Cambridge IGCSE Mathematics 0580: Surface area and volume – Study Guide

This study guide teaches subtopic 5.4, Surface area and volume, from topic 5 (Mensuration) of the Cambridge IGCSE Mathematics 0580 syllabus for 2025, 2026 and 2027. The learning outcome is word-for-word the same in both columns: C5.4 (Core) and E5.4 (Extended) both ask you to carry out calculations and solve problems involving the surface area and volume of a cuboid, prism, cylinder, sphere, pyramid and cone. Core candidates take Paper 1 (non-calculator) and Paper 3 (calculator); Extended candidates take Paper 2 (non-calculator) and Paper 4 (calculator). The examples below mix both kinds of paper. The one Extended-only step on this page is finding a length inside a solid with Pythagoras in three dimensions (E6.6), and it is labelled where it appears.

Use this page with the Cambridge IGCSE Mathematics hub and the printable 0580 checklist. For an overview of the whole of topic 5, see the Mensuration study guide.

What this subtopic covers

0580 reference	What you must be able to do	Tier
C5.4 / E5.4	Surface area and volume of a cuboid and a prism (any solid with a uniform cross-section, such as a cylindrical sector)	Core and Extended
C5.4 / E5.4	Surface area and volume of a cylinder, sphere, pyramid and cone	Core and Extended
C5.4 / E5.4	Give answers in terms of π when asked	Core and Extended
E6.6 (linked)	Find a slant height inside a pyramid from its vertical height	Extended only

Compound solids and parts of solids (a hemisphere, a frustum) belong to the next subtopic, 5.5, and are not taught here.

What the List of formulas gives you

Page 2 of every 0580 paper has a List of formulas. For this subtopic it gives:

Given on the paper	Formula
Curved surface area of a cylinder	A = 2πrh
Curved surface area of a cone (slant edge l)	A = πrl
Surface area of a sphere	A = 4πr²
Volume of a prism (cross-section A, length l)	V = Al
Volume of a pyramid (base area A, height h)	V = (1/3)Ah
Volume of a cylinder	V = πr²h
Volume of a cone	V = (1/3)πr²h
Volume of a sphere	V = (4/3)πr³

What is not given, so you must know it or build it:

• the volume and surface area of a cuboid (a cuboid is a prism, so V = Al still works);
• the surface area of any prism or pyramid (add up the faces);
• the total surface area of a cylinder or cone (add the circular ends to the curved surface);
• the areas of rectangles, trapezia and parallelograms used as faces or cross-sections (C5.2).

On Papers 3 and 4 use the π key or 3.142, and give non-exact answers to 3 significant figures. When a question asks for an exact answer, leave it in terms of π.

Cuboids

For a cuboid with length l, width w and height h:

V  = l × w × h
SA = 2(lw + lh + wh)      (three pairs of equal rectangles)

Worked example 1 (non-calculator). A closed box is a cuboid 8 cm by 5 cm by 3.5 cm. Find its volume and surface area. Then find the surface area if the box has no lid.

V  = 8 × 5 × 3.5 = 40 × 3.5 = 140 cm³

Faces: 8 × 5 = 40, 8 × 3.5 = 28, 5 × 3.5 = 17.5
SA = 2(40 + 28 + 17.5) = 2 × 85.5 = 171 cm²

No lid: remove one 8 × 5 face
SA = 171 − 40 = 131 cm²

Questions also run backwards. If a cuboid has volume 252 cm³ and a base 7 cm by 6 cm, then 252 = 42 × h, so h = 6 cm.

Prisms

A prism has the same cross-section all the way through. The syllabus defines a prism as any solid with a uniform cross-section, so a cuboid, a cylinder and a cylindrical sector all count.

V  = (area of cross-section) × length
SA = 2 × (area of cross-section) + (perimeter of cross-section) × length

Worked example 2 (non-calculator). A triangular prism has a right-angled triangle as its cross-section, with sides 6 cm, 8 cm and 10 cm. The prism is 15 cm long. Find its volume and total surface area.

Cross-section area = ½ × 6 × 8 = 24 cm²
V = 24 × 15 = 360 cm³

Two triangular ends   = 2 × 24 = 48 cm²
Rectangular faces     = (6 + 8 + 10) × 15 = 24 × 15 = 360 cm²
SA = 48 + 360 = 408 cm²

Use the two shorter sides (the ones at the right angle) for the triangle’s base and height, never the 10 cm side.

Worked example 3 (calculator). A block of cheese is a prism whose cross-section is a sector of a circle with radius 9 cm and angle 40°. The block is 20 cm long. Find (a) its volume, in terms of π and to 3 significant figures, and (b) its total surface area.

(a) Sector area = (40/360) × π × 9² = (1/9) × 81π = 9π cm²
    V = 9π × 20 = 180π cm³ = 565.48... = 565 cm³ (3 s.f.)

(b) Two sector ends  = 2 × 9π = 18π
    Arc length       = (40/360) × 2 × π × 9 = 2π
    Curved face      = 2π × 20 = 40π
    Two flat faces   = 2 × (9 × 20) = 360
    SA = 18π + 40π + 360 = 58π + 360 = 542.21... = 542 cm² (3 s.f.)

A cylindrical sector has five faces: two sectors, one curved face and two rectangles. Missing the rectangles is the usual error.

Cylinders

V             = πr²h
Curved SA     = 2πrh
Total SA      = 2πrh + 2πr²   (closed cylinder: two circular ends)

Worked example 4 (non-calculator, answers in terms of π). A closed cylinder has radius 5 cm and height 8 cm. Find its volume, its curved surface area and its total surface area.

V         = π × 5² × 8 = 200π cm³
Curved SA = 2 × π × 5 × 8 = 80π cm²
Ends      = 2 × π × 5² = 50π cm²
Total SA  = 80π + 50π = 130π cm²

Collect like terms (80π + 50π), but do not multiply out π.

Worked example 5 (calculator, working backwards with units). A cylindrical tank has internal radius 18 cm and holds 45 litres when full. Find the height of the tank.

45 litres = 45 000 cm³        (1 litre = 1000 cm³)

π × 18² × h = 45 000
h = 45 000 ÷ (π × 324)
h = 44.209... = 44.2 cm (3 s.f.)

Convert the capacity first, so every quantity is in centimetres.

Spheres

V  = (4/3)πr³
SA = 4πr²

Worked example 6 (non-calculator). A ball has radius 3 cm. Find its volume and surface area in terms of π.

V  = (4/3) × π × 3³ = (4/3) × 27π = 36π cm³
SA = 4 × π × 3² = 36π cm²

The numbers match only because r = 3; the units (cm³, cm²) do not.

Worked example 7 (calculator, finding the radius). A sphere has volume 900 cm³. Find its radius and then its surface area.

(4/3)πr³ = 900
r³ = 900 × 3 ÷ (4π) = 214.86...
r  = ∛214.86... = 5.989... = 5.99 cm (3 s.f.)

SA = 4 × π × 5.989...² = 450.79... = 451 cm² (3 s.f.)

Use the unrounded radius in the second step. The syllabus says that when you use an earlier answer later in a question, you should use the value before it was rounded.

Pyramids

V  = (1/3) × base area × perpendicular height
SA = base area + area of each triangular face

The height in the volume formula is the vertical height from the apex to the base, not the height of a sloping face.

Worked example 8 (non-calculator). A pyramid has a square base of side 16 cm. Its apex is 15 cm vertically above the centre of the base. Each triangular face has a height (slant height) of 17 cm. Find its volume and total surface area.

Base area = 16 × 16 = 256 cm²
V = (1/3) × 256 × 15 = 256 × 5 = 1280 cm³

Each face = ½ × 16 × 17 = 136 cm²
SA = 256 + 4 × 136 = 256 + 544 = 800 cm²

The vertical height (15 cm) goes in the volume; the slant height (17 cm) goes in the face areas.

Extended only. You may be given only the vertical height and have to find the slant height yourself. The apex, the centre of the base and the midpoint of a base edge form a right-angled triangle inside the solid, which is Pythagoras in three dimensions (E6.6):

slant height² = 15² + 8²      (8 = half of 16)
              = 225 + 64 = 289
slant height  = 17 cm

Cones

V         = (1/3)πr²h          (h = perpendicular height)
Curved SA = πrl                (l = slant height)
Total SA  = πrl + πr²          (solid cone: add the circular base)
l² = r² + h²

The radius, perpendicular height and slant height form a right-angled triangle with l as the hypotenuse. If you are given one height and need the other, use Pythagoras on that triangle.

Worked example 9 (calculator). A solid cone has radius 4.5 cm and perpendicular height 9.8 cm. Find its slant height, its volume and its total surface area.

l² = 4.5² + 9.8² = 20.25 + 96.04 = 116.29
l  = 10.783... = 10.8 cm (3 s.f.)

V  = (1/3) × π × 4.5² × 9.8 = 207.81... = 208 cm³ (3 s.f.)

Total SA = π × 4.5 × 10.783... + π × 4.5²
         = 152.45... + 63.61...
         = 216.06... = 216 cm² (3 s.f.)

Keep l = 10.783… in your calculator for the surface area step.

Worked example 10 (non-calculator, working backwards). A cone has radius 6 cm and volume 96π cm³. Find its perpendicular height and its curved surface area in terms of π.

(1/3) × π × 6² × h = 96π
12πh = 96π
h = 8 cm

l² = 6² + 8² = 100,  l = 10 cm
Curved SA = π × 6 × 10 = 60π cm²

Dividing both sides by π early keeps the arithmetic simple.

Solving problems with volume

“Solve problems” in the outcome includes questions where one solid is melted and recast, or a liquid is poured from one container into another. The volume stays the same, so you set two volumes equal or divide one by the other.

Worked example 11 (calculator). A metal cube of side 10 cm is melted down and made into spheres of radius 1.5 cm. No metal is wasted. How many complete spheres can be made?

Cube volume   = 10³ = 1000 cm³
Sphere volume = (4/3) × π × 1.5³ = 4.5π = 14.137... cm³

1000 ÷ 14.137... = 70.73...
Number of complete spheres = 70

Round down: the 71st sphere would need more metal than is left.

Common errors

• Using the slant height in a volume formula. Both the pyramid and the cone use the perpendicular height for volume.
• Giving the curved surface area when the total is asked. The List of formulas gives only the curved part of a cylinder and a cone; add the circular end or ends yourself.
• Forgetting the 1/3. Pyramids and cones are one third of the matching prism or cylinder.
• Missing faces of a prism. Count every face, including the rectangles of a cylindrical sector.
• Mixing units. Convert litres to cm³ (or m to cm) before you substitute.
• Writing π as 3.14 or rounding mid-way. Use the π key or 3.142, and round only the final answer.

Where this leads

Subtopic 5.5 applies these formulas to compound solids and parts of solids. The Mensuration revision notes summarise the formulas for quick recall. Test yourself with the Mensuration practice questions, and revise right-angled triangles with the Pythagoras’ theorem study guide. To check where you stand across the course, try the Core diagnostic or the Extended diagnostic.

Official syllabus

Cambridge International, Cambridge IGCSE Mathematics 0580 syllabus for 2025, 2026 and 2027 (Version 3), Subject content, topic 5 Mensuration: C5.4 and E5.4 Surface area and volume, and the List of formulas for Core (Papers 1 and 3) and Extended (Papers 2 and 4). Linked references: C5.3/E5.3 Circles, arcs and sectors, C5.5/E5.5 Compound shapes and parts of shapes, and E6.6 Pythagoras’ theorem and trigonometry in 3D.