Mass and Density

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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 \(\left[ {a,b} \right].\)

The density of the rod at any point \(x\) is defined by the density function \(\rho \left( x \right).\) Assuming that \(\rho \left( x \right)\) 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.

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.

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:

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