### 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 \(\large{\frac{\text{g}}{\text{cm}}}\normalsize,\) \(x\) 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 \(14200\,\large{\frac{\text{kg}}{\text{m}^3}}\normalsize,\) and its average density near the surface is equal to \(1160\,\large{\frac{\text{kg}}{\text{m}^3}}\normalsize.\) Estimate the mass of the Earth if the density changes linearly and the Earth’s radius is \(6370\,\text{km}.\)### 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 \(14200\,\large{\frac{\text{kg}}{\text{m}^3}}\normalsize,\) and its average density near the surface is equal to \(1160\,\large{\frac{\text{kg}}{\text{m}^3}}\normalsize.\) Estimate the mass of the Earth if the density changes linearly and the Earth’s radius is \(6370\,\text{km}.\)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

\[{\rho \left( 0 \right) = 14200\frac{{\text{kg}}}{{{\text{m}^3}}},\;\;}\kern0pt{\rho \left( R \right) = 1160\frac{{\text{kg}}}{{{\text{m}^3}}},\;\;}\kern0pt{R = 6370\,\text{km} = 6.37\times 10^6\,\text{m,}}\]

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

\[{\frac{{\rho \left( r \right) – \rho \left( 0 \right)}}{{\rho \left( R \right) – \rho \left( 0 \right)}} = \frac{{r – 0}}{{R – 0}},}\;\; \Rightarrow {\rho \left( r \right) = \rho \left( 0 \right) + \frac{{\rho \left( R \right) – \rho \left( 0 \right)}}{R}r }={ 14200 + \frac{{1160 – 14200}}{{6.37 \times {{10}^6}}}r }={ 14200 – 2.047 \times {10^{ – 3}}r.}\]

So, \(a = 14200\) and \(b = 2.047\times{10^{-3}}.\)

Now we can compute the total mass of the Earth:

\[{m = 4\pi \left( {\frac{{a{R^3}}}{3} – \frac{{b{R^4}}}{4}} \right) }={ 4\pi \left[ {\frac{{1.42 \times {{10}^4} \times {{\left( {6.37 \times {{10}^6}} \right)}^3}}}{3} }\right.}-{\left.{ \frac{{2.047 \times {{10}^{ – 3}} \times {{\left( {6.37 \times {{10}^6}} \right)}^4}}}{4}} \right] }={ 4\pi \left[ {122.34 \times {{10}^{22}} – 842.59 \times {{10}^{21}}} \right] }\approx{ 4.79 \times {10^{24}}\,\text{kg}.}\]

This result is about \(20\%\) less than the actual Earth’s mass, which is equal to \(5.97 \times {10^{24}}\,\text{kg}.\) This means that the inner layers are actually more dense than the linear approximation suggests.