Know the booklet formulas: Volume and surface-area formulas for the standard solids are in the formula booklet — but you must know which to use and read the right radius / height.
Tap each solid to see it in 3D with its volume formula — cuboid, cylinder, cone, sphere, hemisphere (½ a sphere) and pyramid.
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What's given vs what you memorise: ✓ in the booklet — you can look it up in the exam.
★ not given — memorise it, or build it from the given parts (add a base circle, halve a sphere, sum the faces…).
| Solid | Volume | Surface area |
|---|---|---|
| Cuboid | ✓ | ★ |
| Cylinder | ✓ | curved ✓; closed adds ★ |
| Cone | ✓ | curved ✓; total adds base ★ |
| Sphere | ✓ | ✓ |
| Hemisphere | ★ | ★ (½ sphere + flat base) |
| Pyramid | ✓ (A = base area) | ★ |
| Prism | ✓ | sum of all faces ★ |
The given 2D areas, drawn — so you can see what b, h, a and r mean. These shapes are the bases / cross-sections that feed the 3D formulas above.
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| Shape | Area — also given | Where you use it |
|---|---|---|
| Parallelogram | cross-section of a slanted prism | |
| Triangle | the base A of a triangular prism / a pyramid | |
| Trapezoid | cross-section of a trough / trapezoidal prism | |
| Circle | a cylinder or cone base, or a hemisphere's flat face | |
| Circle — rim | the distance around the edge (e.g. how far a wheel rolls in one turn) |
Where the 2D area formulas get used: a prism (or cylinder) is its cross-section pushed along a length, so V = cross-section area × length. Tap a cross-section — a parallelogram, triangle or trapezoid builds a prism; a circle builds a cylinder; and a circle's rim (C = 2πr) is how far a wheel rolls in one full turn.
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Worked example — volume of a cylinder
A can has radius 4 cm and height 10 cm.
Find its volume.
Step by step
- Write the cylinder volume formula, where r is the radius and h is the height.
- Substitute r = 4 and h = 10.
- Evaluate and round to 3 s.f.
Final answer
Volume = 160π ≈ 502 cm³ (3 s.f.)
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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 |
Each solid also shows its surface-area formula (the A = … panel) — add up every face, and for a closed cylinder include both circular ends.
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Worked example — surface area of a sphere
Find the surface area of a sphere with radius 5 cm.
Step by step
- Write the sphere surface area formula, where r is the radius.
- Substitute r = 5.
- Evaluate and round to 3 s.f.
Final answer
Surface area = 100π ≈ 314 cm².
Worked example — surface area of a capsule
A capsule is a cylinder of radius r and height h with a hemisphere on each end.
Find a formula for its total surface area S.
Step by step
- The curved side of the cylinder is 2πrh. The two hemispheres are domes only — their flat faces sit against the cylinder, so they aren't exposed.
- Two hemisphere domes together make one whole sphere's surface.
- Add only the exposed surfaces (no flat circles — they are internal joins).
Final answer
S = 2πrh + 4πr².
The capsule: the curved cylinder side (2πrh) plus the two domes, which together make one whole sphere (4πr²). The flat circles where the domes meet the cylinder are internal joins — not part of the outside surface.
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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.
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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.
Just this section's two solids — the cone and the pyramid. Both use a slant height l; tap to compare. (No need for the other solids here.)
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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. The slant height l is the diagonal from tip to base edge, found by Pythagoras on the radius r and vertical height h.
- Substitute r = 3 and h = 4.
- Evaluate.
- (b) Volume. Write the cone volume formula (one third of a cylinder), using the vertical height h.
- Substitute r = 3 and h = 4.
- Evaluate and round to 3 s.f.
- (c) Total surface area. Write the closed-cone formula — base circle plus curved side, where l is the slant height.
- Substitute r = 3 and l = 5.
- Evaluate and round to 3 s.f.
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. Write the cylinder volume formula, where r is the radius and h is the height.
- Substitute r = 4 and h = 12.
- Evaluate and round to 3 s.f.
- Total surface area. Write the closed-cylinder formula — two circular ends plus the curved side.
- Substitute r = 4 and h = 12.
- Evaluate and round to 3 s.f.
Final answer
Volume ≈ 603 cm³; metal area needed ≈ 402 cm².
The can from the question (r = 4, h = 12) — a closed cylinder: V = πr²h and SA = 2πr² + 2πrh.
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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) Write the solid-hemisphere surface area formula — a curved dome 2πr² plus a flat circular base πr², so 3πr².
- Substitute r = 15.
- Evaluate.
- (b) Mass of glaze is proportional to the area covered: write it as the area divided by the coverage rate c (the area covered per gram).
- Substitute A = 2120.6 and c = 200.
- Evaluate and round to 3 s.f.
Final answer
(a) A = 675π ≈ 2120 mm². (b) ≈ 10.6 g of glaze.
The chocolate dome from the question (r = 15 mm) — a solid hemisphere: curved 2πr² + flat base πr² = 3πr².
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Hollow cylinder (a pipe): A pipe is a cylinder with a smaller cylinder removed (the hole). Picture the ring-shaped end: outer radius R, inner radius r.
Volume of metal = big cylinder − hole = .
IB-style question — volume of metal in a pipe
A metal pipe is a hollow cylinder with outer radius 5 cm, inner radius 3 cm and length 20 cm.
Find the volume of metal in the pipe.
Step by step
- The metal is the outer cylinder with the inner cylinder (the hole) taken out — so subtract the two volumes. Both share the same length h, so factorise.
- Substitute R = 5, r = 3 and h = 20.
- Evaluate.
Final answer
V = 320π ≈ 1005 cm³ of metal.
The pipe end is a ring (an annulus): its area is the big circle minus the hole, πR² − πr². That ring runs the full length h, so the volume of metal is π(R² − r²)h.
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