10.1 Volume, Capacity & Optional Packing Extensions

6. Volume by layers, with optional formula extensions

Volume of a Cuboid

A cuboid has six rectangular faces. For Year 5, start by estimating volume with unit cubes. A layer 4 cubes long and 3 cubes wide contains 12 cubes; two such layers contain 24 cubes. Counting equal layers leads to the following formula, which is a later curriculum extension:

$$\text{Volume} = \text{length} \times \text{width} \times \text{height} = l \times w \times h$$

For a cube where all side lengths are equal ($s$):

$$\text{Volume} = s \times s \times s = s^3$$

Relationship Between Volume and Liquid Capacity

Remember the foundational conversion:

$$1\text{ cm}^3 = 1\text{ ml} \quad \text{and} \quad 1{,}000\text{ cm}^3 = 1\text{ litre}$$

If a fish tank has dimensions $50\text{ cm} \times 30\text{ cm} \times 20\text{ cm}$:

$$\text{Volume} = 50 \times 30 \times 20 = 30{,}000\text{ cm}^3 = 30{,}000\text{ ml} = 30\text{ litres}$$

Optional extension: surface area of a cuboid

The total surface area (TSA) is the sum of the areas of all six rectangular exterior faces. Because opposing faces are congruent, they come in three identical pairs:

• Top and bottom faces: $2 \times (l \times w)$
• Front and back faces: $2 \times (l \times h)$
• Left and right faces: $2 \times (w \times h)$

$$\text{Total Surface Area} = 2(lw + lh + wh)$$

For a cube with six identical square faces:

$$\text{Total Surface Area of a Cube} = 6s^2$$

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7. Real-World Applications & The Volume Fallacy

Floor Tiling and Paving Problems

When determining how many rectangular tiles of size $a \times b$ are needed to pave a room of size $L \times W$:

• Ensure both room dimensions and tile dimensions are converted to the same unit (e.g., both in centimetres).
• If tiles can be cut and all offcuts reused without waste, divide room area by tile area and round up to a whole number purchased. Real cutting or pattern constraints can require more tiles.
• To count whole tiles fitting in a fixed, axis-aligned orientation, count complete tiles along each dimension, then multiply. If a length is not an exact multiple, a strip remains uncovered: this does not solve a problem requiring complete coverage without cutting.

The "Volume Fallacy" in 3D Packing Problems

A frequent trap in 11+ exams involves packing small rectangular boxes into a large shipping crate:

To find the true maximum number of boxes that fit when oriented uniformly:

$$\text{Max Boxes} = \left\lfloor \frac{L_{\text{crate}}}{l_{\text{box}}} \right\rfloor \times \left\lfloor \frac{W_{\text{crate}}}{w_{\text{box}}} \right\rfloor \times \left\lfloor \frac{H_{\text{crate}}}{h_{\text{box}}} \right\rfloor$$

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Worked Example 3: Surface Area and Volume of an Open Tank

Problem: An open-topped zinc water tank has dimensions of length $80\text{ cm}$, width $50\text{ cm}$, and height $40\text{ cm}$. For this idealised problem, treat the walls as having negligible thickness, so internal and external dimensions are taken as equal. (a) What is the water capacity of the tank in litres when filled to the brim? (b) What is the total external surface area of the open-topped tank in $\text{cm}^2$?

• Step 1 (Calculate volume and capacity): $$V = 80 \times 50 \times 40 = 160{,}000\text{ cm}^3$$ $$\text{Capacity} = \frac{160{,}000}{1{,}000} = 160\text{ litres}$$
• Step 2 (Calculate external surface area of 5 faces):
  • Bottom face: $80 \times 50 = 4{,}000\text{ cm}^2$.
  • Two long sides: $2 \times (80 \times 40) = 2 \times 3{,}200 = 6{,}400\text{ cm}^2$.
  • Two short sides: $2 \times (50 \times 40) = 2 \times 2{,}000 = 4{,}000\text{ cm}^2$.
  • Total surface area $= 4{,}000 + 6{,}400 + 4{,}000 = 14{,}400\text{ cm}^2$.
• Answer: Capacity $= 160\text{ litres}$; Surface Area $= 14{,}400\text{ cm}^2$.

Estimate with unit cubes before using a formula

Imagine a box holding four cubes along its length, three across its width and two layers in height. One layer contains 4 × 3 = 12 cubes; two layers contain 24. If each cube has edge 1 cm, each occupies 1 cm³ and the model has volume 24 cm³. A box that is twice as high, with the same base, contains twice as many layers and has twice the volume. Changing only the height does not double its perimeter or every face area.

Capacity is the usable inside space of a container. An exterior measurement alone is insufficient when wall thickness matters. For estimates, compare with a familiar 1-litre container or count equal cubes. For a calculation, identify whether the dimensions describe the inside or the outside, and retain cubic units until an explicit capacity conversion is made. These decisions come before multiplying the numbers.