# Calculus

## Applications of Integrals # Mass and Density

## Solved Problems

Click or tap a problem to see the solution.

### 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: $${\rho _0} = {10^7}\frac{{{M_{\odot}}}}{{\text{kpc}}},$$ $$h = {10^4}\,\text{kpc},$$ 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 $$14200\,\frac{\text{kg}}{\text{m}^3},$$ and its average density near the surface is equal to $$1160\,\frac{\text{kg}}{\text{m}^3}.$$ Estimate the mass of the Earth if the density changes linearly and the Earth's radius is $$6370\,\text{km}.$$

### 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: $${\rho _0} = {10^7}\frac{{{M_{\odot}}}}{{\text{kpc}}},$$ $$h = {10^4}\,\text{kpc},$$ 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 $$\frac{\text{kg}}{\text{m}}$$ 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:

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

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 $$\frac{\text{kg}}{\text{m}^5}.$$ Hence,

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

### 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 $${\frac{{\text{kg}}}{{{\text{m}^4}}}}.$$

### Example 10.

The density of the Earth's inner core is about $$14200\,\frac{\text{kg}}{\text{m}^3},$$ and its average density near the surface is equal to $$1160\,\frac{\text{kg}}{\text{m}^3}.$$ 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

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

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:

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

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.