This note covers the volume and surface area of cuboids, prisms and cylinders, and, at Extended level, pyramids, cones, spheres and hemispheres, including compound solids. These calculations are used to find how much a container holds and how much material is needed to cover it.
Key ideas
Volume is the amount of space inside a solid, measured in cubic units such as cm\(^3\) or m\(^3\). Surface area is the total area of all the outside faces, measured in square units.
Prisms. A prism has the same cross-section all the way along its length. This includes cuboids, triangular prisms and cylinders.
- Volume of a prism \(=\) area of cross-section \(\times\) length.
- Surface area of a prism \(=\) two cross-sections plus the area of the rectangular faces around it. The rectangle area is the perimeter of the cross-section times the length.
Cylinders. A cylinder is a prism with a circular cross-section.
- Volume \(V = \pi r^2 h\).
- Curved surface area \(= 2\pi r h\). This is a rectangle when the curved surface is unrolled: its width is the circumference and its height is \(h\).
- Total surface area \(= 2\pi r h + 2\pi r^2\), which includes the two circular ends.
Pyramids, cones and spheres (Extended).
- Pyramid: \(V = \tfrac{1}{3} \times \text{base area} \times \text{perpendicular height}\). The surface area is the base plus all the triangular faces.
- Cone: \(V = \tfrac{1}{3}\pi r^2 h\). Curved surface area \(= \pi r l\), where \(l\) is the slant height. Total surface area adds the circular base \(\pi r^2\). Use Pythagoras to find the slant height: \(l^2 = r^2 + h^2\).
- Sphere: \(V = \tfrac{4}{3}\pi r^3\). Surface area \(= 4\pi r^2\).
- Hemisphere (half a sphere): volume \(\tfrac{2}{3}\pi r^3\). Curved surface area \(2\pi r^2\). If the flat circular face is included, add \(\pi r^2\), giving \(3\pi r^2\).
Compound solids are made by joining solids, such as a cylinder with a hemisphere on top. Find the volume of each part and add. For surface area, include only the faces you can see from the outside. Where two solids join, the touching faces are hidden and are not counted.
Units of volume. \(1\ \text{cm}^3 = 1\ \text{ml}\). \(1000\ \text{cm}^3 = 1\) litre. \(1\ \text{m}^3 = 1000\) litres. Because \(1\ \text{m}^3 = 100^3 = 1\,000\,000\ \text{cm}^3\), always convert lengths first, then calculate the volume.
If you are given the volume and need an unknown length, substitute into the volume formula and rearrange. For a sphere, take a cube root at the end.
A good check on any volume is to compare it with a simple shape that fits around the solid. A cone should have exactly one third of the volume of the cylinder with the same base and height, and a pyramid one third of the matching prism. If your answer is not smaller than that cylinder or prism, look for an error.
Definitions and formulae
| Term or formula | Meaning / when to use |
|---|---|
| Prism: \(V = Al\) | \(A\) is the cross-section area and \(l\) is the length. |
| Cuboid: \(V = lwh\) | Surface area \(= 2(lw + lh + wh)\). |
| Cylinder: \(V = \pi r^2 h\) | Total surface area \(2\pi r h + 2\pi r^2\). |
| Pyramid: \(V = \tfrac{1}{3}Ah\) | \(A\) is the base area and \(h\) is the perpendicular height (Extended). |
| Cone: \(V = \tfrac{1}{3}\pi r^2 h\) | Curved surface area \(\pi r l\) with \(l = \sqrt{r^2 + h^2}\) (Extended). |
| Sphere: \(V = \tfrac{4}{3}\pi r^3\) | Surface area \(4\pi r^2\) (Extended). |
| Conversions | \(1\ \text{cm}^3 = 1\ \text{ml}\); \(1\ \text{m}^3 = 1000\) litres. |
Worked examples
Example 1 A cylindrical tin has radius 4 cm and height 10 cm. Find its volume and its total surface area.
Step 1: volume \(= \pi \times 4^2 \times 10 = 160\pi \approx 502.65\ \text{cm}^3\).
Step 2: curved surface \(= 2\pi \times 4 \times 10 = 80\pi\).
Step 3: two ends \(= 2 \times \pi \times 4^2 = 32\pi\).
Step 4: total surface area \(= 80\pi + 32\pi = 112\pi \approx 351.86\ \text{cm}^2\).
Answer: volume 503 cm\(^3\); total surface area 352 cm\(^2\) (3 s.f.).
Example 2 (Extended) A cone has base radius 9 cm and perpendicular height 12 cm. Find its volume and its total surface area.
Step 1: volume \(= \tfrac{1}{3}\pi \times 9^2 \times 12 = \tfrac{1}{3}\pi \times 972 = 324\pi \approx 1017.88\ \text{cm}^3\).
Step 2: slant height \(l = \sqrt{9^2 + 12^2} = \sqrt{225} = 15\) cm.
Step 3: curved surface area \(= \pi \times 9 \times 15 = 135\pi\).
Step 4: base \(= \pi \times 9^2 = 81\pi\).
Step 5: total surface area \(= 135\pi + 81\pi = 216\pi \approx 678.58\ \text{cm}^2\).
Answer: volume 1020 cm\(^3\); total surface area 679 cm\(^2\) (3 s.f.).
Example 3 (Extended) A storage tank is a cylinder of radius 2 m and height 5 m with a hemisphere of radius 2 m fixed on top. Find the total volume, and say how many litres it can hold.
Step 1: cylinder volume \(= \pi \times 2^2 \times 5 = 20\pi\).
Step 2: hemisphere volume \(= \tfrac{2}{3}\pi \times 2^3 = \tfrac{16\pi}{3}\).
Step 3: total \(= 20\pi + \tfrac{16\pi}{3} = \tfrac{76\pi}{3} \approx 79.587\ \text{m}^3\).
Step 4: convert. \(79.587 \times 1000 = 79\,587\) litres.
Answer: about 79.6 m\(^3\), which is about 79 600 litres (3 s.f.).
Common errors
- Using the slant height instead of the perpendicular height in the volume of a cone or pyramid, or the other way round in the curved surface area of a cone.
- Forgetting the \(\tfrac{1}{3}\) in the volume of pyramids and cones.
- Using the diameter instead of the radius in formulae with \(r\).
- Counting the hidden joining face when finding the surface area of a compound solid.
- Mixing units, such as a radius in cm and a height in m. Convert first.
Exam tip: write the formula, then substitute, then calculate. Keep \(\pi\) in your working (for example \(112\pi\)) until the last step, then round once. Say which units your final answer is in.
Check yourself
- Find the total surface area of a cuboid measuring 5 cm by 4 cm by 3 cm.
Answer
\(2(5 \times 4 + 5 \times 3 + 4 \times 3) = 2(20 + 15 + 12) = 94\ \text{cm}^2\). - (Extended) A pyramid has a square base of side 6 cm and perpendicular height 7 cm. Find its volume.
Answer
\(\tfrac{1}{3} \times 36 \times 7 = 84\ \text{cm}^3\). - (Extended) A sphere has volume \(288\pi\ \text{cm}^3\). Find its radius.
Answer
\(\tfrac{4}{3}\pi r^3 = 288\pi\), so \(r^3 = 288 \times \tfrac{3}{4} = 216\) and \(r = 6\) cm. - A cylindrical glass has radius 5 cm and is filled to a height of 12 cm. How many litres of water does it hold?
Answer
\(V = \pi \times 5^2 \times 12 = 300\pi \approx 942.5\ \text{cm}^3\). Dividing by 1000 gives about 0.942 litres.