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Volume and Surface Area of 3D Solids — Cambridge IGCSE 0580 Maths Extended (2026)
Cuboids, prisms, cylinders, cones, pyramids and spheres. Memorise the formulae from the formula sheet, but understand prism vs pyramid (with 31).
What you’ll learn
Mapped to the Cambridge IGCSE 0580 syllabus (2025-2027).
E6.5 — Calculate volumes and surface areas of cuboids, prisms, cylinders, spheres, pyramids and cones.
E6.5 — Calculate volumes and surface areas of compound shapes.
Prisms (cuboids, cylinders, triangular prisms)
V=area of cross-section×length. Surface area: net.
A prism has the same cross-section throughout its length.
Cuboid (rectangular prism): V=lwh. Surface area =2(lw+lh+wh).
Cylinder (circular prism): V=πr2h. Curved surface area =2πrh. Total surface area (closed cylinder) =2πr2+2πrh.
Triangular prism: V=(21bht)×ℓ where b,ht are the triangle base and height, ℓ is the prism length.
Worked. Cylinder radius 4cm, height 10cm.
V=π×16×10=160π≈503cm3.
Curved SA: 2π×4×10=80π≈251cm2.
Total SA: 2π×16+80π=112π≈352cm2.
Surface area technique. "Unfold" the solid into its NET; sum the areas of every face. For a cylinder: two circles + a rectangle (the curved side unrolled, 2πr wide and h tall).
A prism keeps the same cross-section along its length — volume is cross-section area × length.
Prism: V= cross-section area × length.
Cylinder: V=πr2h, curved SA =2πrh.
Cuboid SA: 2(lw+lh+wh).
Surface area: think NET, sum every face.
Pyramids and cones
V=31×base area×h. The factor 31 separates prism from pyramid.
Total SA (closed cone, with base disc): 65π+25π=90π≈283cm2.
Surface area of a square pyramid. Base + 4 triangular faces. For each triangular face you need its slant height (from base midpoint up the side).
Use the perpendicular height h for volume; the slant height l for the curved surface area of a cone.
Pyramid/cone: V=31× base area ×h.
h = perpendicular height.
Cone slant: l=r2+h2.
Cone curved SA: πrl.
Sphere and hemisphere
V=34πr3. Surface area =4πr2.
Sphere.V=34πr3,SA=4πr2.
Worked. Sphere radius 6cm.
V=34π×216=288π≈905cm3.
SA=4π×36=144π≈452cm2.
Hemisphere (half a sphere).
V=32πr3.
Curved SA =21×4πr2=2πr2. Closed hemisphere also includes the flat circular face: total SA =2πr2+πr2=3πr2.
Sphere: V=34πr3, SA=4πr2.
Hemisphere V: half of sphere.
Closed hemisphere total SA: 3πr2.
Compound solids and frustums
Add or subtract individual solid volumes. A frustum (cone with top sliced) is computed by subtracting the small cone.
Compound solid. A solid built from two or more standard pieces — e.g. a cylinder with a hemispherical cap.
Strategy.
Identify each component (cylinder + hemisphere, etc.).
Compute each volume separately.
Add (joined) or subtract (hollow / drilled).
Worked. Solid is a cylinder of radius 5cm, height 12cm, with a hemispherical cap of radius 5cm on top.
Cylinder V: π×25×12=300π.
Hemisphere V: 32π×125=3250π.
Total: 300π+3250π=3900π+250π=31150π≈1204cm3.
Frustum (truncated cone).Vfrustum=Vbig cone−Vsmall cone.
Use similar-triangle reasoning to find the small cone's height from the frustum dimensions.
Tip. When the question asks for surface area, don't double-count joined faces — they're internal and disappear in the join.
Split a compound solid into standard pieces, then add (joined) or subtract (hollowed) the volumes.
Compound solid: split into known pieces.
Add (joined) or subtract (hollow).
Frustum: big cone minus small cone.
Surface area: ignore internal joined faces.
How it’s examined
Volume / surface area appears every Paper 4 — typically a 6-8 mark compound-solid question. Cambridge gives a formula sheet with sphere, cone, pyramid formulae, but you must recognise WHICH formula to apply. Examiner reports flag forgetting the 31 factor for pyramids/cones and using slant instead of perpendicular height.
A solid cone has base radius 5cm and slant height 13cm. Find the total surface area in terms of π.
Step-by-step solution
Step 1
Total surface area = curved surface + circular base.
SA=πrℓ+πr2
Step 2
Substitute r=5, ℓ=13.
SA=π(5)(13)+π(5)2=65π+25π
Step 3
Add.
SA=90πcm2
Answer
90πcm2
Examiner tip
The examiner report flags candidates who use perpendicular height in πrℓ. Curved surface needs SLANT height; if only the perpendicular height is given, use Pythagoras to find ℓ first.
8Volume of a hemisphere
Extended• hemisphere, volume
▼
Question
Find the volume of a hemisphere of radius 6cm, giving your answer in terms of π.
Step-by-step solution
Step 1
Hemisphere is half a sphere.
V=21×34πr3=32πr3
Step 2
Substitute r=6.
V=32π(216)=144πcm3
Answer
144πcm3
9Find a missing edge given the volume
Extended• Adapted from 0580/22 May/Jun 2023 Q8• cuboid, find edge
▼
Question
A cuboid has volume 336cm3. Its length is 8cm and its width is 6cm. Find its height.
Step-by-step solution
Step 1
V=lwh.
336=8×6×h=48h
Step 2
Divide.
h=48336=7cm
Answer
h=7cm
10Cylinder volume in litres
Extended• cylinder, volume, units
▼
Question
A cylindrical water tank has radius 40cm and height 1.2m. Find the volume in litres, correct to 3 s.f. (1litre=1000cm3.)
Step-by-step solution
Step 1
Convert height to centimetres.
h=1.2m=120cm
Step 2
Apply V=πr2h.
V=π(40)2(120)=192,000πcm3
Step 3
Evaluate.
V≈603,185.8cm3
Step 4
Convert to litres by dividing by 1000.
V≈603litres
Answer
V≈603litres
Examiner tip
The examiner report flags candidates who mix metres and centimetres without converting. Standardise units BEFORE substituting.
11Volume of a compound solid (cylinder + hemisphere)
A toy consists of a cylinder of radius 4cm and height 10cm with a hemisphere of radius 4cm attached to the top. Find the total volume in terms of π.
Step-by-step solution
Step 1
Cylinder volume.
V1=π(4)2(10)=160π
Step 2
Hemisphere volume.
V2=32π(4)3=3128π
Step 3
Total.
V=160π+3128π=3480π+128π=3608πcm3
Answer
V=3608πcm3≈636.7cm3
Examiner tip
The examiner report flags candidates who add a full sphere volume instead of a hemisphere. Read the question — only HALF a sphere is attached.
12Surface area of a compound solid
Challenge• compound solid, surface area
▼
Question
A solid consists of a cylinder of radius 3cm and height 8cm with a cone of radius 3cm and slant height 5cm on top (closed at the bottom). Find the total external surface area in terms of π.
Step-by-step solution
Step 1
External surface comprises: cylinder bottom (circle), cylinder curved side, cone curved side. NO top circle of cylinder (covered by cone base) and NO cone base (interior).
Step 2
Cylinder bottom.
πr2=9π
Step 3
Cylinder curved surface.
2πrh=2π(3)(8)=48π
Step 4
Cone curved surface.
πrℓ=π(3)(5)=15π
Step 5
Total.
SA=9π+48π+15π=72πcm2
Answer
72πcm2
Examiner tip
The examiner report flags candidates who include the top of the cylinder AND the base of the cone — both are internal and must be omitted.
13Liquid-in-container word problem
Challenge• Adapted from 0580/42 Feb/Mar 2024 Q14• volume, rate, real-world
▼
Question
A cylindrical tank of radius 25cm and height 40cm is initially empty. Water flows in at a constant rate of 21litre per second. Find the time, to the nearest second, taken to fill the tank.
Step-by-step solution
Step 1
Volume of the tank.
V=π(25)2(40)=25,000πcm3
Step 2
Evaluate.
V≈78,539.8cm3=78.54litres
Step 3
Time = volume / rate.
t=0.578.54≈157.08s
Step 4
Round to the nearest second.
t≈157s
Answer
t≈157s
Examiner tip
The examiner report flags candidates who confuse cm3 with litres. 1litre=1000cm3 — always convert before dividing by the rate in litres per second.
Key Formulae — Solid Geometry
The formulae you need to memorise for solid geometry on the Cambridge IGCSE 0580 paper, with every variable defined in plain English and a note on when to use it.
Cuboid
V=lwh,SA=2(lw+lh+wh)
l,w,h
length, width, height
When to use
Rectangular box volumes and surface areas.
Cylinder
V=πr2h,SAtotal=2πr2+2πrh
r
radius of circular face
h
height
When to use
Closed cylinder. For an open one, drop one πr2 term.
Cone
V=31πr2h,SA=πr2+πrℓ
r
base radius
h
perpendicular height
ℓ
slant height
When to use
Cone volume needs perpendicular height; surface area needs slant.
Sphere
V=34πr3,SA=4πr2
r
radius
When to use
Sphere or hemisphere (halve as needed).
Pyramid (any base)
V=31×base area×h
h
perpendicular height from apex to base
When to use
Any pyramid, including square- and triangle-based.
Prism
V=cross-sectional area×length
When to use
Any prism — cross-section is constant along its length.
Key Definitions and Keywords — Solid Geometry
Definitions to memorise and the exact keywords mark schemes credit for solid geometry answers — sharpened from recent examiner reports for the 2026 0580 sitting.
Prism
Examiner keyword
A solid with two congruent parallel ends and a constant cross-section.
Cylinder
Examiner keyword
A circular prism — two parallel circular ends with a curved surface joining them.
Slant height
Examiner keyword
The distance from the apex of a cone (or pyramid) to a point on the edge of the base, measured along the surface.
Perpendicular (vertical) height
Examiner keyword
The shortest distance from the apex to the base, perpendicular to the base.
Hemisphere
Half of a sphere. Volume is 32πr3.
Common Mistakes and Misconceptions — Solid Geometry
The traps other students keep falling into on solid geometry questions — taken from recent Cambridge IGCSE 0580 examiner reports and mark schemes — and how to avoid them.
✕Using slant height in V=31πr2h for a cone
0580/42 — recurring
▼
Why it happens
Confusing slant length ℓ with perpendicular height h.
How to avoid it
Volume needs PERPENDICULAR height. Use Pythagoras to recover h from ℓ and r if needed: h=ℓ2−r2.
✕Using cm2 instead of cm3 for volume
▼
Why it happens
Speed.
How to avoid it
Volumes are in cubed units; surface areas are in squared units.
✕Including both circles in surface area for an open cylinder
▼
Why it happens
Defaulting to the closed-cylinder formula.
How to avoid it
Read the question. Open at the top: πr2+2πrh. Closed: 2πr2+2πrh.
✕Rounding π early
▼
Why it happens
Substituting π≈3.14 at step one.
How to avoid it
Keep π symbolic until the final line, or use the calculator's π key. Then round once at the end.
Practice questions
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Solid Geometry — frequently asked questions
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