# Calculus

## Integration of Functions # 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 $$\left[ {a,b} \right].$$

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 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 $$\large{\frac{\text{g}}{\text{cm}}}\normalsize,$$ $$x$$ 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 5

Assuming that the stellar radial distribution within a galaxy obeys the exponential law$\rho \left( r \right) = {\rho _0}{e^{ – \frac{r}{h}}},$estimate the mass of a galaxy with following parameters: $${\rho _0} = {10^7}\large{\frac{{{M_{\odot}}}}{{\text{kpc}}}}\normalsize,$$ $$h = {10^4}\,\text{kpc},$$ where $${M_{\odot}}$$ denotes the solar mass and $$\text{kpc}$$ means a kiloparsec ($$1\,\text{kpc} \approx 3262$$ light-years).

### Example 6

A lamina occupies the region bounded by one arc of the sine curve and the $$x-$$axis. The density at any point of the lamina is proportional to the distance from the point to the $$y-$$axis. Find the mass of the lamina.

### Example 7

A lamina occupies the upper semicircle of radius $$1$$ centered at the origin. Its density is given by the cubic function $$\rho \left( y \right) = {y^3}.$$ Find the mass of the lamina.

### Example 8

A right circular cone has base radius $$R$$ and height $$H.$$ What is the mass of the cone if its density varies along the vertical axis and is given by the function $$\rho \left( y \right) = k{y^2}?$$

### Example 9

A right circular cone with base radius $$R$$ and height $$H$$ is formed by rotating about the $$x-$$axis. The density of the cone is given by the function $$\rho \left( x \right) = kx.$$ Find the mass of the cone assuming that the center of its base is placed in the origin.

### Example 10

The density of the Earth’s inner core is about $$13000\,\large{\frac{\text{kg}}{\text{m}^3}}\normalsize.$$ Suppose that the density of the Earth near the surface is equal to the density of water $$1000\,\large{\frac{\text{kg}}{\text{m}^3}}\normalsize.$$ Estimate the mass of the Earth if the density changes linearly and the Earth’s radius is $$6200\,\text{km}.$$

### 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 $$\large{\frac{\text{g}}{\text{cm}}}\normalsize,$$ $$x$$ 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:

${\rho \left( 0 \right) = 30,\;\;}\kern0pt{\rho \left( 5 \right) = 150.}$

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

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

### Example 5.

Assuming that the stellar radial distribution within a galaxy obeys the exponential law$\rho \left( r \right) = {\rho _0}{e^{ – \frac{r}{h}}},$estimate the mass of a galaxy with following parameters: $${\rho _0} = {10^3}\large{\frac{{{M_{\odot}}}}{{{\text{kpc}^2}}}}\normalsize,$$ $$h = {10^4}\,\text{kpc},$$ where $${M_{\odot}}$$ denotes the solar mass and $$\text{kpc}$$ means a kiloparsec ($$1\,\text{kpc} \approx 3262$$ light-years).

Solution.

We will assume that the galaxy has the form of a thin disc and therefore it is possible to apply the formula

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

Since the exact value of the radius $$R$$ of the galaxy is unknown, we will set it equal to infinity. Calculate the improper integral:

${m = 2\pi \int\limits_0^\infty {r\rho \left( r \right)dr} }={ 2\pi {\rho _0}\int\limits_0^\infty {r{e^{ – \frac{r}{h}}}dr} }={ 2\pi {\rho _0}\lim \limits_{b \to \infty } \int\limits_0^b {r{e^{ – \frac{r}{h}}}dr}.}$

Integrating by parts, we obtain:

${\int {\underbrace r_u\underbrace {{e^{ – \frac{r}{h}}}dr}_{dv}} }={ \left[ {\begin{array}{*{20}{l}} {u = r}\\ {dv = {e^{ – \frac{r}{h}}}dr}\\ {du = dr}\\ {v = – h{e^{ – \frac{r}{h}}}} \end{array}} \right] }={ – hr{e^{ – \frac{r}{h}}} – \int {\left( { – h{e^{ – \frac{r}{h}}}} \right)dr} }={ – hr{e^{ – \frac{r}{h}}} + h\int {{e^{ – \frac{r}{h}}}dr} }={ – hr{e^{ – \frac{r}{h}}} – {h^2}{e^{ – \frac{r}{h}}} }={ – h\left( {r + h} \right){e^{ – \frac{r}{h}}}.}$

Taking limits then yields:

${m = 2\pi {\rho _0}\lim\limits_{b \to \infty } \int\limits_0^b {r{e^{ – \frac{r}{h}}}dr} }={ 2\pi {\rho _0}h\lim\limits_{b \to \infty } \left[ {\left. {\left( { – \left( {r + h} \right){e^{ – \frac{r}{h}}}} \right)} \right|_0^b} \right] }={ 2\pi {\rho _0}h\lim\limits_{b \to \infty } \left[ {h – \frac{{b + h}}{{{e^{\frac{b}{h}}}}}} \right].}$

By L’Hopital’s rule we have that

${\lim\limits_{b \to \infty } \left[ {h – \frac{{b + h}}{{{e^{\frac{b}{h}}}}}} \right] }={ h – \lim\limits_{b \to \infty } \frac{{\left( {b + h} \right)^\prime}}{{\left( {{e^{\frac{b}{h}}}} \right)^\prime}} }={ h – \lim\limits_{b \to \infty } \frac{0}{{\frac{1}{h}{e^{\frac{b}{h}}}}} }={ h.}$

Hence, the mass of the galaxy is given by the equation

$m = 2\pi {\rho _0}{h^2}.$

Substitute the given values:

${m = 2\pi \times {10^3} \times {\left( {{{10}^4}} \right)^2} }={ 2\pi \times {10^{11}} }\approx{ 6.28 \times {10^{11}}\,{M_\odot}},$

that is about $$2$$ times less than the mass of the Milky Way.

### Example 6.

A lamina occupies the region bounded by one arc of the sine curve and the $$x-$$axis. The density at any point of the lamina is proportional to the distance from the point to the $$y-$$axis. Find the mass of the lamina.

Solution.

The general formula for the mass of a region between two curves is

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

Substituting the known functions and limits, we get:

$m = \int\limits_0^\pi {x\sin xdx} .$

To evaluate the integral we use integration by parts:

${m = \int\limits_0^\pi {\underbrace x_u\underbrace {\sin xdx}_{dv}} }={ \left[ {\begin{array}{*{20}{l}} {u = x}\\ {dv = \sin xdx}\\ {du = dx}\\ {v = – \cos x} \end{array}} \right] }={ \left. {\left( { – x\cos x} \right)} \right|_0^\pi – \int\limits_0^\pi {\left( { – \cos x} \right)dx} }={ \left. {\left( { – x\cos x} \right)} \right|_0^\pi + \int\limits_0^\pi {\cos xdx} }={ \left. {\left( {\sin x – x\cos x} \right)} \right|_0^\pi }={ \pi .}$

If the density $${\rho \left( x \right)}$$ is measured in $$\large{\frac{\text{kg}}{\text{m}}}\normalsize$$ and $$x$$ is measured in $$\text{m},$$ then the mass of the lamina is $$m = \pi\,\text{kg}.$$

### Example 7.

A lamina occupies the upper semicircle of radius $$1$$ centered at the origin. Its density is given by the cubic function $$\rho \left( y \right) = {y^3}.$$ Find the mass of the lamina.

Solution.

Here the density function varies along the $$y-$$axis. Therefore we use the following formula to calculate the mass of the lamina:

$m = \int\limits_c^d {\rho \left( y \right)\left[ {f\left( y \right) – g\left( y \right)} \right]dy} .$

Due to symmetry about the $$y-$$axis, we can integrate from $$0$$ to $$1$$ and then multiply the answer by $$2.$$

The circle in the first quadrant is given by the equation $$x = f\left( y \right) = \sqrt {1 – {y^2}} .$$ Hence, the mass of the lamina is expressed by the integral

$m = 2\int\limits_0^1 {{y^3}\sqrt {1 – {y^2}} dy} .$

To evaluate the integral, we change the variable:

${1 – {y^2} = {z^2},}\;\; \Rightarrow {{y^2} = 1 – {z^2},\;\;}\kern0pt{ydy = – zdz,\;\;}\kern0pt{{y^3}dy = \left( {1 – {z^2}} \right)\left( { – zdz} \right) }={ \left( {{z^3} – z} \right)dz.}$

When $$y = 0,$$ $$z = 1,$$ and when $$y = 1,$$ $$z = 0.$$ So, the integral in terms of $$z$$ is written as

${m = 2\int\limits_1^0 {\left( {{z^3} – z} \right)zdz} }={ 2\int\limits_1^0 {\left( {{z^4} – {z^2}} \right)dz} }={ 2\left. {\left[ {\frac{{{z^5}}}{5} – \frac{{{z^3}}}{3}} \right]} \right|_1^0 }={ 2\left[ {0 – \left( {\frac{1}{5} – \frac{1}{3}} \right)} \right] }={ 2\left( {\frac{1}{3} – \frac{1}{5}} \right) }={ \frac{4}{{15}}.}$

### Example 8.

A right circular cone has base radius $$R$$ and height $$H.$$ What is the mass of the cone if its density varies along the vertical axis and is given by the function $$\rho \left( y \right) = k{y^2}?$$

Solution.

Consider a thin slice at a height $$y$$ parallel to the base. The radius $$r$$ of the cross section can be determined from the proportion for similar triangles:

${\frac{r}{R} = \frac{{H – y}}{H},}\;\; \Rightarrow {r = \frac{R}{H}\left( {H – y} \right) .}$

The mass of the slice of thickness $$dy$$ is given by

${dm = \rho \left( y \right)dV = \pi {r^2}\rho \left( y \right)dy }={ \frac{{\pi {R^2}\rho \left( y \right){{\left( {H – y} \right)}^2}}}{{{H^2}}}dy }={ \frac{{\pi k{R^2}{y^2}{{\left( {H – y} \right)}^2}}}{{{H^2}}}dy.}$

To compute the total mass of the cone, we integrate from $$y = 0$$ to $$y = H:$$

${m = \frac{{\pi k{R^2}}}{{{H^2}}}\int\limits_0^H {{y^2}{{\left( {H – y} \right)}^2}dy} }={ \frac{{\pi k{R^2}}}{{{H^2}}}\int\limits_0^H {{y^2}\left( {{H^2} – 2Hy + {y^2}} \right)dy} }={ \frac{{\pi k{R^2}}}{{{H^2}}}\int\limits_0^H {\left( {{H^2}{y^2} – 2H{y^3} + {y^4}} \right)dy} }={ \frac{{\pi k{R^2}}}{{{H^2}}}\left. {\left( {\frac{{{H^2}{y^3}}}{3} – \frac{{H{y^4}}}{2} + \frac{{{y^5}}}{5}} \right)} \right|_0^H }={ \pi k{R^2}{H^3}\left( {\frac{1}{3} – \frac{1}{2} + \frac{1}{5}} \right) }={ \frac{{\pi k{R^2}{H^3}}}{{30}}.}$

Let’s check the answer using dimensional analysis. The density is expressed by the function $$\rho \left( y \right) = k{y^2}.$$ So, if the variable $$y$$ is measured in metres and the mass is measured in kilograms, then the coefficient $$k$$ is measured in $$\large{\frac{\text{kg}}{\text{m}^5}}\normalsize.$$ Hence,

$\require{cancel}{{m} = {\frac{{\pi k{R^2}{H^3}}}{{30}}}}={ \left[ {\frac{{kg}}{{{m^5}}}} \right]\left[ {{m^2}} \right]\left[ {{m^3}} \right] }={ \frac{{\left[ {kg} \right]\cancel{\left[ {{m^5}} \right]}}}{{\cancel{\left[ {{m^5}} \right]}}} }={ \left[ {kg} \right].}$

### Example 9.

A right circular cone with base radius $$R$$ and height $$H$$ is formed by rotating about the $$x-$$axis. The density of the cone is given by the function $$\rho \left( x \right) = kx.$$ Find the mass of the cone assuming that the center of its base is placed in the origin.

Solution.

We calculate the mass of the cone by the formula

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

The equation of the straight line $$y = f\left( x \right)$$ is expressed as follows:

${y = f\left( x \right) }={ R – \frac{R}{H}x }={ \frac{R}{H}\left( {H – x} \right).}$

Substituting the density function $$\rho \left( x \right) = kx$$ and integrating from $$x = 0$$ to $$x = H$$, we have

${m = \pi \int\limits_0^H {kx\frac{{{R^2}}}{{{H^2}}}{{\left( {H – x} \right)}^2}dx} }={ \frac{{\pi k{R^2}}}{{{H^2}}}\int\limits_0^H {x{{\left( {H – x} \right)}^2}dx} }={ \frac{{\pi k{R^2}}}{{{H^2}}}\int\limits_0^H {x\left( {{H^2} – 2Hx + {x^2}} \right)dx} }={ \frac{{\pi k{R^2}}}{{{H^2}}}\int\limits_0^H {\left( {{H^2}x – 2H{x^2} + {x^3}} \right)dx} }={ \frac{{\pi k{R^2}}}{{{H^2}}}\left. {\left( {\frac{{{H^2}{x^2}}}{2} – \frac{{2H{x^3}}}{3} + \frac{{{x^4}}}{4}} \right)} \right|_0^H }={ \pi k{R^2}{H^2}\left( {\frac{1}{2} – \frac{2}{3} + \frac{1}{4}} \right) }={ \frac{{5\pi k{R^2}{H^2}}}{{12}}.}$

Note that if the cone’s radius and height are measured in metres and the mass in kilograms, then the coefficient $$k$$ is measured in $${\large{\frac{{\text{kg}}}{{{\text{m}^4}}}}\normalsize}.$$

### Example 10.

The density of the Earth’s inner core is about $$13000\,\large{\frac{\text{kg}}{\text{m}^3}}\normalsize.$$ Suppose that the density of the Earth near the surface is equal to the density of water $$1000\,\large{\frac{\text{kg}}{\text{m}^3}}\normalsize.$$ Estimate the mass of the Earth if the density changes linearly and the Earth’s radius is $$6200\,\text{km}.$$

Solution.

First, let’s derive the equation for calculating the mass of a ball with a linear density distribution.

If we take an arbitrary thin layer of thickness $$dr$$ at a distance $$r$$ from the center, its volume is given by

$dV = 4\pi {r^2}dr.$

The mass of the layer is

${dm = \rho \left( r \right)dV }={ 4\pi \rho \left( r \right){r^2}dr.}$

Hence, the total mass of the ball is given by the integral

$m = 4\pi \int\limits_0^R {\rho \left( r \right){r^2}dr} .$

Assuming that the density decreases linearly, we write it in the form $$\rho \left( r \right) = a – br,$$ where $$a$$ and $$b$$ are positive coefficients that can be found from the boundary conditions. Then the mass of the Earth is expressed as follows:

${m = 4\pi \int\limits_0^R {\left( {a – br} \right){r^2}dr} }={ 4\pi \int\limits_0^R {\left( {a{r^2} – b{r^3}} \right)dr} }={ 4\pi \left. {\left( {\frac{{a{r^3}}}{3} – \frac{{b{r^4}}}{4}} \right)} \right|_0^R }={ 4\pi \left( {\frac{{a{R^3}}}{3} – \frac{{b{R^4}}}{4}} \right).}$

Determine the coefficients $$a$$ and $$b.$$ Given that

${\rho \left( 0 \right) = 13000\frac{{\text{kg}}}{{{\text{m}^3}}},\;\;}\kern0pt{\rho \left( R \right) = 1000\frac{{\text{kg}}}{{{\text{m}^3}}},\;\;}\kern0pt{R = 6200\,\text{km} = 6.2\times10^6\,\text{m,}}$

and using the two-point form of a straight line equation, we get

${\frac{{\rho \left( r \right) – \rho \left( 0 \right)}}{{\rho \left( R \right) – \rho \left( 0 \right)}} = \frac{{r – 0}}{{R – 0}},}\;\; \Rightarrow {\rho \left( r \right) = \rho \left( 0 \right) + \frac{{\rho \left( R \right) – \rho \left( 0 \right)}}{R}r }={ 13000 + \frac{{1000 – 13000}}{{6.2 \times {{10}^6}}}r }={ 13000 – 1.935 \times {10^{ – 3}}r.}$

So, $$a = 13000$$ and $$b = 1.935\times{10^{-3}}.$$

Now we can compute the total mass of the Earth:

${m = 4\pi \left( {\frac{{a{R^3}}}{3} – \frac{{b{R^4}}}{4}} \right) }={ 4\pi \left[ {\frac{{1.3 \times {{10}^4} \times {{\left( {6.2 \times {{10}^6}} \right)}^3}}}{3} }\right.}-{\left.{ \frac{{1.935 \times {{10}^{ – 3}} \times {{\left( {6.2 \times {{10}^6}} \right)}^4}}}{4}} \right] }={ 4\pi \left[ {103.28 \times {{10}^{22}} – 714.81 \times {{10}^{21}}} \right] }\approx{ 4 \times {10^{24}}\,\text{kg}.}$

This result is about $$30\%$$ less than the actual Earth’s mass, which is equal to $$6 \times {10^{24}}\,\text{kg}.$$ This means that the inner layers are actually more dense than the linear approximation suggests.