From Density Along a Rod to Total Mass
A rod may be made of material whose density varies from one point to another. If the density is greater in some sections than in others, multiplying the rod’s length by a single density value will not generally give its mass. Instead, we account for the density at each position and add its contribution along the rod.
When density is given by a piecewise linear function, the calculation is especially manageable. We divide the rod at the points where the linear rule changes, find the mass on each section, and add those masses. Geometrically, each section contributes the area of a trapezoid whose parallel sides are the density values at the section’s endpoints.
Setting Up the Mass Integral
Suppose a rod occupies the interval from \(x=a\) to \(x=b\), where position is measured in metres, and its density is \(\rho(x)\) kilograms per metre. A short piece of rod of length \(\Delta x\) near position \(x\) has approximate mass \(\rho(x)\Delta x\). Adding these contributions over the rod and taking the limit leads to the total-mass formula
The units confirm the interpretation: density has units \(\mathrm{kg/m}\), and \(dx\) has units \(\mathrm{m}\), so the integral has units \(\mathrm{kg}\). When the density changes its linear rule at section boundaries, split the integral at every such boundary. For boundaries \(a=x_0<x_1<\cdots<x_n=b\), this gives
A value assigned to the density at a single boundary does not affect the integral. The density can therefore have a jump there: the section to the left may end at one density value, while the section to the right begins at another. Each section is integrated using its own linear rule and endpoint values.
Why a Linear Section Gives a Trapezoid
On a section from \(x=a\) to \(x=b\), let the density be linear, with endpoint values \(\rho(a)\) and \(\rho(b)\). Its rule can be written by linearly interpolating between those values:
Integrating this expression explains why only the endpoint values are needed. Set \(L=b-a\), the length of the section. The integral of the constant part is \(L\rho(a)\). The remaining part integrates to \(\frac{\rho(b)-\rho(a)}{L}\cdot\frac{L^2}{2}\). Combining the terms gives the section-mass formula:
This is also the area formula for a trapezoid: its parallel sides have lengths \(\rho(a)\) and \(\rho(b)\), and the distance between them is \(L\). Density is the vertical quantity in this graph, so its graph’s area has the units of mass. If density is nonnegative, the section mass is the ordinary positive area under its graph.
The formula is exact for a linear section, not an approximation. It works for increasing, decreasing, and constant densities. For a physical rod, density should be nonnegative throughout. Since a linear function stays between its endpoint values on a section, nonnegative endpoint densities guarantee nonnegative density throughout that section.
Worked Example: Two Sections with a Shared Density
Worked Example: Two Sections with a Shared Density
A \(4\)-metre rod has density, in \(\mathrm{kg/m}\), given by \(\rho(x)=1+x\) for \(0\le x\le2\), and \(\rho(x)=4-\frac{x}{2}\) for \(2\le x\le4\). Find its total mass.
Split the mass integral at the point \(x=2\), where the rule changes:
For the first section, integrate and evaluate at its two boundaries:
For the second section:
Adding the section masses gives \(M=4+5=9\ \mathrm{kg}\). To check geometrically, the first section has length \(2\ \mathrm{m}\) and endpoint densities \(1\) and \(3\ \mathrm{kg/m}\), so its trapezoid area is \(2(1+3)/2=4\ \mathrm{kg}\). The second has length \(2\ \mathrm{m}\) and endpoint densities \(3\) and \(2\ \mathrm{kg/m}\), giving \(2(3+2)/2=5\ \mathrm{kg}\). The geometric total, \(4+5=9\ \mathrm{kg}\), agrees with the integrals.
Worked Example: A Jump at a Section Boundary
Worked Example: A Jump at a Section Boundary
A \(5\)-metre rod has density \(4-x\) on \(0\le x\le3\), and density \(2+(x-3)\) on \(3\le x\le5\), with density measured in \(\mathrm{kg/m}\). The density is \(1\ \mathrm{kg/m}\) just to the left of \(x=3\), and \(2\ \mathrm{kg/m}\) at the start of the second section. Find the mass.
Calculate each integral separately. On the first section, the endpoint densities are \(4\) and \(1\ \mathrm{kg/m}\):
On the second section, \(2+(x-3)=x-1\). Its endpoint densities are \(2\) and \(4\ \mathrm{kg/m}\), so the integral is
Therefore \(M=7.5+6=13.5\ \mathrm{kg}\). The trapezoid check gives \(3(4+1)/2=7.5\ \mathrm{kg}\) for the first section and \(2(2+4)/2=6\ \mathrm{kg}\) for the second, again totaling \(13.5\ \mathrm{kg}\). The jump at \(x=3\) creates no extra mass: a single position has zero length and contributes nothing to the integral.
Worked Example: Three Sections and a Geometric Bound
Worked Example: Three Sections and a Geometric Bound
A \(4\)-metre rod is divided into sections with boundaries \(0,1,3,4\) metres. The endpoint densities, in \(\mathrm{kg/m}\), are \(0.5\) and \(1.1\) on the first section, \(1.1\) and \(2.1\) on the second, and \(2.1\) and \(0.9\) on the third. Each section has linear density. Find the total mass and check it against bounds from the minimum and maximum densities.
The first section has length \(1\ \mathrm{m}\). Its mass is
The second section has length \(3-1=2\ \mathrm{m}\), so its mass is
The third section has length \(4-3=1\ \mathrm{m}\). Its mass is
Adding the contributions gives \(M=0.8+3.2+1.5=5.5\ \mathrm{kg}\). To check, use the minimum and maximum endpoint density on each linear section. The section masses must lie between the section length times these densities. Thus the total mass is at least \(1(0.5)+2(1.1)+1(0.9)=3.6\ \mathrm{kg}\), and at most \(1(1.1)+2(2.1)+1(2.1)=7.4\ \mathrm{kg}\). Since \(3.6\le5.5\le7.4\), the calculated mass lies within the bounds. The exact trapezoid calculations above provide the sharper value.
A Common Mistake: Using Only the Rod’s Endpoints
A tempting shortcut is to average the density at the two ends of the entire rod and multiply by the total length. This is correct only when the density is linear across the whole rod. A piecewise linear density may rise or fall within the rod, so its interior values cannot be discarded.
Another useful check is to compare the result with lower and upper estimates. On a linear section, the density stays between its endpoint values, so section length times the smaller endpoint density is a lower bound for its mass, and section length times the larger endpoint density is an upper bound. If a proposed mass lies outside the sum of these bounds, check the section lengths, density values, units, and arithmetic.
Summary
For a rod with piecewise linear density, total mass is found by splitting the integral at every change in the density rule, integrating each section, and adding the results. On a section of length \(L\) with endpoint densities \(\rho_1\) and \(\rho_2\), the exact contribution is \(L(\rho_1+\rho_2)/2\), the area of the corresponding trapezoid. Sectionwise density bounds and the trapezoid geometry provide direct checks on the result.
Check Your Understanding
Use the section boundaries and density rules stated in each question. Include units when reporting mass.
- A linear section is \(3\) metres long, with endpoint densities \(2\) and \(6\ \mathrm{kg/m}\). What is its mass?
- Why should a mass integral be split where the linear density rule changes?
- A section has length \(2\) metres and endpoint densities \(1\) and \(3\ \mathrm{kg/m}\). Give a lower bound, an upper bound, and the exact mass.
- Can a density jump at a single boundary add a positive amount to the rod’s mass? Explain.
- When is multiplying the total rod length by the average of the two end densities a valid way to find mass?
