This note covers volume and surface area of common solids (prisms, pyramids, cylinders, cones, spheres and combinations of them), the use of right-angled trigonometry in three dimensions, and unit conversions for capacity. These skills are needed for packaging, tanks, buildings and any design problem involving real objects.
Key ideas
Volume measures the space inside a solid, in cubic units. Surface area is the total area of all the outer faces, in square units. Most formulae are given in the formula booklet, but you must know which to use and how to rearrange them.
- Prism (including cuboid and cylinder): volume = (area of cross-section) \(\times\) (length). For a cylinder, \(V = \pi r^2 h\).
- Pyramid and cone: volume = \(\dfrac{1}{3} \times\) (area of base) \(\times\) (vertical height). For a cone, \(V = \dfrac{1}{3}\pi r^2 h\).
- Sphere: \(V = \dfrac{4}{3}\pi r^3\) and surface area \(= 4\pi r^2\). A hemisphere has half of each, but its flat circular face is extra if the solid is closed.
- Cylinder surface area: curved surface \(2\pi rh\), plus \(\pi r^2\) for each flat end.
- Cone surface area: curved surface \(\pi rl\), where \(l\) is the slant height, plus \(\pi r^2\) for the base. Find \(l\) from \(l^2 = r^2 + h^2\).
- Pyramid surface area: base area plus the area of each triangular face. The height of each face is the slant height, not the vertical height of the pyramid.
Composite solids. Break the object into standard solids. For volume, add (or subtract) the volumes. For surface area, add only the faces that are on the outside. Where two solids join, the faces that touch are hidden and must not be counted.
Units. Convert lengths before you find areas and volumes. Useful conversions: \(1\ \text{m}^3 = 1000\ \text{L}\), \(1\ \text{L} = 1000\ \text{cm}^3\) and \(1\ \text{cm}^3 = 1\ \text{mL}\). Note that \(1\ \text{m}^3 = 1\,000\,000\ \text{cm}^3\), because the length factor 100 is cubed.
Three-dimensional trigonometry. To find lengths and angles in a solid, pick out right-angled triangles inside it. Pythagoras and \(\sin\), \(\cos\), \(\tan\) then do the work. Redraw each triangle flat, and mark the right angle.
- The angle between a line and a plane is the angle between the line and its shadow on the plane (its projection). For a line from the apex of a pyramid to a base corner, this is the angle between the edge and the line from the corner to the centre of the base.
- The angle between two planes is found by drawing a line in each plane perpendicular to the line where they meet. For a pyramid face and its base, use the vertical height and the perpendicular distance from the centre of the base to the edge.
- The space diagonal of a cuboid with sides \(a\), \(b\), \(c\) is \(\sqrt{a^2 + b^2 + c^2}\). It comes from using Pythagoras twice.
When a problem is in a real context, give the final answer with sensible units and accuracy, and check that it is reasonable. For example, a tank holding a few litres is not 5000 m³.
Definitions and formulae
| Term or formula | Meaning / when to use |
|---|---|
| \(V = \pi r^2 h\) | Volume of a cylinder. |
| \(V = \dfrac{1}{3}\pi r^2 h\) | Volume of a cone, \(h\) the perpendicular height. |
| \(V = \dfrac{1}{3}Ah\) | Volume of any pyramid, base area \(A\). |
| \(V = \dfrac{4}{3}\pi r^3\) | Volume of a sphere. |
| \(A = 4\pi r^2\) | Surface area of a sphere. |
| \(A = 2\pi rh + 2\pi r^2\) | Total surface area of a closed cylinder. |
| \(A = \pi rl + \pi r^2\) | Total surface area of a closed cone, \(l\) the slant height. |
| \(l = \sqrt{r^2 + h^2}\) | Slant height of a cone. |
| \(d = \sqrt{a^2 + b^2 + c^2}\) | Space diagonal of a cuboid. |
| \(1\ \text{m}^3 = 1000\ \text{L}\) | Capacity conversion. |
Worked examples
Example 1 A grain store is a cylinder of radius 3 m and height 8 m with a hemisphere of the same radius on top. The base is flat and closed. Find (a) the volume in litres, (b) the total outside surface area.
(a) Cylinder: \(\pi(3)^2(8) = 72\pi\) m³. Hemisphere: \(\dfrac{1}{2}\times\dfrac{4}{3}\pi(3)^3 = \dfrac{2}{3}\pi(27) = 18\pi\) m³.
Total volume \(= 72\pi + 18\pi = 90\pi = 282.74\ldots\) m³. In litres: \(282.74 \times 1000 = 282\,743\) L.
(b) The outside surface has three parts. Curved side of cylinder: \(2\pi(3)(8) = 48\pi\). Flat base: \(\pi(3)^2 = 9\pi\). Curved surface of hemisphere: \(\dfrac{1}{2}(4\pi)(3)^2 = 18\pi\).
The top of the cylinder is joined to the hemisphere, so it is not counted. Total \(= 48\pi + 9\pi + 18\pi = 75\pi = 235.6\ldots\) m².
Answer: (a) 283 000 L (3 s.f.); (b) 236 m².
Example 2 A tent is a square-based pyramid. The base has side 6 m and the vertical height is 4 m. Find (a) the volume, (b) the area of fabric in the four sloping faces, (c) the angle between a sloping edge and the base, (d) the angle between a sloping face and the base.
(a) Base area \(= 36\) m². \(V = \dfrac{1}{3}(36)(4) = 48\) m³.
(b) The slant height of a face runs from the middle of a base edge to the apex. The distance from the centre of the base to the middle of an edge is 3 m. So \(l = \sqrt{3^2 + 4^2} = 5\) m.
Each face has area \(\dfrac{1}{2}(6)(5) = 15\) m². Four faces: \(60\) m².
(c) The base diagonal is \(\sqrt{6^2 + 6^2} = \sqrt{72}\), so the centre of the base is \(\dfrac{\sqrt{72}}{2} = 4.243\) m from a corner.
In the right-angled triangle with the vertical height (4 m) and this distance, \(\tan\theta = \dfrac{4}{4.243}\), so \(\theta = 43.3^\circ\).
(d) The right-angled triangle uses the vertical height (4 m) and the distance 3 m from the centre to the middle of an edge. \(\tan\phi = \dfrac{4}{3}\), so \(\phi = 53.1^\circ\).
Answer: (a) 48 m³; (b) 60 m²; (c) 43.3°; (d) 53.1°.
Example 3 A spherical water tank has an internal diameter of 2.4 m. (a) Find its capacity in litres. (b) It is filled at 25 litres per minute. How long does it take to fill, in hours and minutes? (c) A cylindrical tank of radius 0.8 m is to hold the same volume. How tall must it be?
(a) The radius is 1.2 m. \(V = \dfrac{4}{3}\pi(1.2)^3 = 7.238\ldots\) m³ \(= 7238\) L.
(b) Time \(= \dfrac{7238.2}{25} = 289.5\) minutes. This is \(\dfrac{289.5}{60} = 4.825\) hours, which is 4 hours 50 minutes to the nearest minute.
(c) Use \(V = \pi r^2 h\) with \(V = 7.2382\) m³ and \(r = 0.8\): \(h = \dfrac{7.2382}{\pi(0.64)} = 3.600\) m.
Answer: (a) 7240 L; (b) about 4 h 50 min; (c) 3.60 m.
Common errors
- Using the slant height in a volume formula (for a cone or pyramid) instead of the vertical height.
- Counting the faces where two solids join when finding the surface area of a composite solid.
- Using the diameter instead of the radius in a sphere, cylinder or cone formula.
- Converting units incorrectly, for example treating \(1\ \text{m}^3\) as 100 L instead of 1000 L.
- Finding an angle in a 3D diagram using a triangle that does not have a right angle, or with the wrong two sides.
Exam tip: Draw each right-angled triangle you use as a separate flat sketch, with its sides labelled and the right angle marked. Then choose the correct ratio from the sketch.
Check yourself
- A cone has radius 5 cm and vertical height 12 cm. Find its volume and its total surface area.
Answer
\(V = \dfrac{1}{3}\pi(25)(12) = 100\pi = 314\) cm³. The slant height is \(\sqrt{25 + 144} = 13\) cm. Surface area \(= \pi(5)(13) + \pi(25) = 90\pi = 283\) cm². - Find the volume and surface area of a sphere of radius 2.5 cm.
Answer
\(V = \dfrac{4}{3}\pi(2.5)^3 = 65.4\) cm³. \(A = 4\pi(2.5)^2 = 78.5\) cm². - A cuboid measures 8 cm by 6 cm by 5 cm (height). Find the length of the space diagonal and the angle it makes with the base.
Answer
The base diagonal is \(\sqrt{64 + 36} = 10\) cm. The space diagonal is \(\sqrt{100 + 25} = 11.2\) cm. The angle with the base satisfies \(\tan\theta = \dfrac{5}{10}\), so \(\theta = 26.6^\circ\). - A cylindrical cup has radius 4 cm and height 12 cm. How many millilitres does it hold?
Answer
\(V = \pi(4)^2(12) = 192\pi = 603\) cm³, and \(1\ \text{cm}^3 = 1\) mL, so it holds 603 mL.