Solving 3D Measurement and Unit Conversion Problems

Welcome back. Last lesson dealt with two-dimensional measurement: perimeter measures an outside distance, area measures a flat region, and the trapezoidal rule estimates an irregular area. This lesson extends that same careful unit logic into three dimensions.

You will learn to distinguish surface area, volume, capacity, and mass; choose the right formula for prisms and cylinders; convert safely between cubic units and litres; and solve mixed practical problems. These are often high-value exam questions because the calculation is usually straightforward once the quantity and units are correctly identified.

---

Four quantities that sound similar but answer different questions

Before choosing a formula, decide what the question is actually asking for.

Quantity	What it measures	Typical context	Common units
Surface area	Total area of the outside faces	Paint, wrapping, metal sheet	${\text{cm}}^2$, ${\text{m}}^2$
Volume	Space occupied by a three-dimensional object	Concrete, sand, a solid block	${\text{cm}}^3$, ${\text{m}}^3$
Capacity	Volume a container can hold internally	Water tank, bottle, pool	$\text{mL}$, $\text{L}$, $\text{kL}$
Mass	Amount of matter	Weight of contents or material	$\text{g}$, $\text{kg}$, $\text{t}$

A useful interpretation rule is:

• If you are covering an object, use surface area.
• If you are filling an object, use volume or capacity.
• If you are finding how heavy a known volume of material is, use density and mass.

The units themselves offer a built-in check:

$$\text{surface area uses square units}$$ $$\text{volume uses cubic units}$$ $$\text{capacity usually uses litres or millilitres}$$

---

Volume: build from a cross-section

A prism has the same cross-section all the way along its length. A rectangular box, triangular prism, and trapezoidal prism are all prisms.

For every right prism:

$$V=A_{\text{cross section}}\times L$$

where $L$ is the length of the prism.

For a rectangular prism, the cross-section is a rectangle, so this becomes:

$$V=lwh$$

For a cylinder, the cross-section is a circle:

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

Here, $r$ is the radius and $h$ is the perpendicular height.

Worked example: volume of a triangular prism

A triangular prism has a triangular cross-section with base $8\text{ cm}$, perpendicular height $5\text{ cm}$, and prism length $14\text{ cm}$.

First find the cross-sectional area:

$$A_{\text{triangle}}=\frac{1}{2}bh$$ $$A_{\text{triangle}}=\frac{1}{2}\times 8\times 5=20{\text{ cm}}^2$$

Now multiply by the prism length:

$$V=20\times 14$$ $$V=280{\text{ cm}}^3$$

The volume is:

$$\boxed{280{\text{ cm}}^3}$$

The intermediate area is in ${\text{cm}}^2$. Multiplying it by a length in $\text{cm}$ produces ${\text{cm}}^3$, which is a useful way to check that the answer is genuinely a volume.

For a compound prism, find the cross-sectional area by splitting it into familiar two-dimensional shapes, just as you did in the previous lesson, then multiply the total cross-sectional area by the prism length.

---

Surface area: count exposed faces, not dimensions

Surface area is the total area of every exposed outer face. Unlike volume, it is not about filling the object; it is about the material needed to cover it.

For a rectangular prism:

$$SA=2lw+2lh+2wh$$

This works because there are three pairs of matching rectangular faces.

For any right prism, an efficient general formula is:

$$SA=2A_{\text{cross section}}+P_{\text{cross section}}L$$

where $P_{\text{cross section}}$ is the perimeter of the cross-section. The first part counts the two ends; the second part represents the rectangular faces around the sides.

For a closed cylinder:

$$SA=2\pi r^2+2\pi rh$$

The two $\pi r^2$ terms are the circular ends. The $2\pi rh$ term is the curved side.

Open containers: do not include faces that are missing

Questions often describe a cylinder as open, open at one end, or open at the top. In those cases, the missing circular face must not be counted.

For an open-top cylindrical container:

$$SA=\pi r^2+2\pi rh$$

There is one circular base and one curved surface, but no lid.

For composite solids, a joined face is internal rather than exposed. Do not count it in the final surface area. This is a common source of over-counting.

Worked example: material for an open cylindrical container

A container is open at the top, with radius $4\text{ cm}$ and height $20\text{ cm}$. Find the area of material needed.

The base has area:

$$A_{\text{base}}=\pi r^2$$ $$A_{\text{base}}=\pi (4{)}^2=16\pi$$

The curved side has area:

$$A_{\text{curved}}=2\pi rh$$ $$A_{\text{curved}}=2\pi (4)(20)=160\pi$$

So the total material area is:

$$SA=16\pi +160\pi$$ $$SA=176\pi$$ $$SA\approx 552.9{\text{ cm}}^2$$ $$\boxed{552.9{\text{ cm}}^2}$$

Notice that the answer uses square units, even though the object itself is three-dimensional.

---

Unit conversion: why square and cubic units change so quickly

A factor of $100$ applies when converting metres to centimetres:

$$1\text{ m}=100\text{ cm}$$

But area uses two dimensions:

$$1{\text{ m}}^2=(100\text{ cm}{)}^2=10000{\text{ cm}}^2$$

And volume uses three dimensions:

$$1{\text{ m}}^3=(100\text{ cm}{)}^3=1000000{\text{ cm}}^3$$

That exponent is not decoration. It determines the conversion factor.

For capacity, memorise these exact relationships:

$$1{\text{ cm}}^3=1\text{ mL}$$ $$1000{\text{ cm}}^3=1000\text{ mL}=1\text{ L}$$ $$1{\text{ m}}^3=1000\text{ L}=1\text{ kL}$$

So a volume of $280{\text{ cm}}^3$, for example, has a capacity of:

$$280\text{ mL}$$

or:

$$0.280\text{ L}$$

Two conversion checks

Convert $0.35{\text{ m}}^2$ to ${\text{cm}}^2$:

$$0.35\times 10000=3500{\text{ cm}}^2$$

Convert $2.4{\text{ m}}^3$ to litres:

$$2.4\times 1000=2400\text{ L}$$

The first conversion uses $10000$ because it is area. The second uses $1000$ because $1{\text{ m}}^3=1000\text{ L}$.

---

Mass and density: connecting “how much space” to “how heavy”

Mass questions often supply a material’s density. Density tells you the mass in each unit of volume.

$$m=\rho V$$

where:

• $m$ is mass;
• $\rho$ is density;
• $V$ is volume.

The units must match. For example:

• density in $\text{g/mL}$ requires volume in $\text{mL}$, giving mass in $\text{g}$;
• density in ${\text{kg/m}}^3$ requires volume in ${\text{m}}^3$, giving mass in $\text{kg}$.

Do not automatically treat $1\text{ L}$ as $1\text{ kg}$. That is approximately true for water under ordinary conditions, but not for oil, concrete, syrup, fuel, or most other materials.

Integrated worked example: surface area, volume, capacity, and mass

A thin-walled sealed cylindrical container has radius $3.5\text{ cm}$ and height $12\text{ cm}$. It is filled with a liquid of density $1.30\text{ g/mL}$.

1. Find the external surface area

$$SA=2\pi r^2+2\pi rh$$ $$SA=2\pi (3.5{)}^2+2\pi (3.5)(12)$$ $$SA=108.5\pi$$ $$SA\approx 340.9{\text{ cm}}^2$$

2. Find the volume and capacity

$$V=\pi r^{2}h$$ $$V=\pi (3.5{)}^2(12)$$ $$V=147\pi$$ $$V\approx 461.8{\text{ cm}}^3$$

Because $1{\text{ cm}}^3=1\text{ mL}$:

$$\text{capacity}\approx 461.8\text{ mL}$$ $$\text{capacity}\approx 0.462\text{ L}$$

3. Find the mass of liquid

The volume is already in $\text{mL}$, matching the density units:

$$m=\rho V$$ $$m=1.30\times 461.8$$ $$m\approx 600.3\text{ g}$$

So the liquid’s mass is:

$$\boxed{600.3\text{ g}}$$

In a full exam response, label each part clearly. This prevents a correct volume result from being mistaken for the requested mass or capacity.

---

Working backwards from volume

Some questions give a volume or capacity and ask for an unknown dimension. Start with the relevant formula, substitute known values, then rearrange.

For example, an aquarium has internal length $0.80\text{ m}$, width $0.30\text{ m}$, and capacity $144\text{ L}$. Find its internal depth.

Convert the capacity first:

$$144\text{ L}=0.144{\text{ m}}^3$$

Use the rectangular-prism formula:

$$V=lwh$$ $$0.144=0.80\times 0.30\times h$$ $$0.144=0.24h$$ $$h=\frac{0.144}{0.24}$$ $$h=0.60\text{ m}$$

The aquarium depth is:

$$\boxed{0.60\text{ m}}$$

---

Exam routine for three-dimensional measurement

Use this sequence before calculating:

1. Underline what is required: surface area, volume, capacity, mass, or an unknown dimension.
2. Choose a formula that matches both the solid and the required quantity.
3. Check dimensions: use radius rather than diameter in circle formulas, and perpendicular height where required.
4. Convert units before calculation if they do not match.
5. Account for the physical object: open top, missing face, internal capacity, hollow section, or joined solids.
6. Round only at the end unless instructed otherwise.
7. Write a final answer with the correct unit.

A quick plausibility check is also valuable: if a tank is described in litres but you produce square metres, you have likely calculated surface area instead of capacity.

---

Key takeaways

Surface area measures the exposed outside of a solid and uses square units. Volume measures three-dimensional space and uses cubic units. Capacity is the internal volume of a container, usually expressed in millilitres or litres.

For prisms:

$$V=A_{\text{cross section}}L$$

For cylinders:

$$V=\pi r^{2}h$$ $$SA=2\pi r^2+2\pi rh$$

The key conversion facts are:

$$1{\text{ cm}}^3=1\text{ mL}$$ $$1000\text{ mL}=1\text{ L}$$ $$1{\text{ m}}^3=1000\text{ L}$$

Finally, when density is provided, connect mass and volume using:

$$m=\rho V$$

Next, the course moves into data analysis: identifying data types and choosing sampling methods that produce useful, fair conclusions.