From Varying Density to One Mean Value
A rod may be heavier in some regions than in others. Its linear density describes how much mass is present per unit length at each position. When we need one number to represent the whole rod, we use its mean density.
The mean is not generally the density at every point. Instead, it is the constant density that a uniform rod of the same length would need in order to have the same total mass. This interpretation connects a local description, which varies along the rod, to a single overall value.
Definition and Conditions
Let a rod occupy the interval from \(x=a\) to \(x=b\), where \(a<b\). Its length is \(L=b-a\). Let \(\rho(x)\) denote its linear density at position \(x\), measured in units of mass per unit length. For example, if mass is measured in kilograms and length in metres, \(\rho(x)\) has units of kilograms per metre.
Assume that \(\rho\) is integrable on \([a,b]\), so that the total mass is given by the density integral. In typical physical models, the density is continuous and nonnegative. The total mass \(M\) and the mean density \(\bar{\rho}\) are then defined by:
The units confirm the calculation: mass divided by length has units of mass per unit length, just like \(\rho(x)\). The total mass \(M\) is measured in units of mass, not density. Keeping these quantities distinct is important when interpreting a result.
Why the Formula Gives the Mean
There are two ways to see why dividing by length is the right operation. First, the average value of an integrable function on an interval of positive length is its integral divided by the interval length. Since the rod’s mass is the integral of its density, the mean density is exactly this average value applied to \(\rho\).
A second argument gives the physical interpretation directly. A uniform rod with constant linear density \(c\) and length \(L\) has mass \(cL\). To find the constant density that gives the same mass as the original rod, set \(cL=M\). Because \(L>0\), this equation has the unique solution \(c=M/L\).
Integrate the varying density over the rod: \(M=\int_a^b\rho(x)\,dx\).
A uniform density \(c\) over length \(L\) gives mass \(cL\).
Solving \(cL=M\) gives \(c=M/L=\bar{\rho}\), the mean density.
This reasoning also shows what the mean does and does not claim. It preserves the total mass and the rod’s length, but it does not say that the actual density is constant or that it equals the mean everywhere.
Worked Examples
Worked Example: A Density That Increases Linearly
A rod extends from \(x=0\) to \(x=4\,\text{m}\). Its density is \(\rho(x)=2\,\text{kg/m}+(3\,\text{kg/m}^2)x\). Find its total mass and mean density, then compare the mean with the density at the ends.
Integrate the density to obtain the mass:
Now divide by the rod’s length, \(4\,\text{m}\):
The endpoint densities are \(\rho(0)=2\,\text{kg/m}\) and \(\rho(4\,\text{m})=14\,\text{kg/m}\). Thus the mean, \(8\,\text{kg/m}\), lies between the endpoint values. It represents the constant density of a \(4\,\text{m}\) rod with the same \(32\,\text{kg}\) mass.
For a check, differentiating \(2\,\text{kg/m}\,x+\frac{3\,\text{kg/m}^2}{2}x^2\) gives \(2\,\text{kg/m}+(3\,\text{kg/m}^2)x\), the original density. Also, \((8\,\text{kg/m})(4\,\text{m})=32\,\text{kg}\), the calculated mass.
Worked Example: A Density That Increases Quadratically
A \(2\,\text{m}\) rod has density \(\rho(x)=1\,\text{kg/m}+(1\,\text{kg/m}^3)x^2\) for \(0\le x\le2\,\text{m}\). Find its mean density and compare it with the density at the two ends.
First compute the mass:
The mean density is therefore:
At the left end, \(\rho(0)=1\,\text{kg/m}\). At the right end, \(\rho(2\,\text{m})=1\,\text{kg/m}+4\,\text{kg/m}=5\,\text{kg/m}\). The mean density is between these values, but it is not their arithmetic average.
To check the mass calculation, differentiate \(1\,\text{kg/m}\,x+\frac{1\,\text{kg/m}^3}{3}x^3\). The result is the given \(\rho(x)\). The uniform-rod check also works: \((7/3\,\text{kg/m})(2\,\text{m})=14/3\,\text{kg}\).
Worked Example: A Density That Rises and Falls
A rod occupies \(0\le x\le2\,\text{m}\) and has density \(\rho(x)=\left(4+2\cos\left(\frac{\pi x}{1\,\text{m}}\right)\right)\,\text{kg/m}\). Find the mean density and compare it with the local density at the ends and midpoint.
Integrate over the whole rod:
Divide by the length \(2\,\text{m}\): \(\bar{\rho}=8\,\text{kg}/(2\,\text{m})=4\,\text{kg/m}\). At each end, the density is \(6\,\text{kg/m}\), while at the midpoint \(x=1\,\text{m}\), it is \(2\,\text{kg/m}\). The density varies on both sides of the mean, but its mean is still determined by total mass divided by length.
Differentiating the displayed antiderivative gives \(4\,\text{kg/m}+2\,\text{kg/m}\cos\left(\frac{\pi x}{1\,\text{m}}\right)\), the original density. Finally, \((4\,\text{kg/m})(2\,\text{m})=8\,\text{kg}\), confirming the total mass.
Comparing the Mean with Local Density
For a continuous density on a closed interval, the density has a minimum and a maximum there. Its mean lies between them. In particular, if \(m\le \rho(x)\le M_{\max}\) throughout the rod, integrating these inequalities gives:
Since \(b-a>0\), dividing through by the length yields \(m\le\bar{\rho}\le M_{\max}\). The mean summarizes the whole rod; it does not specify where the density is high or low. Different density profiles can have the same integral and therefore the same mean.
A Common Mistake: Averaging the Endpoint Values
A frequent error is to take the arithmetic average of the densities at the two ends instead of averaging density over the entire interval. For the quadratic example, the endpoint average would be:
This is not the rod’s mean density. The correct value is \(\frac{7}{3}\,\text{kg/m}\), found from the total mass divided by length. The endpoint calculation uses only two local values and ignores how the density behaves between them. It agrees with the true mean in some special cases, but it is not the general formula.
Another mistake is to report the total mass as the mean density. In the first example, \(32\,\text{kg}\) is the mass, while \(8\,\text{kg/m}\) is the mean density. Dividing by the length both gives the correct quantity and changes the units appropriately.
Why Mean Density Is Useful
Mean density provides a compact comparison between rods of different lengths or density profiles. If two rods have the same mean density, each has the same mass per unit length on average, even if one is uniform and the other varies substantially. Conversely, knowing the mean and the length immediately gives the total mass through \(M=\bar{\rho}(b-a)\).
The calculation therefore has a clear sequence: model the local density, integrate to find total mass, then divide by length. The integral accounts for every position along the rod; the division converts the accumulated mass into one representative density.
Check Your Understanding
Answer each question using the definition of mean density and the distinction between local density and total mass.
- A rod has length \(5\,\text{m}\) and total mass \(20\,\text{kg}\). What is its mean linear density, including units?
- Why does the formula for mean density require division by the rod’s length?
- If a continuous density ranges from \(2\,\text{kg/m}\) to \(9\,\text{kg/m}\), what bounds must its mean density satisfy?
- Why is the arithmetic average of the endpoint densities not generally the rod’s mean density?
- A rod’s density varies along its length. What property does a uniform rod with the same length and mean density share with it?
