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 $[a, b] .$

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_{a}^{b} ρ (x) d x .

Mass of a Thin Disk

Suppose that $ρ (r)$ represents the radial density of a thin disk of radius $R .$

Then the mass of the disk is given by

m = 2 π \int_{0}^{R} r ρ (r) d r .

Mass of a Region Bounded by Two Curves

Suppose a region is enclosed by two curves $y = f (x),$ $y = g (x)$ 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_{a}^{b} ρ (x) [f (x) - g (x)] d x,

where $f (x) \geq g (x)$ on the interval $[a, b],$ and $ρ (x)$ 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 $A (x) .$

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

The mass of the solid is

m = \int_{a}^{b} ρ (x) A (x) d x .

Mass of a Solid of Revolution

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

If $ρ (x)$ 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 = π \int_{a}^{b} ρ (x) f^{2} (x) d x .

Solved Problems

Click or tap a problem to see the solution.

Example 1

A rod with a linear density given by $ρ (x) = x^{3} + x$ 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 $L = 10 cm$ have its mass distributed according to the density function $ρ (x) = 50 e^{- \frac{x}{10}},$ where $ρ (x)$ is measured in $\frac{g}{cm},$ $x$ is measured in $cm .$ 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 $5 km$ 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 $ρ_{0}$ and decreases linearly to zero at the edge of the dish.

Example 1.

A rod with a linear density given by $ρ (x) = x^{3} + x$ 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_{a}^{b} ρ (x) d x = \int_{0}^{2} (x^{3} + x) d x = {(\frac{x^{4}}{4} + \frac{x^{2}}{2}) |}_{0}^{2} = 6.

If $ρ$ is measured in kilograms per meter and $x$ is measured in meters, then the mass is $m = 6 kg .$

Example 2.

Solution.

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

m = \int_{a}^{b} ρ (x) d x = \int_{0}^{10} 50 e^{- \frac{x}{10}} d x = - 500 {e^{- \frac{x}{10}} |}_{0}^{10} = 500 (1 - \frac{1}{e}) = \frac{500 (e - 1)}{e} \approx 316 g .

Example 3.

Solution.

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

ρ (0) = 30, ρ (5) = 150.

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

\frac{ρ - 30}{150 - 30} = \frac{x - 0}{5 - 0}, \Rightarrow \frac{ρ - 30}{120} = \frac{x}{5}, \Rightarrow ρ - 30 = 24 x, \Rightarrow ρ (x) = 24 x + 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_{a}^{b} ρ (x) d x = 4 \int_{0}^{5} (24 x + 30) d x = {4 (12 x^{2} + 30 x) |}_{0}^{5} = 4 (300 + 150) = 1800 cars .

Example 4.

Determine the total amount of bacteria in a circular petri dish of radius $R$ if the density at the center is $ρ_{0}$ and decreases linearly to zero at the edge of the dish.

Solution.

The density of bacteria varies according to the law

ρ (r) = ρ_{0} (1 - \frac{r}{R}),

where $0 \leq r \leq R .$

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

N = 2 π \int_{0}^{R} r ρ (r) d r .

This yields:

N = 2 π ρ_{0} \int_{0}^{R} r (1 - \frac{r}{R}) d r = 2 π ρ_{0} \int_{0}^{R} (r - \frac{r^{2}}{R}) d r = 2 π ρ_{0} {(\frac{r^{2}}{2} - \frac{r^{3}}{3 R}) |}_{0}^{R} = 2 π ρ_{0} (\frac{R^{2}}{2} - \frac{R^{2}}{3}) = \frac{2 π ρ_{0} R^{2}}{6} = \frac{π ρ_{0} R^{2}}{3} .