Applications of Integrals

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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].

A thin rod with a density function rho(x).
Figure 1.

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 \(R.\)

A thin disk with a radial density function rho(r).
Figure 2.

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 \(x = b.\)

Mass of a lamina with non-uniform density distribution occupying a region bounded by two curves.
Figure 3.

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 \(A\left( x \right).\)

Solid with one-dimensional density function
Figure 4.

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.

Solid of revolution with a density function rho(x)
Figure 5.

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

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.


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.


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.


Night traffic jam
Figure 6.

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.


Bacteria colony in a petri dish
Figure 7.

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.

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