Volume and Surface Area

Mathematics revision notes · Cambridge (CIE) · 0580

1. Which formulas are printed

Given and not given

Volume measures the space a solid fills, in cubed units such as cm³, and surface area measures the total area of its outside, in squared units such as cm²
The paper prints some of these formulas and not others, and knowing which is which saves memorising the wrong things

Printed in the List of formulas	Not printed — learn these
Volume of a prism, V = Al	Volume of a cuboid, V = lwh
Volume of a cylinder, V = πr²h	Total surface area of a cylinder
Volume of a pyramid, V = ⅓Ah	Total surface area of a cone
Volume of a cone, V = ⅓πr²h	Surface area of a cuboid, prism or pyramid
Volume of a sphere, V = 4/3 πr³	—
Curved surface area of a cylinder, A = 2πrh	—
Curved surface area of a cone, A = πrl	—
Surface area of a sphere, A = 4πr²	—

Notice the pattern: every volume you need is printed (a cuboid's is the prism formula, V = Al), but only the curved surface areas are
Any surface area involving flat faces has to be assembled from those faces yourself
The letter l in the cone formula is the sloping edge, not the perpendicular height, and these papers use the words "sloping edge" rather than "slant height"

2. Volume

Prisms and cylinders

A prism is any solid with the same cross-section all the way along, and its volume is that cross-sectional area times its length
On this specification "prism" is defined broadly, so a cylinder is a prism with a circular cross-section, and even a cylindrical sector counts
So a triangular prism whose cross-section has area 24 cm² and whose length is 15 cm has volume 24 × 15 = 360 cm³
A cuboid is a prism with a rectangular cross-section, and its volume is length × width × height

A cuboid with its hidden edges dashed and its three dimensions named: length along the front bottom edge, width along the bottom edge running back, and height up the side

For a cylinder the cross-section is a circle, which is why V = πr²h

A cylinder with the radius r marked across the top face from its centre and the height h marked up the side, the hidden part of the base drawn as a dashed curve — Source: Volume by Save My Exams

Worked example

A volume in m³, cm³ and litres

A water tank is a cuboid 1.5 m long, 0.8 m wide and 0.6 m deep. Find its volume in m³ and in cm³, and the number of litres it holds.

Solution:

Volume = 1.5 × 0.8 × 0.6 = 0.72 m³
One metre is 100 cm, so 1 m³ is 100 × 100 × 100 = 1 000 000 cm³
Volume = 0.72 × 1 000 000 = 720 000 cm³
One litre is 1000 cm³, so divide by 1000
Capacity = 720 000 ÷ 1000 = 720 litres
Check by changing the lengths to centimetres first: 150 × 80 × 60 = 720 000 cm³
The length factor of 100 is cubed because three lengths are multiplied, so multiplying 0.72 by 100 would be wrong

Pyramids, cones and spheres

A pyramid has volume ⅓ × base area × perpendicular height, and the height must run straight down from the apex to the base

A square-based pyramid with a red arrow running straight down from the apex to the centre of the base, marked h, which is the perpendicular height rather than the length of a sloping edge — Source: Volume by Save My Exams

A cone is a pyramid on a circular base, so its volume is ⅓πr²h, again using the perpendicular height
The sloping edge of a cone is longer than its perpendicular height, and Pythagoras converts between them since the radius, the height and the sloping edge form a right-angled triangle

A cone with the perpendicular height h drawn from the apex down to the centre of the base, a small square marking the right angle where it meets the radius r — Source: Volume by Save My Exams

A sphere has volume 4/3 πr³, and a hemisphere is half of that

A sphere with its radius r drawn from the centre out to the surface, the equator shown as a curve with its hidden half dashed — Source: Volume by Save My Exams

Worked example

A cone from its sloping edge

A cone has base radius 5 cm and sloping edge 13 cm. Find its volume in terms of π.

A cone standing on its circular base. The sloping edge from the rim up to the apex is labelled 13 cm and the base radius is labelled 5 cm. A dashed vertical line runs from the centre of the base to the apex and a dashed radius runs out to the rim, meeting at a marked right angle; the vertical height carries no label. The figure is marked NOT TO SCALE.

Solution:

The volume formula needs the perpendicular height, but the sloping edge is given
The radius, the height and the sloping edge form a right-angled triangle, with the sloping edge as the hypotenuse
By Pythagoras, h² = 13² − 5² = 169 − 25 = 144
So h = 12 cm
Volume = ⅓ × π × 5² × 12 = ⅓ × π × 25 × 12
Volume = 100π cm³

Worked example

Finding a radius from a volume

A cylinder has radius 4 cm and height 18 cm. A sphere has the same volume as the cylinder. Find the radius of the sphere.

Two solids side by side: on the left an upright cylinder with a dashed radius of 4 cm across its top and a height of 18 cm, and on the right a sphere with its equator drawn and a dashed radius labelled r cm. The figure is marked NOT TO SCALE.

Solution:

Volume of the cylinder = π × 4² × 18 = π × 16 × 18 = 288π cm³
Call the sphere's radius r and set its volume equal to that: 4/3 πr³ = 288π
Divide both sides by π: 4/3 r³ = 288
Multiply by 3 and divide by 4: r³ = 288 × 3 ÷ 4 = 216
Take the cube root: r = ∛216 = 6 cm
Check: 4/3 × π × 6³ = 4/3 × 216π = 288π
Keeping π as a symbol lets it cancel, so no calculator is needed

3. Surface area

Building it from the faces

Surface area is the total of every face and curved surface on the outside of the solid
For a solid with flat faces, work out each face separately and add them, using the net if it helps
- A cuboid 7 cm by 4 cm by 3 cm has faces in three matching pairs: 2(7 × 4) + 2(7 × 3) + 2(4 × 3) = 56 + 42 + 24 = 122 cm²
- A square-based pyramid has one square and four identical triangles

The net of a cube laid flat as six squares in a cross, so the surface area is just the total area of those six faces — Source: Equations & problem-solving by Save My Exams

For a cylinder, add the curved surface to the two circular ends: 2πrh + 2πr²
For a cone, add the curved surface to the circular base: πrl + πr²

A cone labelled with its radius r, perpendicular height h and slant height l along the sloping edge, beside its net: a sector of slant radius l joined to a circle of radius r, so the curved surface uses l and not h — Source: Surface Area by Save My Exams

Only the curved parts are printed, so these totals must be assembled
Check the units: surface area is always in squared units even though the solid is three-dimensional

Worked example

Total surface area of a cylinder

A solid cylinder has radius 5 cm and height 12 cm. Find its total surface area in terms of π.

An upright solid cylinder drawn with a full ellipse for the visible top rim and a dashed far half on the bottom rim. A dashed radius on the top face is labelled 5 cm and the vertical side is labelled 12 cm. The figure is marked NOT TO SCALE.

Solution:

A solid cylinder has three surfaces: two circular ends and the curved side
The curved surface area is the one that is printed on the paper: 2πrh
Curved surface = 2 × π × 5 × 12 = 120π
Each circular end has area πr² = π × 5² = 25π
There are two of them, giving 2 × 25π = 50π
Total surface area = 120π + 50π = 170π cm²
The total for a cylinder is not printed, only the curved part, so the two ends have to be added by hand
If the cylinder were open at one end, only one circle would be added

4. Compound solids and parts of solids

Joining and cutting

A compound solid is two shapes joined, and a part of a solid is a piece cut from one, such as a hemisphere or a frustum

A cone sitting on top of a hemisphere so the two share a circular join: the slant edge of the cone is 12 cm and the radius of the hemisphere is 5 cm, and the flat circle where they meet is inside the solid rather than on its surface — Source: Surface Area by Save My Exams

Volumes simply add or subtract, so a cylinder with a hemisphere on top has the two volumes added
Surface areas do not simply add, because the faces where the two solids meet are inside the finished solid and are no longer surfaces
Work along the outside of the solid, listing each surface you would actually touch, then add only those
A frustum is a cone or pyramid with its top cut off parallel to the base, so its volume is the whole solid minus the small one removed, and the syllabus names it directly

A frustum drawn as a cone with its tip cut off: the base radius is 20 cm, the radius of the flat top is 10 cm, the whole cone is 30 cm tall from the dashed apex down to the base, and the dashed removed tip is 15 cm of that, so the frustum itself is 15 cm tall — Source: Problem-solving with volumes by Save My Exams

Exam tip

Find the missing length of the cut-off part first

A frustum question usually leaves out one length of the cut-off piece (the small cone's radius or height, or the small pyramid's base side). Each length of the small solid is the same fraction k of the matching large one, so find the missing length from that ratio first and write it down: it earns marks of its own. Then subtract the small volume from the large one, or multiply the large volume by 1 − k³.

A step-shaped solid 10 cm deep whose cross-section is an L: 7 cm up the left side, 4 cm across the top step, then down to a 2 cm lower step, over a 9 cm base — Source: Problem-solving with volumes by Save My Exams

Answers may be asked for in terms of π, so keep π symbolic throughout and collect at the end

Worked example

Surface area from a net

The net of a solid square-based pyramid is shown. Find its total surface area.

The net of a square-based pyramid: a square of side 15 cm in the middle with an identical triangle folded out from each of its four edges, and the perpendicular height of one triangle marked 23 cm — Source: Surface area by Save My Exams

Solution:

A net shows every face laid flat, so the surface area is just the total area of the shapes on it
There is one square base and four identical triangles
Square: 15 × 15 = 225 cm²
Each triangle has base 15 cm and perpendicular height 23 cm
One triangle: ½ × 15 × 23 = 172.5 cm²
Four triangles: 4 × 172.5 = 690 cm²
Total surface area = 225 + 690 = 915 cm²
The 23 cm is the height of the triangle, not the height of the pyramid — the two are different lengths, joined by Pythagoras
No formula for this is printed on the paper; it has to be built from the faces

Worked example

Surface area of a joined solid

A solid is made by joining a hemisphere of radius 5 cm to the top of a cylinder of radius 5 cm and height 14 cm. Find its total surface area in terms of π.

A solid made of an upright cylinder with a dome sitting on its top rim. A dashed radius on the join circle is labelled 5 cm and the straight side of the cylinder is labelled 14 cm. The circle where the two solids meet is drawn, showing that it is inside the finished solid. The figure is marked NOT TO SCALE.

Solution:

Walk round the outside and list the surfaces that are actually on show
The curved surface of the cylinder: 2π × 5 × 14 = 140π cm²
The flat circular base of the cylinder: π × 5² = 25π cm²
The curved surface of the hemisphere, which is half a sphere: ½ × 4π × 5² = 50π cm²
The top circle of the cylinder is not counted, because the hemisphere sits on it and it is no longer on the outside
Total = 140π + 25π + 50π
Total surface area = 215π cm²

Worked example

Equal total surface areas

A solid cone has base radius 3 cm. A solid hemisphere has radius 5 cm. The two solids have the same total surface area. Find the sloping edge of the cone.

Two solids side by side: on the left a cone standing on its base, with a dashed base radius of 3 cm and one sloping edge labelled l cm and no height drawn; on the right a hemisphere resting on its flat face, with a dashed radius of 5 cm on that face. The figure is marked NOT TO SCALE.

Solution:

On its own, a solid hemisphere has two surfaces: half the surface of a sphere and a flat circle
Hemisphere: ½ × 4π × 5² + π × 5² = 50π + 25π = 75π cm²
The cone has a curved surface, πrl with l the sloping edge, and a circular base
Cone: π × 3 × l + π × 3² = 3πl + 9π
Set the totals equal and divide by π: 3l + 9 = 75
3l = 66, so l = 22 cm
Check: 3π × 22 + 9π = 66π + 9π = 75π
Both flat circles count here, because each one is on the outside of its solid, unlike the hidden join in the example above