IB Maths AI HL Geometry of 3D Shapes Paper 1 & 2 Cylinder, cone, sphere, pyramid ~10 min read

Volume & Surface Area

All the standard 3D shape formulas live in the formula booklet, so the work isn’t memorising them — it’s choosing the right one and applying it cleanly. For composite shapes (like an ice-cream cone topped with a scoop), split into known pieces, compute each, then add. For surface area, be careful which faces are exposed and which are hidden at joins.

📘 What you need to know

Prism / cuboid / cylinder volume: V = (cross-section area) × (length). For cuboid: V = lwh; cylinder: V = πr²h.
Pyramid / cone volume: V = (1/3)×(base area)×(height). Cone: V = (1/3)πr²h.
Sphere volume: V = (4/3)πr³. Sphere surface area: A = 4πr².
Cylinder surface area (closed): A = 2πr² + 2πrh — two circular ends plus the curved side.
Cone surface area (closed): A = πr² + πrl — circular base plus curved face, where l is the slant height.
Cone Pythagoras: slant height l, vertical height h, base radius r satisfy l² = h² + r².
Composite shapes: split into known pieces, add volumes. For surface area, exclude any faces hidden at the joins.

The shape formulas at a glance

Most volume formulas in this section live in the formula booklet under “prior learning” or “geometry & trigonometry”, so the work isn’t memorisation. The pattern to notice is that prisms (cuboid, cylinder, any uniform cross-section) use V = base area × length, while pyramids and cones (anything tapering to a point) use V = (1/3) × base area × height. The factor 1/3 is the key difference. Spheres get their own formulas, since they have no flat bases.

The four shapes you’ll see most often. Cylinder and cone use the same base shape (circle), but cone gets the (1/3) factor. Sphere and pyramid have their own formulas; for pyramids, Pythagoras gives the slant height needed for surface area.

Key 3D formulas cylinder: V = πr²h · cone: V = (1/3)πr²h · sphere: V = (4/3)πr³ cylinder SA = 2πr²+2πrh · cone SA = πr²+πrl · sphere SA = 4πr²

Composite shapes and Pythagoras

For shapes built from two or more known pieces (a cylinder topped with a hemisphere, a cone with a sphere scoop, a pyramid on a cuboid), compute each piece’s volume separately and add. For surface area, identify which faces are exposed — flat circles at joins are hidden and don’t count, while curved surfaces and any other free faces do. The cone slant height l rarely appears directly in a problem; instead you’re often given the vertical height h and base radius r, and you find l from l² = h²+r². Similarly, in pyramid surface-area problems, Pythagoras inside the pyramid gives the slant from apex to the midpoint of a base edge.

Composite SA checklist: walk around the outside of the composite shape and list each exposed face. Each circular cross-section at a join is hidden — don’t double-count by adding both the cylinder top and the hemisphere flat face. Flat bottoms that touch the ground are usually still exposed (they’re the outside of the shape).

🧭 Recipe — volume & surface area

Identify the shape(s): cylinder, cone, sphere, pyramid, prism. For composites, list each piece.
Pull the formulas from the booklet and write them down with the right symbols.
Substitute carefully: keep π symbolic where possible for exact-form answers.
For surface area, list each exposed face and add. Exclude faces hidden at joins.
Use Pythagoras inside the shape when you need a slant height: cone slant l² = h²+r²; pyramid slant by similar in-shape triangle.

Worked examples

WE 1

Cylinder volume

A cylinder has radius 4 cm and height 10 cm. Find its volume in exact form and to 3 s.f.

apply V = πr²h V = π(4²)(10) = π(16)(10) = 160π V = 160π cm³ (exact) decimal 160π ≈ 502.65 V ≈ 503 cm³ (3 s.f.)

WE 2

Cone total surface area

A cone has base radius 6 cm and slant height 10 cm. Find the total surface area (including the circular base) in exact form and to 3 s.f.

apply A = πr² + πrl A = π(6²) + π(6)(10) = 36π + 60π A = 96π cm² (exact) decimal 96π ≈ 301.59 A ≈ 302 cm² (3 s.f.)

WE 3

Sphere: given surface area, find volume

A sphere has surface area 100π cm². Find its volume in exact form.

use A = 4πr² to find r 4πr² = 100π ⇒ r² = 25 ⇒ r = 5 apply V = (4/3)πr³ V = (4/3)π(5³) = (4/3)π(125) V = 500π/3 cm³ (exact) decimal V ≈ 524 cm³ (3 s.f.)

WE 4

Composite: cylinder + hemisphere (water tank)

A water tank is made up of a cylinder of radius 5 m and height 12 m with a hemispherical dome of the same radius on top. (a) Find the total volume in exact form and to 3 s.f. (b) Find the total external surface area (curved side, dome, and flat bottom) in exact form and to 3 s.f.

(a) volume of cylinder V_cyl = π(25)(12) = 300π volume of hemisphere V_hemi = (1/2)(4/3)π(125) = 250π/3 add V_total = 300π + 250π/3 = 1150π/3 V ≈ 1200 m³ (3 s.f.) (b) external surfaces cyl curved: 2π(5)(12) = 120π hemi curved: 2π(25) = 50π flat bottom: π(25) = 25π (top of cyl + hemi flat side are hidden at join) A = 195π ≈ 613 m² (3 s.f.)

WE 5

Square pyramid: volume, slant, surface area

A right pyramid has a square base of side 6 cm and a vertical height of 4 cm above the centre of the base. Find: (a) its volume; (b) the slant height from the apex to the midpoint of a base edge; (c) its total surface area.

(a) V = (1/3)(base area)(h) = (1/3)(36)(4) V = 48 cm³ (b) slant from apex to midpoint of edge horizontal distance to mid-edge = 6/2 = 3 vertical height = 4 slant = √(3² + 4²) = √25 slant = 5 cm (3-4-5 triple!) (c) SA = base + 4 triangles each triangle: ½(6)(5) = 15 total = 36 + 4(15) = 36 + 60 SA = 96 cm²

WE 6

Composite: ice-cream cone + hemisphere scoop

An ice-cream cone has base radius 5 cm and slant height 13 cm. A hemispherical scoop of ice cream of radius 5 cm sits on top of the cone. Find: (a) the vertical height of the cone; (b) the total volume of cone + scoop in exact form and to 3 s.f.; (c) the total external surface area (curved cone + curved hemisphere) in exact form and to 3 s.f.

(a) cone height by Pythagoras l² = h² + r² ⇒ h² = 169 − 25 = 144 h = 12 cm (5-12-13 triple) (b) volume of cone V_cone = (1/3)π(25)(12) = 100π volume of hemisphere V_hemi = (1/2)(4/3)π(125) = 250π/3 add V_total = 100π + 250π/3 = 550π/3 V ≈ 576 cm³ (3 s.f.) (c) external SA cone curved: πrl = π(5)(13) = 65π hemi curved: 2πr² = 50π SA = 115π ≈ 361 cm² (3 s.f.) cone base and hemi flat side are joined and not external.

💡 Top tips

Formula booklet is your friend — volume formulas are listed; you don’t need to memorise them. But you must know how to use them correctly.
Keep π symbolic for exact-form answers; decimal only at the final step.
Hemisphere = half of sphere: volume (1/2)(4/3)πr³ = (2/3)πr³; curved SA (1/2)(4πr²) = 2πr².
For cone problems, the slant l is usually found via Pythagoras: l² = h²+r². Look for clean (3,4,5), (5,12,13), (8,15,17) triples.
For pyramid slant heights, draw the right triangle inside the pyramid: vertical leg = height, horizontal leg = half the base edge (for triangle face slant) or half the diagonal (for edge-to-apex).

⚠ Common mistakes

Confusing height h with slant l in a cone — volume uses h, surface area uses l.
Forgetting the 1/3 factor for cone or pyramid volume.
Including hidden faces in composite SA — faces at joins are not external.
Missing the flat base when “total surface area” of a cone or cylinder is asked — remember to add πr².
Using diameter instead of radius: if the problem gives a diameter, halve it first.
Mixing up volume and SA units: cm³ vs cm²; check at the end.

Chapter complete — you now have both 3D Coordinate Geometry (midpoint and distance in 3D) and Volume & Surface Area (formulas for cylinders, cones, spheres, pyramids, and composites). Together they cover the geometry of 3D shapes for AI HL.

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