Volumes and surface areas

Volume is the amount of space a solid occupies; it is measured in cubic units: cm³, dm³ (a litre), m³. Volume problems come up all the time in practice: how many litres a container holds, how much concrete a building needs, how many parts can be cast from a block of metal. The formulas fall into two large families: “cylinder-like” solids (prism, cylinder) — V = S·h, and “pointed” solids (pyramid, cone) — V = ⅓ S·h. Understand this link and you will remember half of the formulas automatically. This lesson continues “Polyhedra” and “Solids of revolution”.

§1. Basic properties of volume

Volume has the following properties:

• congruent solids have equal volumes;
• if a solid is split into parts, its volume is the sum of the volumes of the parts;
• a cube with edge 1 has volume 1 (the unit).

The second property is the main tool for complicated solids: split the solid into simple pieces or, conversely, subtract the surplus from a larger solid.
When converting units, the length unit is cubed:

Worked example

A cylindrical hole of radius 1 is drilled through a cube of edge 4, through the centres of opposite faces. Find the volume of what remains.

1. Volume of the cube: 4³ = 64.
2. The hole is a cylinder of radius 1 and height 4: π · 1² · 4 = 4π.
3. By subtracting volumes: V = 64 − 4π.

§2. Prism and cylinder

Why “base × height”? Slice a prism into very thin layers along its height: each layer equals the base, and the number of layers is proportional to the height. The base may be any shape — a triangle, a hexagon or a disc — and the formula is the same. In an oblique prism h is not the lateral edge but the perpendicular distance between the bases.
This idea is Cavalieri’s principle: if every plane parallel to the base cuts two solids in sections of equal area, the solids have equal volumes. That is why an oblique prism has the same volume as a right prism with the same base and height — just as a slanted stack of coins keeps its volume.

Worked example

A regular triangular prism has base side 6 and height 10. Find its volume.

1. The base is an equilateral triangle: S = a²√3/4 = 36√3/4 = 9√3.
2. V = S·h = 9√3 · 10 = 90√3.

Worked example

A cylindrical barrel has base radius 3 dm and height 10 dm. How many litres does it hold (π ≈ 3.14)?

1. V = πR²h = 3.14 · 9 · 10 = 282.6 dm³.
2. 1 dm³ = 1 l, so about 282.6 litres.

§3. Pyramid and cone

A prism (cylinder) with the same base and height has 3 times the volume of the pyramid (cone). You can see this by experiment: water from a conical vessel fills a cylinder with the same base and height in exactly three pourings. Geometrically, segments from a cube’s centre to its vertices split the cube into six equal pyramids — which is where the ⅓ comes from.
For a regular tetrahedron with edge a: V = a³√2/12.

Worked example

A regular square pyramid has base side 6 and height 4. Find its volume.

1. The base is a square: S = 36.
2. V = ⅓ · 36 · 4 = 48.

Worked example

A cone has R = 3, h = 4. Find its volume.

1. Base area πR² = 9π.
2. V = ⅓ · 9π · 4 = 12π.

§4. Frustums of a pyramid and a cone

A pyramid whose top is cut off by a plane parallel to the base is a frustum. With base areas S₁ and S₂ and height h:

The second formula follows from the first with S₁ = πR², S₂ = πr². If you forget it, compute the frustum as “large pyramid minus small pyramid”.

Worked example

A frustum of a cone has base radii 3 and 1 and height 3. Find its volume.

1. R² + Rr + r² = 9 + 3 + 1 = 13.
2. V = ⅓ π · 3 · 13 = 13π.

§5. The ball

Archimedes’ result: a ball inscribed in a cylinder has 2/3 of the cylinder’s volume. Check: for a ball of radius R the cylinder has radius R and height 2R, V_cyl = 2πR³, (4/3)πR³ : 2πR³ = 2/3.
In water-level problems, when a body is submerged the volume of displaced water equals the volume of the body (Archimedes’ principle).

Worked example (water level)

A ball of radius 3 is fully submerged in water in a cylindrical vessel of radius 6. By how much does the water level rise?

1. Volume of the ball: (4/3)π · 27 = 36π.
2. The raised layer of water is a cylinder of radius 6 and height Δh: π · 36 · Δh.
3. 36π · Δh = 36π → Δh = 1.

§6. Similar solids and recasting

If every linear dimension of a solid changes by a factor k:

Reason: an area is a product of two dimensions, a volume of three. For example, doubling the radius of a ball multiplies its volume by 8 and its surface area by 4.
When metal is melted and recast (or liquid is poured), the volume is conserved; only the shape changes.

Worked example

A cube with edge 3 is melted and recast into cubes with edge 1. How many cubes are obtained, and by what factor does the surface area grow?

1. Volume is conserved: 27 : 1 = 27 cubes.
2. Original surface: 6 · 9 = 54; new surface: 27 · 6 = 162.
3. The surface grows by a factor of 162 : 54 = 3.

Check yourself

1. The volume of a cube with edge 4.

2. The total surface area of a cube with volume 27.

3. The volume of a cylinder with R = 2, h = 3.

4. A cone has volume 12π and radius 3. Find its height.

5. The volume of a pyramid with base area 15 and height 8.

6. Two similar cones have heights in the ratio 2 : 3. The ratio of volumes?

7. The volume of a ball of radius 3.

8. The volume of a regular tetrahedron with edge 6.

Common mistakes

• Forgetting the ⅓ for pyramids and cones.
• Writing R² in the volume of a ball — it must be R³.
• Thinking volumes of similar solids change by k or k² — they change by k³.
• Converting units wrongly: 1 m³ = 1000 dm³, not 100; 1 dm³ = 1 l.
• Using the lateral edge instead of the height in an oblique prism.
• Substituting a side instead of the base area (a square base needs a², not a).