# Mass and Density

## Mass of a Thin Rod

We can use integration for calculating mass based on a density function.

Consider a thin wire or rod that is located on an interval [a, b].

The density of the rod at any point x is defined by the density function ρ (x). Assuming that ρ (x) is an integrable function, the mass of the rod is given by the integral

$m = \int\limits_a^b {\rho \left( x \right)dx} .$

## Mass of a Thin Disk

Suppose that $$\rho \left( r \right)$$ represents the radial density of a thin disk of radius $$R.$$

Then the mass of the disk is given by

$m = 2\pi \int\limits_0^R {r\rho \left( r \right)dr} .$

## Mass of a Region Bounded by Two Curves

Suppose a region is enclosed by two curves $$y = f\left( x \right),$$ $$y = g\left( x \right)$$ and by two vertical lines $$x = a$$ and $$x = b.$$

If the density of the lamina which occupies the region only depends on the $$x-$$coordinate, the total mass of the lamina is given by the integral

$m = \int\limits_a^b {\rho \left( x \right)\left[ {f\left( x \right) - g\left( x \right)} \right]dx} ,$

where $$f\left( x \right) \ge g\left( x \right)$$ on the interval $$\left[ {a,b} \right],$$ and $${\rho \left( x \right)}$$ is the density of the material changing along the $$x-$$axis.

## Mass of a Solid with One-Dimensional Density Function

Consider a solid $$S$$ that extends in the $$x-$$direction from $$x = a$$ to $$x = b$$ with cross sectional area $$A\left( x \right).$$

Suppose that the density function $$\rho \left( x \right)$$ depends on $$x$$ but is constant inside each cross section $$A\left( x \right).$$

The mass of the solid is

$m = \int\limits_a^b {\rho \left( x \right)A\left( x \right)dx} .$

## Mass of a Solid of Revolution

Let $$S$$ be a solid of revolution obtained by rotating the region under the curve $$y = f\left( x \right)$$ on the interval $$\left[ {a,b} \right]$$ around the $$x-$$axis.

If $$\rho \left( x \right)$$ is the density of the solid material depending on the $$x-$$coordinate, then the mass of the solid can be calculated by the formula

$m = \pi \int\limits_a^b {\rho \left( x \right){f^2}\left( x \right)dx} .$

## Solved Problems

Click or tap a problem to see the solution.

### Example 1

A rod with a linear density given by $\rho \left( x \right) = {x^3} + x$ lies on the $$x-$$axis between $$x = 0$$ and $$x = 2.$$ Find the mass of the rod.

### Example 2

Let a thin rod of length $$L = 10\,\text{cm}$$ have its mass distributed according to the density function $\rho \left( x \right) = 50{e^{ - \frac{x}{{10}}}},$ where $$\rho \left( x \right)$$ is measured in $$\frac{\text{g}}{\text{cm}},$$ $$x$$ is measured in $$\text{cm}.$$ Calculate the total mass of the rod.

### Example 3

Suppose that the density of cars in traffic congestion on a highway changes linearly from 30 to 150 cars per km per lane on a $$5\,\text{km}$$ long stretch. Estimate the total number of cars on the highway stretch if it has $$4$$ lanes.

### Example 4

Determine the total amount of bacteria in a circular petri dish of radius $$R$$ if the density at the center is $${\rho_0}$$ and decreases linearly to zero at the edge of the dish.

### Example 1.

A rod with a linear density given by $\rho \left( x \right) = {x^3} + x$ lies on the $$x-$$axis between $$x = 0$$ and $$x = 2.$$ Find the mass of the rod.

Solution.

We need to integrate the following:

$m = \int\limits_a^b {\rho \left( x \right)dx} = \int\limits_0^2 {\left( {{x^3} + x} \right)dx} = \left. {\left( {\frac{{{x^4}}}{4} + \frac{{{x^2}}}{2}} \right)} \right|_0^2 = 6.$

If $$\rho$$ is measured in kilograms per meter and $$x$$ is measured in meters, then the mass is $$m = 6\,\text{kg}.$$

### Example 2.

Let a thin rod of length $$L = 10\,\text{cm}$$ have its mass distributed according to the density function $\rho \left( x \right) = 50{e^{ - \frac{x}{{10}}}},$ where $$\rho \left( x \right)$$ is measured in $$\frac{\text{g}}{\text{cm}},$$ $$x$$ is measured in $$\text{cm}.$$ Calculate the total mass of the rod.

Solution.

To find the mass of the rod we integrate the density function:

$m = \int\limits_a^b {\rho \left( x \right)dx} = \int\limits_0^{10} {50{e^{ - \frac{x}{{10}}}}dx} = - 500\left. {{e^{ - \frac{x}{{10}}}}} \right|_0^{10} = 500\left( {1 - \frac{1}{e}} \right) = \frac{{500\left( {e - 1} \right)}}{e} \approx 316\,\text{g}.$

### Example 3.

Suppose that the density of cars in traffic congestion on a highway changes linearly from 30 to 150 cars per km per lane on a $$5\,\text{km}$$ long stretch. Estimate the total number of cars on the highway stretch if it has $$4$$ lanes.

Solution.

First we derive the equation for the density function $$\rho \left( x \right).$$ Since the function is linear, it is defined by two points:

$\rho \left( 0 \right) = 30,\;\; \rho \left( 5 \right) = 150.$

Using the two-point form of a straight line equation, we have

$\frac{{\rho - 30}}{{150 - 30}} = \frac{{x - 0}}{{5 - 0}},\;\; \Rightarrow \frac{{\rho - 30}}{{120}} = \frac{x}{5},\;\; \Rightarrow \rho - 30 = 24x,\;\; \Rightarrow \rho \left( x \right) = 24x + 30.$

Now, to estimate the amount of cars on the highway stretch, we integrate the density function and multiply the result by $$4:$$

$N = 4\int\limits_a^b {\rho \left( x \right)dx} = 4\int\limits_0^5 {\left( {24x + 30} \right)dx} = \left. {4\left( {12{x^2} + 30x} \right)} \right|_0^5 = 4\left({ 300 + 150 }\right) = 1800\,\text{cars}.$

### Example 4.

Determine the total amount of bacteria in a circular petri dish of radius $$R$$ if the density at the center is $${\rho_0}$$ and decreases linearly to zero at the edge of the dish.

Solution.

The density of bacteria varies according to the law

$\rho \left( r \right) = {\rho _0}\left( {1 - \frac{r}{R}} \right),$

where $$0 \le r \le R.$$

To find the total number of bacteria in the dish, we use the formula

$N = 2\pi \int\limits_0^R {r\rho \left( r \right)dr} .$

This yields:

$N = 2\pi {\rho _0}\int\limits_0^R {r\left( {1 - \frac{r}{R}} \right)dr} = 2\pi {\rho _0}\int\limits_0^R {\left( {r - \frac{{{r^2}}}{R}} \right)dr} = 2\pi {\rho _0}\left. {\left( {\frac{{{r^2}}}{2} - \frac{{{r^3}}}{{3R}}} \right)} \right|_0^R = 2\pi {\rho _0}\left( {\frac{{{R^2}}}{2} - \frac{{{R^2}}}{3}} \right) = \frac{{2\pi {\rho _0}{R^2}}}{6} = \frac{{\pi {\rho _0}{R^2}}}{3}.$

See more problems on Page 2.