Math Core

Area and volume formulas become powerful when you attach them to real quantities. How heavy is a steel beam? How crowded is a city? Will a block of wood float? Each of these questions is answered by a density: an amount of something packed into each unit of area or volume.

Density as a rate

A density is a ratio that compares a quantity to the space it fills.

Definition

Density

Density is the amount of a quantity per unit of area or volume.

\text{density} = \frac{\text{quantity}}{\text{area or volume}}

Mass density compares mass to volume, in units like grams per cubic centimeter ( $\text{g/cm}^3$ ) or kilograms per cubic meter.
Population density compares a count of people (or trees, or animals) to area, in units like people per square mile.

Because density is a ratio, the same equation can be solved three ways. If you know any two of density, quantity and size, you can find the third:

d = \frac{m}{V} \qquad m = dV \qquad V = \frac{m}{d}

Always keep track of units. They tell you whether to multiply or divide: $\text{g/cm}^3 \times \text{cm}^3 = \text{g}$ , and the cubic centimeters cancel.

Mass density

Every pure material has a characteristic density. Water is about $1 \text{ g/cm}^3$ , aluminum about $2.7$ , steel about $7.85$ , and gold about $19.3$ . That makes density a useful fingerprint.

Worked example: Is it gold?

A shiny rectangular bar measures 10 cm by 4 cm by 2 cm and has a mass of 840 g. Pure gold has a density of about $19.3 \text{ g/cm}^3$ . Could the bar be solid gold?

The volume is $10 \cdot 4 \cdot 2 = 80 \text{ cm}^3$ , so

d = \frac{840 \text{ g}}{80 \text{ cm}^3} = 10.5 \text{ g/cm}^3

That's barely more than half the density of gold. A solid gold bar that size would have a mass of about $19.3 \cdot 80 = 1544$ g. The bar is not solid gold.

Worked example: The mass of a ball bearing

A steel ball bearing has a radius of 2 cm. Steel has a density of $7.85 \text{ g/cm}^3$ . Find the mass of the ball to the nearest tenth of a gram.

First find the volume, then multiply by the density:

\begin{aligned} V &= \tfrac{4}{3}\pi(2)^3 = \tfrac{32}{3}\pi \approx 33.51 \text{ cm}^3 \\ m &= dV \approx 7.85 \cdot 33.51 \approx 263.1 \text{ g} \end{aligned}

Common mistake

Don't round too early. Keep the exact volume, $\tfrac{32}{3}\pi$ , or at least several decimal places, until the final step. Rounding the volume to 34 first would give $266.9$ g, which is off by almost 4 grams.

Density also predicts whether an object floats. An object that is less dense than water (less than $1 \text{ g/cm}^3$ ) floats in it, and one that is denser sinks.

Population density

Population density uses area instead of volume. It lets you compare places of very different sizes on an equal footing.

Worked example: A round town

A town is roughly a circle with a radius of 3 miles. About 42,000 people live there. Find its population density to the nearest whole number.

\begin{aligned} A &= \pi(3)^2 = 9\pi \approx 28.27 \text{ mi}^2 \\ \text{density} &= \frac{42{,}000}{9\pi} \approx 1485 \text{ people per square mile} \end{aligned}

Modeling with geometric solids

Real objects are never perfect cylinders or spheres. Modeling means choosing a simple shape that is close enough to answer the question, measuring what you can, and being honest about what you left out.

A good model follows a cycle:

Choose a shape. A tree trunk is roughly a cylinder. An orange is roughly a sphere.
Measure. Use dimensions you can actually get. You can wrap a tape measure around a tree, but you can't measure its radius directly.
Compute with the formulas.
Check whether the answer is reasonable, and whether the model over- or underestimates.

Worked example: How heavy is the log?

A tree trunk is 12 m tall, and its circumference is 2.5 m. The wood has a density of about $600 \text{ kg/m}^3$ . Model the trunk as a cylinder to estimate its mass.

From the circumference, $2\pi r = 2.5$ , so $r = \dfrac{2.5}{2\pi} \approx 0.398$ m.

\begin{aligned} V &= \pi r^2 h = \pi\left(\dfrac{2.5}{2\pi}\right)^2(12) \approx 5.97 \text{ m}^3 \\ m &= dV \approx 600 \cdot 5.97 \approx 3580 \text{ kg} \end{aligned}

A real trunk tapers toward the top, so it is narrower on average than the base. The cylinder model probably overestimates the mass. A cone or a frustum would be a closer model if you could measure the top.

Tip

Before computing, estimate. A cylinder about 0.8 m across and 12 m tall has a volume a little under $1 \times 1 \times 12$ , so somewhere around 5 or 6 cubic meters. If your calculator says 60, look for an error.

Practice

Practice 1

A metal block has a mass of 540 g and a volume of 200 cubic centimeters. What is its density, in grams per cubic centimeter?