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5b · E5.4–E5.5

Surface area and volume

Topic 5 Mensuration · Core and Extended · Papers 1–4

🎯What you need to be able to do

  • Find the volume and surface area of cuboids, prisms and cylinders.
  • Use the given formulas for spheres, pyramids and cones.
  • Solve problems with compound solids and parts of solids, such as a hemisphere, and give answers in terms of \( \pi \).
  • Find the volume and surface area of a frustum EXTENDED.
  • Rearrange a formula to find a missing length from a volume or area.

📚The mathematics

Prisms and cylinders

A prism is any solid with a uniform cross-section, and its volume is cross-sectional area × length. A cylinder is a prism with a circular cross-section, and so is a slice of a cylinder (a “cylindrical sector”). Surface area is the area of all the faces: for a closed cylinder, two circles plus the curved surface \( 2\pi rh \) — the curved surface unrolls into a rectangle \( 2\pi r \) long and \(h\) high.

Formulas given in the exam

cylinder: curved SA \( 2\pi rh \), volume \( \pi r^{2}h \)
cone: curved SA \( \pi rl \), volume \( \tfrac{1}{3}\pi r^{2}h \)
sphere: SA \( 4\pi r^{2} \), volume \( \tfrac{4}{3}\pi r^{3} \)
pyramid: volume \( \tfrac{1}{3} \times \) base area \( \times h \)

Two things the list does not do for you: it gives only the curved surface, so add the flat faces yourself; and the cone’s \(l\) is the slant height, which you often find with Pythagoras from \(r\) and the vertical height \(h\).

Left: a cylinder of radius 4 cm and height 10 cm, with its curved surface shown unrolled into a rectangle 8 pi by 10. Right: a cone of radius 6 cm and vertical height 8 cm, with the slant height l = 10 cm found from the right-angled triangle inside it.
The worked example’s solids. The cone’s slant height comes from Pythagoras: \( l = \sqrt{6^{2} + 8^{2}} = 10 \).

Parts of solids and compound solids

A hemisphere is half a sphere: volume \( \tfrac{2}{3}\pi r^{3} \), curved surface \( 2\pi r^{2} \), plus a flat circle \( \pi r^{2} \) if it is solid. For a compound solid add the volumes, but for the surface area only count the faces on the outside.

EXTENDED A frustum is a cone with its top cut off parallel to the base: its volume is the large cone minus the small cone. The two cones are similar, which gives the missing dimensions (topic 4b).

A cone of radius 6 cm and height 12 cm with its top 6 cm, a small cone of radius 3 cm, cut off by a horizontal plane. The remaining frustum is shaded; its volume is the large cone minus the small cone, 144 pi minus 18 pi = 126 pi cubic centimetres.
Frustum = large cone \( - \) small cone \( = 144\pi - 18\pi = 126\pi \) cm³.

✏️Worked example

(a) (Non-calculator.) A closed cylinder has radius 4 cm and height 10 cm. Find its volume and total surface area in terms of \( \pi \). [3] (b) (Non-calculator.) A cone has radius 6 cm and vertical height 8 cm. Find its volume and curved surface area in terms of \( \pi \). [3] (c) (Calculator.) EXTENDED The cone in the diagram above has height 12 cm and radius 6 cm; the top 6 cm is cut off. Find the volume of the frustum. [3]

(a) \( V = \pi(4^{2})(10) = 160\pi \) cm³. SA \( = 2\pi(4^{2}) + 2\pi(4)(10) = 32\pi + 80\pi = 112\pi \) cm².

(b) \( V = \tfrac{1}{3}\pi(36)(8) = 96\pi \) cm³. \( l = \sqrt{36 + 64} = 10 \), so curved SA \( = \pi(6)(10) = 60\pi \) cm².

(c) The small cone is similar with scale factor \( \tfrac{1}{2} \), so its radius is 3 cm. Volume \( = \tfrac{1}{3}\pi(6^{2})(12) - \tfrac{1}{3}\pi(3^{2})(6) = 144\pi - 18\pi = 126\pi = 396 \) cm³ (3 s.f.).

Check it. Halving every length divides the volume by \( 2^{3} = 8 \): \( 144\pi \div 8 = 18\pi \) ✓. So the frustum is \( \tfrac{7}{8} \) of the whole cone, even though it is only half the height.
Using the vertical height for \(l\). \( \pi rh = 48\pi \) is the most common wrong answer to (b). The curved surface uses the slant height.

📝Practise

Questions in the style of the current papers. EXTENDED marks Extended-only content.

1. (Non-calculator.) A triangular prism is 15 cm long. Its cross-section is a triangle with base 6 cm and height 4 cm. Find its volume. [2]
Cross-section \( \tfrac{1}{2} \times 6 \times 4 = 12 \) cm²; volume \( 12 \times 15 = 180 \) cm³.
2. (Non-calculator.) Find the volume of a cylinder of radius 3 cm and height 7 cm, in terms of \( \pi \). [1]
\( \pi \times 9 \times 7 = 63\pi \) cm³.
3. (Non-calculator.) A pyramid has a square base of side 8 cm and a vertical height of 9 cm. Find its volume. [2]
\( \tfrac{1}{3} \times 64 \times 9 = 192 \) cm³.
4. (Calculator.) A sphere has radius 4.5 cm. Find its surface area and its volume. [2]
SA \( = 4\pi(4.5)^{2} = 254 \) cm²; \( V = \tfrac{4}{3}\pi(4.5)^{3} = 382 \) cm³ (3 s.f.).
5. (Calculator.) Find the total surface area of a solid hemisphere of radius 5 cm. [3]
Curved \( 2\pi(25) = 50\pi \); flat circle \( 25\pi \); total \( 75\pi = 236 \) cm².
6. (Calculator.) A cuboid tank measures 40 cm by 30 cm by 25 cm. How many litres does it hold when full? [2]
\( 40 \times 30 \times 25 = 30\,000 \) cm³ \( = 30 \) litres.
7. (Calculator.) EXTENDED A cone has volume 300 cm³ and height 10 cm. Find its radius. [3]
\( \tfrac{1}{3}\pi r^{2}(10) = 300 \Rightarrow r^{2} = \dfrac{90}{\pi} = 28.6\ldots \Rightarrow r = 5.35 \) cm.
8. (Calculator.) EXTENDED A solid metal sphere of radius 3 cm is melted and recast as a cylinder of radius 2 cm. Find the height of the cylinder. [3]
Volume is unchanged: \( \tfrac{4}{3}\pi(27) = 36\pi \). \( \pi(4)h = 36\pi \Rightarrow h = 9 \) cm.

🔗Go deeper — other people’s work

These are external resources, not mine. If one stops working, tell me and everything above it on this page still stands.

  • GeoGebra 3D — unfold a cylinder or cone into its net
  • Corbettmaths — volume and surface area of cones, spheres and frustums