From Density to Total Mass
A rod can have different amounts of mass in different parts. If one region is made of a denser material, for example, equal-length pieces from that region and a lighter region will not have equal masses. To find the mass of the whole rod, we need to account for how its density varies along its length.
The relevant quantity is linear density: mass per unit length. When the density depends on position, the total mass is found by accumulating the density over the rod’s interval. The definite integral expresses that accumulation.
Set Up the Physical Model
Place the rod along a coordinate axis. Suppose it extends from position \(x=a\) to position \(x=b\), where \(a\le b\). The variable \(x\) records position along the rod, and \(\rho(x)\) gives the linear density at that position.
The bounds are the positions of the rod’s two ends, not necessarily zero and its length. For instance, a rod occupying positions from \(2\) m to \(6\) m has length \(4\) m, but its mass integral has bounds \(2\) and \(6\).
The formula requires a density model on the interval that can be integrated. In physical applications, density should also be nonnegative throughout the rod: a negative value would not represent a physical mass per unit length. A model that becomes negative on the stated interval must be reconsidered rather than interpreted as ordinary rod density.
Why the Integral Gives Mass
We can justify the formula by dividing the rod into short segments. On a very short segment, the density changes little, so its mass is approximately the density at a point in the segment multiplied by the segment’s length. Adding those approximate masses gives an approximation to the total mass.
Divide \([a,b]\) into subintervals with lengths \(\Delta x_i\). Choose a sample position \(x_i^*\) in each subinterval.
On a segment that is sufficiently short, density is nearly constant, so its mass is approximately \(\rho(x_i^*)\Delta x_i\).
The sum \(\sum_i \rho(x_i^*)\Delta x_i\) approximates the mass of the entire rod.
As the largest segment length tends to zero, these Riemann sums converge to the definite integral \(\int_a^b \rho(x)\,dx\), which is the total mass.
This argument explains why the density is multiplied by a small length before the contributions are added. Density alone is mass per length, not mass. Each product \(\rho(x_i^*)\Delta x_i\) has units of mass, and the integral retains those units.
Worked Examples
Worked Example: A Density That Increases Along the Rod
A rod occupies \(0\le x\le 4\) meters and has density \(\rho(x)=(2\ \text{kg/m})+(3\ \text{kg/m}^2)x\). Find its mass.
The endpoints give the integration bounds \(0\) m and \(4\) m. Integrate the density with respect to position:
Check the antiderivative by differentiating with respect to \(x\): \(\frac{d}{dx}\big((2\ \text{kg/m})x+(\frac{3}{2}\ \text{kg/m}^2)x^2\big)=(2\ \text{kg/m})+(3\ \text{kg/m}^2)x=\rho(x)\). The endpoint evaluation uses \(x=4\) m and \(x=0\) m, and each term in the result has units of kilograms. Thus the mass is \(32\) kg.
Worked Example: A Density That Decreases Along the Rod
A rod occupies \(2\le x\le 6\) meters. Its density is \(\rho(x)=(5\ \text{kg/m})-(0.5\ \text{kg/m}^2)x\). Find the mass.
The model remains positive over the rod: its density is \(4\) kg/m at \(x=2\) m and \(2\) kg/m at \(x=6\) m. Use the actual endpoint positions as bounds:
Differentiate the antiderivative: \(\frac{d}{dx}\big((5\ \text{kg/m})x-(0.25\ \text{kg/m}^2)x^2\big)=(5\ \text{kg/m})-(0.5\ \text{kg/m}^2)x\), as required. Subtracting the lower-end value from the upper-end value gives \(12\) kg, a positive mass with the correct units.
Worked Example: A Curved Density Profile
A rod occupies \(0\le x\le 2\) meters and has density \(\rho(x)=(1\ \text{kg/m})+(1\ \text{kg/m}^3)x^2\). Find its mass.
Although the density is not linear in \(x\), it is a position-dependent linear density, so the same mass formula applies:
For a check, differentiating the antiderivative gives \((1\ \text{kg/m})+(1\ \text{kg/m}^3)x^2\), exactly the density function. Evaluating it at the endpoints gives \(\frac{14}{3}\) kg minus \(0\) kg. The units and positive sign are consistent with a mass.
Common Pitfall: Using One Density Value for the Whole Rod
A frequent error is to choose a density value at one point and multiply it by the full rod length, as if the density were constant. That is generally not the same as accumulating a varying density along the rod.
For the rod in the third example, the midpoint is \(x=1\) m, and \(\rho(1\text{ m})=2\) kg/m. Treating this midpoint density as if it applied everywhere gives the incorrect calculation
The integral gives \(\frac{14}{3}\) kg, not \(4\) kg. The midpoint calculation is an approximation here, not the exact mass. The correction is to integrate the density over the full interval. Multiplying one density value by the length is exact when the density is constant; for a varying density, it is not generally exact.
Another related check is dimensional: multiplying a density in kg/m by a length in m gives kg, but correct units alone do not guarantee the value is correct. The density must also be accumulated over the actual interval using the model provided.
A Reliable Setup for Mass Problems
Before evaluating an integral, make the physical quantities explicit. This short sequence helps prevent errors in bounds, units, and interpretation.
Write the interval of positions occupied by the rod. Use those positions as the limits of integration.
Confirm that the function gives mass per length, and that its position variable uses the same length unit as the endpoints.
Set up \(M=\int_a^b\rho(x)\,dx\), then evaluate the definite integral.
Confirm that the mass has units of mass, is nonnegative, and comes from the stated endpoint positions.
What the Result Represents
The definite integral produces the total mass of the rod, not the density at a particular position. Local density describes how much mass is concentrated per unit length near a point; integrating combines all those local contributions over the rod.
This distinction is useful whenever an object is not uniform. A model may assign different densities to different positions, and the integral converts that spatial description into one total quantity. In each case, the essential setup is the same: identify the interval, use density per unit length, and integrate with respect to position.
Check Your Understanding
Answer these questions using the definition and examples above.
- A rod occupies positions from \(1\) m to \(5\) m. What are the bounds for its mass integral?
- If density is measured in grams per centimeter and position in centimeters, what units should the integral have?
- For \(\rho(x)=(3\ \text{kg/m})+(2\ \text{kg/m}^2)x\) on \(0\le x\le 2\) m, set up the mass integral without evaluating it.
- Why does a negative density value somewhere on the rod signal a problem with a physical density model?
- When is multiplying a single density value by the rod’s length guaranteed to give its exact mass?
