has side 7 cm. Find its surface area.
2. Find the volume of a cylinder of radius 2.5 cm and height 8 cm, in terms of π.
3. Find the volume of a sphere of diameter 10 cm, in terms of π.
4. A cone has radius 6 cm and height 8 cm. Find the curved surface area in terms of π.
5. Find the volume of a pyramid with a 6 cm by 5 cm base and height 9 cm.
6. Find the total surface area of a solid hemisphere of radius 2 cm, in terms of π.
7. A cuboid has volume 240 cm³ and a base measuring 4 cm by 6 cm. Find its height.
8. How many cm³ are there in 2 m³?
9. A cylinder has volume 250π cm³ and radius 5 cm. Find its height.
10. A cuboid of base 8 cm by 8 cm and height 5 cm has a square-based pyramid of height 3 cm and the same base on top. Find the total volume.

Answers

1. 6 × 7² = 294 cm².
2. π × 2.5² × 8 = 50π cm³.
3. Radius 5, so (4/3) × π × 125 = 500π/3 cm³ (≈ 523.6).
4. l = √(36 + 64) = 10, so π × 6 × 10 = 60π cm².
5. (1/3) × 30 × 9 = 90 cm³.
6. 3 × π × 2² = 12π cm².
7. 240 ÷ 24 = 10 cm.
8. 2 × 1 000 000 = 2 000 000 cm³.
9. 250π = π × 25 × h, so h = 10 cm.
10. 320 + (1/3) × 64 × 3 = 320 + 64 = 384 cm³.

Where marks are usually lost

• Using the diameter in place of the radius, especially when the radius is given as “half the width”.
• Putting the slant height into a volume formula.
• Leaving out the base of a cone or an end of a cylinder in a “total surface area” question, or adding one that is not there.
• Counting the hidden joining face of a composite solid.
• Missing the hypotenuse when finding the perimeter of a triangular cross-section.
• Converting units wrongly: 1 m³ is 1 000 000 cm³, not 100 cm³ or 1000 cm³.
• Rounding π to 3.14 or rounding mid-way and finishing a few units out. Examiners advise keeping exact values until the end.
• Not showing the substitution. If you write only a final decimal and it is wrong, there is little to credit.
• Writing the wrong unit, or none, on the final answer.

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.