6.3 Volume, Surface Area, Capacity, and Packing Problems

Three-dimensional measurement problems in school settings fall into three distinct categories: finding the interior space a container occupies (volume and liquid capacity), finding the total exterior or interior panelling required to build or line a solid (surface area), and finding how many rigid items fit inside a storage carton (3D packing).

Volume of Prisms and Cylinders

Any prism or cylinder has a uniform cross-sectional base of area $A_{\text{base}}$ projected along a perpendicular height or length $h$. Its volume is always:

$$V = A_{\text{base}} \times h$$

• Rectangular Prism (boxes, classrooms, swimming pools, raised garden beds): The base is a rectangle of area $l \times w$, giving:

$$V = l \times w \times h$$

• Triangular Prism (pitched roof spaces, sports wedges, camping tents): The cross-sectional base is a triangle of area $\frac{1}{2}b_{\text{tri}}h_{\text{tri}}$ extruded along length $L$, giving:

$$V = \left(\frac{1}{2} b_{\text{tri}} h_{\text{tri}}\right) \times L$$

• Cylinder (rainwater tanks, pipes, cylindrical storage drums): The circular base has area $\pi r^2$, giving:

$$V = \pi r^2 h$$

The Unit Consistency Rule and Bulk Landscaping Orders

Before multiplying $l \times w \times h$, check that every dimension is expressed in the same linear unit. In school grounds projects—such as ordering bulk garden soil, bark mulch, or ready-mix concrete—length and width are usually given in metres ($\text{m}$) while depth is given in centimetres ($\text{cm}$) or millimetres ($\text{mm}$). Always convert depth to metres first ($1\text{ cm} = 0.01\text{ m}$ and $1\text{ mm} = 0.001\text{ m}$). For instance, if a school builds $4$ raised rectangular garden beds each $5\text{ m}$ long by $1.6\text{ m}$ wide and fills them with compost to a depth of $25\text{ cm}$ ($0.25\text{ m}$) at $72 per $\text{m}^3$:

$$V_{\text{total}} = 4 \times (5\text{ m} \times 1.6\text{ m} \times 0.25\text{ m}) = 4 \times 2.0\text{ m}^3 = 8.0\text{ m}^3, \quad \text{Cost} = 8.0 \times 72 = 576\text{ dollars}$$

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Converting Volume to Liquid Capacity and Flow Rates

Once you have calculated geometric volume in $\text{m}^3$ or $\text{cm}^3$, convert directly into litres ($\text{L}$) using the cubic bridge identities:

• From cubic metres to litres: Multiply by $1{,}000$ because $1\text{ m}^3 = 1{,}000\text{ L} = 1\text{ kL}$.
• From cubic centimetres to litres: Divide by $1{,}000$ because $1{,}000\text{ cm}^3 = 1{,}000\text{ mL} = 1\text{ L}$.

Pump and Hose Flow-Rate Problems

When a tank or swimming pool of liquid capacity $C$ (in litres) is filled or drained at a constant flow rate $R$ (in $\text{L/min}$ or $\text{L/hour}$):

$$\text{Time} = \frac{\text{Capacity in Litres }(C)}{\text{Flow Rate }(R)}$$

Always read carefully to check whether the tank starts partially full (e.g., already $20\%$ full) or is filled to less than $100\%$ capacity (e.g., filled to $90\%$ depth or $15\text{ cm}$ below the top rim).

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Surface Area of Closed vs. Open-Top Prisms and Cylinders

The surface area ($SA$) of a three-dimensional solid is the sum of the areas of all its exposed faces, measured in square units ($\text{cm}^2$ or $\text{m}^2$).

• Closed Rectangular Prism (6 faces): A sealed cardboard box or wooden block has a top and bottom ($2lw$), front and back ($2lh$), and left and right ends ($2wh$):

$$SA_{\text{closed}} = 2(lw + lh + wh)$$

• Open-Top Rectangular Prism (5 faces): A tiled swimming pool, open storage tub, or lidless glass aquarium has only one base ($lw$) plus the four vertical walls:

$$SA_{\text{open}} = lw + 2lh + 2wh$$

• Four Vertical Walls Only (4 faces): When painting the interior walls of a classroom (excluding both floor and ceiling):

$$SA_{\text{walls}} = 2lh + 2wh = 2h(l + w)$$

• Cylindrical Surface Area: Unrolling the curved side wall of a cylinder of radius $r$ and height $h$ forms a rectangle of width $2\pi r$ (the circumference) and height $h$, giving curved lateral area $2\pi rh$. Adding the circular base ($\pi r^2$) and top lid ($\pi r^2$) gives $SA_{\text{closed cylinder}} = 2\pi r^2 + 2\pi rh$.

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The 3D Box-Packing Trap: Dimension-by-Dimension Floor Division

Suppose you are packing rigid small boxes of dimensions $l \times w \times h$ into a larger shipping carton of internal dimensions $L \times W \times H$. Never divide the total carton volume by the small-box volume ($V_{\text{carton}} \div V_{\text{box}}$)! Volume division assumes that leftover empty strips along the edges can be melted together to fit fractional boxes.

Instead, for a specified box orientation, divide each dimension separately, round down to the nearest whole integer (using the floor function $\lfloor \cdot \rfloor$), and multiply the three whole numbers:

$$N_{\text{boxes}} = \left\lfloor \frac{L}{l} \right\rfloor \times \left\lfloor \frac{W}{w} \right\rfloor \times \left\lfloor \frac{H}{h} \right\rfloor$$

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Worked Examples in School Contexts

Worked Example 1: Cylindrical Rainwater Tank and Pump Time

A school installs a cylindrical rainwater tank with an internal diameter of $2.8\text{ m}$ and a height of $2.5\text{ m}$. Using $\pi \approx \frac{22}{7}$, find (a) the maximum capacity of the tank in litres, and (b) how many hours and minutes a transfer pump delivering $40\text{ L/min}$ takes to fill the tank from empty to $75\%$ capacity.

1. Find radius and full volume in $\text{m}^3$: Diameter $d = 2.8\text{ m} \implies r = 1.4\text{ m}$.

$$V = \pi r^2 h = \frac{22}{7} \times 1.4 \times 1.4 \times 2.5 = 22 \times 0.2 \times 1.4 \times 2.5 = 15.4\text{ m}^3$$

1. Convert to litres: $15.4\text{ m}^3 \times 1{,}000 = 15{,}400\text{ L}$.
2. Find $75\%$ capacity and pump time: $0.75 \times 15{,}400\text{ L} = 11{,}550\text{ L}$. Dividing by $40\text{ L/min}$ gives $11{,}550 \div 40 = 288.75\text{ minutes} = 4\text{ hours } 48.75\text{ minutes}$ (or $4\text{ hours } 48\text{ minutes } 45\text{ seconds}$).

Worked Example 2: Packing Maths Kits into a Storage Tub

A teacher packs rigid maths resource boxes measuring $15\text{ cm}$ long by $12\text{ cm}$ wide by $10\text{ cm}$ high into a plastic tub with internal dimensions of $60\text{ cm}$ long by $50\text{ cm}$ wide by $35\text{ cm}$ high, keeping all boxes oriented parallel to the tub's length, width, and height. How many boxes fit?

1. Length count: $\lfloor 60 \div 15 \rfloor = 4$ boxes.
2. Width count: $\lfloor 50 \div 12 \rfloor = \lfloor 4.167 \rfloor = 4$ boxes ($2\text{ cm}$ of unused gap remains).
3. Height count: $\lfloor 35 \div 10 \rfloor = \lfloor 3.5 \rfloor = 3$ layers ($5\text{ cm}$ of headspace remains).
4. Total boxes: $4 \times 4 \times 3 = 48\text{ boxes}$ (whereas naive volume division $105{,}000 \div 1{,}800 = 58.3$ would falsely suggest $58$ boxes!).