# 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

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

Integrating by parts, we obtain:

Taking limits then yields:

By L'Hopital's rule we have that

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

Substitute the given values:

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

Substituting the known functions and limits, we get:

To evaluate the integral we use integration by parts:

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:

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

To evaluate the integral, we change the variable:

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

### 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:

The mass of the slice of thickness \(dy\) is given by

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

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,

### 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

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

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

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

The mass of the layer is

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

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:

Determine the coefficients \(a\) and \(b.\) Given that

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

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

Now we can compute the total mass of the Earth:

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.