Library International Mathematics 0607 Volume & Surface Area
O Level · International Mathematics 0607

Volume & Surface Area

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Cambridge (CIE) IGCSE — International Maths: Extended

Volume & Surface Area

The big idea: volume tells you how much space is trapped inside a 3D shape, and surface area tells you how much "wrapping paper" you'd need to cover the outside — and almost every formula you need is just area × length, or a fraction of a shape you already know.

Summary — everything in this chapter, at a glance
  • Volume = amount of 3D space a shape fills. Measured in cubic units (cm³, m³...).
  • Cuboid volume: V = lwh — the only volume formula you must memorise; everything else is given in the exam.
  • Any prism (constant cross-section): V = A × l (cross-sectional area × length).
  • Cylinder is really just a prism with a circular cross-section: V = πr²h.
  • Pyramid and cone volumes are both "⅓ × base area × height" — they shrink to a point, so you divide by 3.
  • Sphere volume: V = (4/3)πr³.
  • Real exam shapes are rarely "pure" — they're often compound (add volumes), a fraction of a shape (e.g. hemisphere = half sphere), or a frustum (big cone minus small cone).
  • Surface area = sum of the areas of every face. For flat-faced solids, sketch the net and add up each face's area.
  • Cylinder curved surface: A = 2πrh. Cone curved surface: A = πrl (l = slant height, not perpendicular height!). Sphere: A = 4πr².
Topic 1 — Volume of Standard Solids

Cubes and Cuboids

A cuboid is just a fancy name for a rectangular box — think of a shoebox or a brick. A cube is a special cuboid where all three edges (length, width, height) are exactly equal, like a dice.

To find how much space is inside, you're really asking: "how many 1cm × 1cm × 1cm cubes could I pack inside this box?" Picture building up layers — one flat layer covers the base (length × width), and then you stack up h of those layers. That's exactly why the formula multiplies all three dimensions together.

Formula
V = l × w × h
Volume = length × width × height
NOT given in the exam — memorise this one!
Watch the wording
You'll sometimes see "depth" used instead of "height", or "breadth" used instead of "width". Don't let the vocabulary throw you — it's still just three perpendicular edges multiplied together.
Practice Question
A cuboid-shaped water tank has length 1.2 m, width 80 cm, and height 60 cm. Find its volume in cm³.

Prisms — the master idea behind everything else

A prism is any 3D shape that has the exact same 2D cross-section running all the way through it, like a loaf of bread — slice it anywhere along its length and you get an identical shape. A cuboid is actually just a prism with a rectangular cross-section!

This is the single most useful idea in the whole chapter: volume of a prism = area of the cross-section × the length of the prism. It doesn't matter what shape the cross-section is — a triangle, an L-shape, a trapezium, even a weird compound shape — as long as you can find its area, you can find the volume.

Formula
V = A × l
Volume = cross-sectional area × length
Given in the exam

If you're given the volume and length, you can rearrange to find the cross-sectional area: A = V ÷ l.

Practice Question
A prism has a triangular cross-section with base 6 cm and height 4 cm, and the prism itself is 15 cm long. Find its volume.

Cylinders

A cylinder (think of a soup can) is just a prism whose cross-section happens to be a circle. Since the area of a circle is πr², and volume of a prism is area × length, you just swap "length" for "height" and you get the cylinder formula.

Formula
V = πr²h
Volume = π × radius² × height
Given in the exam
Worked Example (from the notes)
A cylinder has radius 8 cm and height 20 cm. Find its volume to 3 s.f.

V = π × 8² × 20 = π × 64 × 20 = 4021.238... = 4020 cm³ (3 s.f.)
Practice Question
A cylindrical can has diameter 10 cm and height 15 cm. Find its volume, correct to 3 significant figures.

Pyramids and Cones — the "⅓" shapes

Both pyramids and cones taper to a single point (an apex) instead of continuing at a constant cross-section like a prism. Imagine three identical pyramids fitting together to exactly fill a cuboid of the same base and height — that's genuinely where the comes from. So the pattern is simple: take the "prism version" of the shape's volume formula, and multiply by ⅓.

A cone is just a pyramid with a circular base, so its formula is the pyramid formula with A = πr² substituted in.

Pyramid
V = ⅓ × A × h
Volume = ⅓ × base area × perpendicular height
Given in the exam
Cone
V = ⅓ × πr²h
Volume = ⅓ × π × radius² × perpendicular height
Given in the exam
The #1 mistake here
The h in these formulas must be the perpendicular height (straight up from the base to the apex) — never the slant height (the length of the sloping edge). Exam diagrams often give you both, so double check which one you're plugging in!
Practice Question
A cone has base radius 6 cm and perpendicular height 10 cm. Find its volume to 3 s.f.

Spheres

A sphere (a perfectly round ball) has just one formula to know for volume. Unlike the shapes above, it doesn't come from a simple "base × height" idea — but you're always given it in the exam, so you just need to be confident substituting into it.

Formula
V = (4/3)πr³
Volume = four-thirds × π × radius cubed
Given in the exam
Memory tip
Cube the radius, not the diameter! If you're given the diameter, halve it first, then cube it.
Practice Question
Find the volume of a sphere with radius 9 cm, giving your answer to 3 s.f.
Topic 2 — Problem-Solving with Volumes

In the exam, shapes are almost never a "pure" cuboid or cone sitting on their own. Instead, they tend to fall into one of three disguises. Spotting which disguise you're dealing with is honestly half the battle — once you know which category a question falls into, the maths itself is usually just the formulas from Topic 1.

Disguise 1: It's secretly a prism

If the 3D shape has an L-shaped, cross-shaped, or otherwise "compound" 2D cross-section running through it, treat it exactly like a prism: find the area of that (possibly weird) cross-section by splitting it into rectangles/triangles, then multiply by the length.

Worked Example (from the notes)
A step-shaped prism has a cross-section that can be split into a 7×4 rectangle and a (9−4)×2 rectangle, and the prism is 10 cm long.

Cross-sectional area = (7 × 4) + [(9 − 4) × 2] = 28 + 10 = 38 cm²
Volume = 38 × 10 = 380 cm³

Disguise 2: It's a fraction of a standard shape

Sometimes you only get part of a shape — most commonly a hemisphere (exactly half a sphere — think of a bowl). Just calculate the volume of the "full" version of the shape, then take the fraction you need.

Hemisphere
V = ½ × (4/3)πr³ = (2/3)πr³
A hemisphere is exactly half a sphere

A trickier version of this is a frustum — a cone or pyramid with its pointed top sliced off (imagine a lampshade, or a bucket). To find its volume, you find the volume of the full, un-sliced cone/pyramid, then subtract the volume of the small cone/pyramid that was removed from the top.

Frustum
V = V(large cone) − V(small cone)
Worked Example (from the notes)
A frustum comes from a cone of height 30 cm, radius 20 cm, with the top 15 cm (radius 10 cm) sliced off.

Large cone: V = ⅓ × π × 20² × 30 = 4000π = 12 566.37... cm³
Small cone: V = ⅓ × π × 10² × 15 = 500π = 1570.80... cm³
Frustum = 4000π − 500π = 3500π = 10 995.57... ≈ 11 000 cm³ (3 s.f.)

Disguise 3: It's a compound object

This is when two (or more) standard 3D shapes are stuck together — like an ice cream cone with a hemisphere of ice cream on top, or a pencil (cylinder + cone). Just find the volume of each separate solid using the formulas from Topic 1, then add them together.

Exam strategy
Before calculating anything, jot down a one-line plan, e.g. "find volume of cone, find volume of hemisphere, add together." This stops you rushing into the wrong operation under time pressure, and it also means you can pick up method marks even if your final number is slightly off.
Practice Question
A toy is made from a hemisphere of radius 4 cm sitting on top of a cylinder of radius 4 cm and height 10 cm. Find the total volume, to 3 s.f.
Topic 3 — Surface Area

Flat-faced solids: cubes, cuboids, prisms, pyramids

Surface area is simply the total area you'd need to wrap around the outside of a shape — it's a 2D idea (area) being applied to every face of a 3D object. For any solid made of flat faces, the strategy is always the same:

  • Imagine (or sketch) the net — the shape "unfolded" flat.
  • Work out the area of each individual face.
  • Add every face's area together.

For example, a square-based pyramid (with the apex directly above the centre of the base) unfolds into a net of one square base plus four identical isosceles triangles. You'd calculate the area of the square, calculate the area of one triangle, then add the square to four lots of the triangle.

Good news
You need to remember the rectangle area formula yourself (length × width), but the formula for the area of a triangle is always given to you in the exam.
Practice Question
A closed cuboid box has dimensions 5 cm × 4 cm × 3 cm. Find its total surface area.

Cylinders

A cylinder has two flat circular faces (top and bottom) and one curved surface wrapped around the middle. If you unroll that curved surface, it flattens out into a rectangle — its height is the same as the cylinder's height, and its width is exactly the circumference of the circular base (2πr), because that's the distance it has to wrap around.

Curved surface area
A = 2πrh
Given in the exam
Total surface area (with both circular ends)
A(total) = 2πrh + 2πr²
NOT given — you build this yourself from the curved area + 2 circles
Read the question carefully
Some exam questions describe an open cylinder (like a tin with no lid, or a pipe) — in that case you only add one circle, not two. Always check whether the shape is open, closed, or open at both ends before deciding how many circles to include.
Practice Question
A closed cylindrical tin has radius 7 cm and height 12 cm. Find its total surface area to 3 s.f.

Cones

A cone has one flat circular base and one curved surface. If you unroll the curved surface, it becomes a sector (a "pizza slice" shape) of radius equal to the slant height, lnot the perpendicular height. This is the same trap as with volume, but even more important here, because the curved surface area formula only works with the slant height.

Curved surface area
A = πrl
l = slant height (the sloping edge), not the perpendicular height
Given in the exam
Total surface area (with the base)
A(total) = πrl + πr²
NOT given — build it from curved area + base circle
If you're only given the height
If a question gives you the perpendicular height instead of the slant height, use Pythagoras' theorem first: l² = r² + h², since the radius, perpendicular height, and slant height form a right-angled triangle inside the cone.
Practice Question
A cone has radius 6 cm and slant height 10 cm (no base included — it's open). Find its curved surface area.

Spheres and Hemispheres

A sphere has just one continuous curved surface — no flat faces at all, since it's perfectly round all the way around.

Sphere
A = 4πr²
Given in the exam

A hemisphere is exactly half of a sphere sliced through the middle. That means it has half the curved surface area of a full sphere — plus a brand new flat circular face where it was sliced (which a full sphere doesn't have at all). Don't forget this flat circle — it's the single most commonly forgotten piece in the whole surface area topic!

Hemisphere (total)
A = 2πr² + πr²
Half the sphere's curved area (2πr²) + the flat circular base (πr²)
NOT given — you must build this yourself
Worked Example (from the notes)
A toy = a cone (radius 5 cm, slant height 12 cm) sitting on top of a hemisphere of the same radius. Find the total surface area to 3 s.f.

Cone's curved area: A = πrl = π × 5 × 12 = 60π
Hemisphere's curved area: A = (4πr²)/2 = (4π×5²)/2 = 50π
(No flat circle is added here — it's hidden inside the toy where the cone meets the hemisphere!)
Total = 60π + 50π = 110π = 345.575... ≈ 346 cm² (3 s.f.)
Practice Question
Find the total surface area of a solid hemisphere (flat face included) with radius 6 cm, to 3 s.f.
What to Memorise

Everything tagged Given is printed on your formula sheet — don't waste energy memorising it, just practice using it. Everything tagged Not given you genuinely need in your head.

Cuboid Volume
V = lwh
Memorise
Prism Volume
V = A × l
Given
Cylinder Volume
V = πr²h
Given
Pyramid Volume
V = ⅓Ah
Given
Cone Volume
V = ⅓πr²h
Given
Sphere Volume
V = (4/3)πr³
Given
Hemisphere Volume
V = (2/3)πr³
Build it: ½ of sphere
Cylinder Curved S.A.
A = 2πrh
Given
Cylinder Total S.A.
A = 2πrh + 2πr²
Memorise
Cone Curved S.A.
A = πrl
Given
Cone Total S.A.
A = πrl + πr²
Memorise
Sphere Surface Area
A = 4πr²
Given
Hemisphere Total S.A.
A = 2πr² + πr²
Memorise
Frustum Volume
V(big) − V(small)
Method to remember
Concepts Checklist
Exam Tips — Common Mistakes & Mark-Scheme Traps
Slant height vs. perpendicular height
This is the #1 cone mistake. Volume always needs the perpendicular height. Curved surface area always needs the slant height. Mixing them up is one of the most common ways students lose marks — always label which one you've been given, or work out the missing one using Pythagoras (l² = r² + h²) before you substitute.
Radius vs. diameter
If a question gives the diameter, you must halve it to get the radius before plugging into any formula. Examiners deliberately give diameters to catch students out — always check which one you've been handed.
Mismatched units
Always check every length is in the same unit before you calculate — mixing metres and centimetres in the same formula is an extremely common error. Convert everything first, then calculate.
Forgetting hidden/extra faces
In compound shapes, faces where two solids join are usually hidden and should NOT be included in the surface area (like where the cone meets the hemisphere in the toy example). But don't forget genuinely new faces that appear, like the flat circular base of a hemisphere that a full sphere doesn't have. Always sketch or picture the actual outer surface before adding areas.
Premature rounding
Keep your answer in terms of π (or use full calculator display) throughout your working, and only round at the very last step, to the number of significant figures asked for. Rounding too early stacks up small errors that examiners can penalise.
What examiners look for
  • A clear, labelled method — write down which formula you're using before you substitute numbers
  • Correct substitution shown as a step (not just a jump straight to the final answer)
  • Units included in your final answer (cm³ for volume, cm² for surface area)
  • Answers rounded to the exact degree of accuracy the question asks for — no more, no less
  • For "show that" style questions, keeping exact values (in terms of π) rather than rounding decimals
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