All volume formulas are in the IB booklet: You do not need to memorise these — but you must be fluent in setting them up quickly.
The most common solids in IB Paper 1 and 2 are: cuboid, cylinder, cone, sphere, and pyramid.
| Solid | Volume formula | Notes |
|---|---|---|
| Cuboid | V = l × w × h | Length × width × height |
| Prism (any base) | V = Across × length | Cross-section area × length (triangular, trapezoidal, hexagonal…) |
| Cylinder | V = πr²h | π × radius² × height |
| Cone | V = ⅓πr²h | One-third of a cylinder with same r and h |
| Sphere | V = 4/3 πr³ | Depends only on radius |
| Hemisphere | V = ⅔πr³ | Half a sphere — half the volume |
| Pyramid | V = ⅓ × Abase × h | One-third × base area × perpendicular height |
[Diagram: math-solid-volume] - Available in full study mode
Worked example — volume of a cylinder
A can has radius 4 cm and height 10 cm.
Find its volume.
Step by step
- Use the cylinder volume formula.
- Calculate.
Final answer
Volume = 160π ≈ 502 cm³ (3 s.f.)
Surface area = sum of all faces: Surface area is the total area of the outer surface.
For each solid, think about all faces separately and add them up.
| Solid | Surface area formula | Notes |
|---|---|---|
| Cuboid (closed) | SA = 2(lw + lh + wh) | 3 pairs of rectangular faces |
| Cylinder (closed) | SA = 2πr² + 2πrh | 2 circles + curved side |
| Sphere | SA = 4πr² | All one curved surface |
| Hemisphere (solid) | SA = 3πr² | Curved 2πr² + flat base πr² |
| Cone (closed) | SA = πr² + πrl | Base circle + curved side; l = slant height |
[Diagram: math-solid-volume] - Available in full study mode
Worked example — surface area of a sphere
Find the surface area of a sphere with radius 5 cm.
Step by step
- Use the sphere surface area formula.
- Round to 3 s.f.
Final answer
Surface area = 100π ≈ 314 cm².
Open vs closed containers: IB questions sometimes ask for an open cylinder (no lid).
In that case, remove one circular end: SA = πr² + 2πrh.
Surface area — decide which faces actually count: Before you add areas, picture the real object and count only the surfaces that exist:
• Open top / no lid → drop one circle (an open cylinder is ).
• Solid hemisphere → the curved dome plus its flat circular base: . (A hollow dome with no base is just .)
• Composite solid (e.g. a cylinder topped by a hemisphere) → count only the exposed faces; the circle where the pieces join is internal, so it is not painted on either side.
Know your predicted grade
Take timed mock exams and get detailed feedback on every answer. See exactly where you're losing marks.
Slant height vs vertical height: For a cone, the slant height l is the distance along the curved surface from tip to base edge.
The vertical height h goes straight down from the tip.
Use Pythagoras: l² = r² + h².
Pyramids work the same way: A right pyramid has the same slant-height idea. Its slant height is the height of a triangular face — from the apex down to the midpoint of a base edge.
For a square base of side b, the vertical height h, the slant height l and half the base (b ÷ 2) form a right triangle:
.
Its volume is V = ⅓ × base area × height.
[Diagram: math-solid-volume] - Available in full study mode
Worked example — cone volume and surface area
A cone has base radius 3 cm and vertical height 4 cm.
Find its (a) slant height, (b) volume, (c) total surface area.
Step by step
- (a) Slant height.
- (b) Volume.
- (c) Total surface area.
Final answer
l = 5 cm, V ≈ 37.7 cm³, SA ≈ 75.4 cm².
Worked example — material needed for a can
A tin of food is a closed cylinder with radius 4 cm and height 12 cm.
Find the volume and the minimum area of metal sheet needed to make it.
Step by step
- Volume.
- Total surface area (closed cylinder).
Final answer
Volume ≈ 603 cm³; metal area needed ≈ 402 cm².
[Diagram: math-solid-volume] - Available in full study mode
Units matter: Volume is always in cubic units (cm³, m³).
Surface area is in square units (cm², m²).
Mixing them up in an exam loses marks.
IB-style question — a chocolate dome (surface area → mass)
A chocolate is a solid hemisphere of radius 15 mm.
(a) Find the total surface area of one chocolate.
(b) The whole surface is coated in an edible glaze. 1 gram of glaze covers 200 mm². Find the mass of glaze needed for one chocolate, correct to 3 significant figures.
Step by step
- (a) A solid hemisphere has a curved dome AND a flat circular base, so add both.
- Substitute r = 15.
- (b) Mass of glaze is proportional to the area covered: divide the area by the coverage rate.
Final answer
(a) A = 675π ≈ 2120 mm². (b) ≈ 10.6 g of glaze.
[Diagram: math-solid-volume] - Available in full study mode