Cambridge IGCSE Mathematics 0580

Surface area and volume

The volume of any prism or cylinder is the area of its cross-section times its length; cones and pyramids are a third of the prism with the same base and height; a sphere is $\tfrac{4}{3}\pi r^3$ . Surface area is the sum of the faces — and for the curved solids, Cambridge gives you those formulas on the paper. Knowing which are given and which are not is worth as much as the arithmetic.

Surface area and volume is the single heaviest named skill in Cambridge IGCSE Maths 0580: 158 marks across the 42 papers Quanta has mapped, on 24 of them. The marks concentrate in compound solids — a cone on a cylinder, a hemisphere on a prism — where the method is always the same: break it into standard pieces, and be careful about which faces survive when they are joined.

Updated 15 September 2026

What the formula sheet gives you

The 0580 papers print a short list of formulas. The curved solids are on it; the everyday ones are not:

Solid	Volume	Surface area
Cuboid	lwh	2(lw + lh + wh) — know it
Prism	(cross-section area) × length	2 × cross-section + perimeter × length
Cylinder	πr²h	curved 2πrh; closed 2πrh + 2πr²
Cone	⅓πr²h (given)	curved πrl (given), l = slant height
Pyramid	⅓ × base area × h (given)	base + triangular faces
Sphere	4/3 πr³ (given)	4πr² (given)

So the ones to memorise are the cuboid, the cylinder and the general prism — the simple ones students assume will be given. Everything with a $\tfrac{1}{3}$ or a sphere in it is printed for you.

Prisms and cylinders

A prism is any solid with the same cross-section all the way through. Whatever that cross-section is — triangle, trapezium, L-shape, circle — find its area, then multiply by the length:

V_{\text{prism}} = A_{\text{cross-section}} \times \ell \qquad V_{\text{cylinder}} = \pi r^2 h

For the surface area, think of it as wrapping paper: two copies of the cross-section, plus one rectangle whose width is the perimeter of the cross-section and whose height is the length. For a cylinder that rectangle is the curved surface, $2\pi r h$ — the circumference times the height.

The word open changes the answer: an open cylinder (a pipe) has no circular ends, a cylindrical tank open at the top has one. Read the question for it, because it is usually the only difference between full marks and one.

Cones, pyramids and spheres

V_{\text{cone}} = \tfrac{1}{3}\pi r^2 h \qquad A_{\text{curved}} = \pi r l \qquad V_{\text{sphere}} = \tfrac{4}{3}\pi r^3 \qquad A_{\text{sphere}} = 4\pi r^2

The trap in every cone question is the difference between the vertical height $h$ (used for volume) and the slant height $l$ (used for curved surface area). They are two sides of a right-angled triangle with the radius:

l^2 = r^2 + h^2

A question that gives you one and needs the other expects Pythagoras and will not say so. For a hemisphere, halve the sphere volume — but its surface area is $2\pi r^2$ curved plus $\pi r^2$ for the flat circular face, if that face is exposed.

Compound solids

Most of the marks. A shape made of two standard solids — a cone on a cylinder, a hemisphere on a cuboid, a cylinder with a hole bored through it. The method:

Volume: add or subtract. A solid made of two pieces is the sum; a hole is a subtraction. Nothing disappears.
Surface area: ask which faces you can see. Where two solids join, the touching faces are inside and do not count. This is the difference between adding two surface areas (wrong) and adding the exposed parts (right).

For a cone sitting on a cylinder of the same radius, the visible surface is: the cone’s curved surface, the cylinder’s curved surface, and the cylinder’s base circle. The cylinder’s top circle and the cone’s base circle are pressed together and are not part of the surface at all.

Worked example

A solid is made of a cylinder of radius 5 cm and height 12 cm with a cone of the same radius and vertical height 9 cm on top. Find (a) the total volume, (b) the total surface area. Give answers to 3 significant figures.

(a) Volume. Cylinder: $\pi \times 5^2 \times 12 = 300\pi$ . Cone: $\tfrac{1}{3}\pi \times 5^2 \times 9 = 75\pi$ .

V = 300\pi + 75\pi = 375\pi = 1178.097\ldots = 1180 \text{ cm}^3

(b) Surface area. First the slant height, which the question does not give:

l = \sqrt{5^2 + 9^2} = \sqrt{106} = 10.2956\ldots

Now the three visible surfaces:

\text{cone curved} = \pi r l = \pi \times 5 \times 10.2956 = 161.72

\text{cylinder curved} = 2\pi r h = 2\pi \times 5 \times 12 = 376.99

\text{base circle} = \pi r^2 = \pi \times 25 = 78.54

A = 161.72 + 376.99 + 78.54 = 617.25 = 617 \text{ cm}^2

The circle where the cone meets the cylinder appears nowhere: it is inside the solid. Adding it would give 696 cm², and it is the single most common wrong answer to this shape.

Units and similar solids

Converting units

Lengths, areas and volumes convert by different factors, because the conversion applies once per dimension:

1\,\text{m} = 100\,\text{cm} \qquad 1\,\text{m}^2 = 100^2 = 10\,000\,\text{cm}^2 \qquad 1\,\text{m}^3 = 100^3 = 1\,000\,000\,\text{cm}^3

Capacity joins in: $1\,\text{cm}^3 = 1\,\text{ml}$ and $1\,000\,\text{cm}^3 = 1\,\text{litre}$ . A question giving a tank in metres and asking for litres is testing exactly this.

Similar solids

If two solids are similar with length ratio $k$ , their areas are in the ratio $k^2$ and their volumes in the ratio $k^3$ . So doubling every length multiplies surface area by 4 and volume by 8. Extended questions run it backwards: given a volume ratio of 27:8, the length ratio is 3:2, because $\sqrt[3]{27} : \sqrt[3]{8}$ .

Common mistakes

1.Slant height used for volume, or vertical height for curved area
Volume needs $h$ ; curved surface area needs $l$ . Convert between them with $l^2 = r^2 + h^2$ before starting.
2.Counting the joined faces in a compound solid
Where two pieces meet, neither face is on the surface. Sketch the solid and mark only what you could paint.
3.Diameter used as radius
Questions give diameters deliberately. Halve first, and write down $r = \ldots$ so it is on the page.
4.Rounding π or the intermediate values
Keep $375\pi$ or the full calculator value until the end. Rounding the slant height to 10.3 shifts the final area by enough to lose the accuracy mark.
5.Converting volume with the length factor
$1\,\text{m}^3$ is a million cm³, not a hundred. The factor is cubed.
6.Wrong unit on the answer
Volume is cm³, surface area cm². Papers award a mark for the unit, and it is the cheapest one on the paper.

Common questions

Which volume formulas are given in the IGCSE 0580 exam?

The cone, pyramid and sphere formulas are printed on the paper, along with the cone’s curved surface area and the sphere’s surface area. The cuboid, general prism and cylinder are not — you need those from memory.

What is the difference between slant height and vertical height?

The vertical height runs from the apex straight down to the centre of the base and is used for volume; the slant height runs from the apex down the sloping surface to the edge of the base and is used for curved surface area. They are linked by $l^2 = r^2 + h^2$ .

How do you find the surface area of a compound solid?

Add only the faces that are visible from outside. Faces where two solids are joined lie inside the shape and are excluded — which is why the answer is less than the sum of the two separate surface areas.

How do you convert cm³ to m³?

Divide by 1 000 000, because there are $100^3$ cubic centimetres in a cubic metre. Areas use $100^2 = 10\,000$ , and lengths just 100.

Practise surface area and volume against real mark schemes

Quanta has real Cambridge IGCSE Maths 0580 past-paper questions, broken into skill checkpoints, marked criterion by criterion the way examiners mark — and it tracks which skills you’re missing. Free for individual students.

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