MathematicsFoundation20 min read

Surface Area and Volume

Solids, and the difference between covering one and filling it

This topic appears in:

01

Volume fills, surface area covers

The two quantities answer different questions and are measured in different units. Volume is how much fits inside, in cubic units. Surface area is how much material would wrap around the outside, in square units.

Deciding which a question wants is the first step, and the wording gives it away. "How much water will it hold" is volume; "how much paint to cover it" is surface area. So is "how much card is needed to make it" — a net question in disguise.

Compare the cylinder and the cone at the same size. The cone holds exactly one third of the cylinder that would contain it — which is worth remembering as a check rather than a separate formula.

02

Prisms: one rule covers all of them

A prism is a solid with the same cross-section all the way along its length. A cuboid is a prism with a rectangular cross-section; a cylinder is a prism with a circular one; a triangular prism has a triangular one.

That gives a single formula for all of them: volume = area of cross-section × length. Rather than memorising a separate formula for each solid, find the area of the face that repeats and multiply by how far it extends.

any prism:V = area of cross-section × lengthcuboid:V = l w hA = 2(lw + lh + wh)cylinder:V = πr²hA = 2πr² + 2πrhcurved surface of a cylinder = 2πrh(it unrolls into a rectangle)the cuboid and cylinder formulas are not given in the exam and must be known

Why the curved surface of a cylinder is 2πrh

Cut the label off a tin and unroll it. It is a rectangle: its height is the height of the tin, and its width is the distance all the way round — the circumference, 2πr. So the curved area is 2πrh, and adding the two circular ends of πr² each gives the total. Deriving it this way makes the formula impossible to misremember.

03

Cones, pyramids and spheres

These are not prisms — the cross-section shrinks as you move up — so the prism rule does not apply. Their formulas are given on the exam paper, but knowing what each symbol means is not.

The point that costs marks is the difference between the vertical height h and the slant height l of a cone. The volume uses h, the perpendicular distance from base to apex. The curved surface area uses l, measured along the sloping side. They are connected by Pythagoras, and a question giving one while requiring the other is entirely standard.

pyramid: V = ⅓ × base area × heightcone:V = ⅓ πr²hcurved surface = πrlsphere:V = 4⁄3 πr³surface area = 4πr²l² = r² + h²(slant height by Pythagoras)the cone volume uses the VERTICAL height h; the curved surface uses the SLANT height l
r
radius of the base
h
perpendicular height from base to apex
l
slant height along the sloping surfacealways the longest of the three
Worked example

A cone has base radius 5 cm and vertical height 12 cm. Find its volume and its total surface area. Take π = 3.142.

  1. Volume = ⅓πr²h = ⅓ × 3.142 × 25 × 12 = 314.2 cm³.The vertical height goes into the volume.
  2. For the curved surface, first find the slant height: l² = 5² + 12² = 169, so l = 13.A 5-12-13 triangle — examiners choose these numbers so the root is exact.
  3. Curved surface = πrl = 3.142 × 5 × 13 = 204.2 cm².The slant height, not the vertical one.
  4. Base = πr² = 3.142 × 25 = 78.6 cm²."Total" surface area includes the base; "curved" does not.
  5. Total = 204.2 + 78.6 = 282.8 cm².

Volume 314.2 cm³; total surface area 282.8 cm²

Read whether the base is included

A cone-shaped hat has no base, so only the curved surface is wanted. A solid cone standing on a table has one. Questions say "curved surface area" or "total surface area", and the two differ by πr² — enough to lose the final mark. The same applies to an open cylinder such as a pipe or a tin without a lid.

04

Composite solids and the effect of scaling

A shape made of two solids joined together is handled by adding volumes, exactly as with areas. Surface area needs more care: where the two meet, the joining faces are inside the solid and are not part of the surface.

And the scaling rule from similar figures applies to solids too. Multiply every length by k and the surface area multiplies by while the volume multiplies by . That is why a small model needs far less material than its scale suggests, and why large animals have proportionally thicker legs than small ones.

Before you leave this chapter

  1. Volume fills (cubic units); surface area covers (square units). Read which is wanted.
  2. Every prism: volume = cross-section area × length.
  3. A cylinder's curved surface unrolls into a rectangle 2πr by h.
  4. Cone volume uses the vertical height; curved surface uses the slant height, with l² = r² + h².
  5. Lengths × k gives areas × k² and volumes × k³.
05

Nets, and why they make surface area easy

A net is the flat shape that folds up into a solid. Drawing one turns a surface-area question into an ordinary area question, because every face becomes a flat shape whose area you already know how to find.

A cuboid unfolds into six rectangles, in three matching pairs. A cylinder unfolds into two circles and one rectangle whose width is the circumference. A cone unfolds into a circle and a sector. Sketching the net first is the surest way to avoid counting a face twice or missing one entirely.

SolidNet consists ofTotal surface area
Cuboid6 rectangles in 3 pairs2(lw + lh + wh)
Cylinder2 circles + 1 rectangle2πr² + 2πrh
Cone1 circle + 1 sectorπr² + πrl
Triangular prism2 triangles + 3 rectangles2(½bh) + the three faces
Square pyramid1 square + 4 trianglesb² + 4(½ × b × slant)

Counting the faces is the whole method

Most lost marks in surface area come from a missing or duplicated face, not from arithmetic. Sketch the net, label each piece with its dimensions, find each area, and add. It takes a minute longer than trying to hold the solid in your head and it is very much more reliable — particularly for an open container, where one face is deliberately absent.

Practice questions

6 questions · 20 marks · full working on every one

Try each one on paper first, then open the working. The marks are shown where they are actually awarded, because that is where they are actually lost.

Short questions

3 · 6 marks

Two marks each, in the style of the short-question section of the paper. Answer in two or three lines.

SQ1[2 marks]
A cuboid measures 8 cm by 5 cm by 3 cm. Find its volume and surface area.
Model answer

Volume = 8 × 5 × 3 = 120 cm³. Surface area = 2(40 + 24 + 15) = 158 cm².

Examiner tip. The three distinct faces are 8×5, 8×3 and 5×3, and each occurs twice. Doubling the sum of the three is quicker than adding six areas.

SQ2[2 marks]
State the difference between the vertical height and the slant height of a cone.
Model answer

The vertical height is the perpendicular distance from the centre of the base to the apex, used in the volume. The slant height is measured along the sloping surface from the rim to the apex, used in the curved surface area. They are related by l² = r² + h².

Examiner tip. The slant height is always the longest of the three, being the hypotenuse. If your l is smaller than h, the Pythagoras was the wrong way round.

SQ3[2 marks]
A sphere has radius 6 cm. Find its volume, taking π = 3.142.
Model answer

V = 4⁄3 πr³ = 4⁄3 × 3.142 × 216 = 904.9 cm³.

Examiner tip. Cube the radius before multiplying, and note that 6³ = 216 rather than 6 × 3. Cubing is the step most often mishandled.

Solved numericals

2 · 8 marks

Full working, one step per line, with the marks shown where they are awarded.

N1[4 marks]
A cylindrical tank has radius 1.5 m and height 4 m. Find its volume in m³ and its capacity in litres. Take π = 3.142 and 1 m³ = 1000 litres.
Full working
  1. V = πr²h = 3.142 × 1.5² × 4square the radius, do not double it[1]
  2. = 3.142 × 2.25 × 4 = 28.278[1]
  3. Capacity = 28.278 × 1000[1]
  4. = 28 278 litres, about 28 300 litres to 3 s.f.[1]

28.3 m³, about 28 300 litres

Examiner tip. Squaring 1.5 gives 2.25, not 3. Doubling instead of squaring is the commonest arithmetic slip in cylinder questions.

N2[4 marks]
A solid cone has base radius 8 cm and slant height 17 cm. Find its vertical height and its volume, taking π = 3.142.
Full working
  1. By Pythagoras h² = l² − r² = 289 − 64 = 225the slant height is the hypotenuse, so it is squared and the radius subtracted[1]
  2. h = 15 cman 8-15-17 triple[1]
  3. V = ⅓πr²h = ⅓ × 3.142 × 64 × 15the volume uses the vertical height[1]
  4. = 1005.4 cm³[1]

Height 15 cm; volume 1005 cm³

Examiner tip. Subtract to find h and add to find l. Getting the direction wrong gives an h larger than the slant height, which is geometrically impossible and worth noticing.

Long questions

1 · 6 marks

Theory and numerical together, as they appear in the long-question section.

LQ1[6 marks]
A silo consists of a cylinder of radius 3 m and height 10 m with a hemisphere of the same radius on top.
  1. Find the total volume, taking π = 3.142.
  2. Find the external surface area, excluding the flat base.
  3. A scale model is built with all lengths one tenth of the real ones. State the ratio of the model's surface area and volume to the real silo's.
Mark scheme
  1. Cylinder = πr²h = 3.142 × 9 × 10 = 282.78[1]
  2. Hemisphere = ½ × 4⁄3 πr³ = ⅔ × 3.142 × 27 = 56.56 m³; total = 339.3half a sphere[1]
  3. Curved cylinder = 2πrh = 2 × 3.142 × 3 × 10 = 188.5[1]
  4. Hemisphere = ½ × 4πr² = 2 × 3.142 × 9 = 56.6 m²; total = 245.1the circle where they join is internal and excluded[1]
  5. Surface area ratio = (1/10)² = 1 : 100[1]
  6. Volume ratio = (1/10)³ = 1 : 1000[1]

(a) 339.3 m³ (b) 245.1 m² (c) areas 1 : 100, volumes 1 : 1000

Examiner tip. The join between the two solids is inside the silo, so neither the top of the cylinder nor the flat face of the hemisphere counts towards the external surface.