From Small Pieces to Total Mass
A body need not have the same density everywhere. A rod may be heavier per unit length in some locations than in others, and a solid may have regions with different amounts of material per unit volume. To find the total mass, we add the masses of small pieces throughout the body. When the pieces are made arbitrarily small, this sum becomes an integral.
The type of density depends on the kind of piece being described. A thin rod is measured by length, a thin sheet by area, and a three-dimensional body by volume. The density units reveal which kind of integral is appropriate.
Mass is measured in kilograms. It is not the same as weight, which is a force and is measured in newtons. The integrals in this tutorial calculate mass.
Three Kinds of Density
| Body being modeled | Density | Mass element | Total mass |
|---|---|---|---|
| Rod or wire | Linear density \(\lambda\), in \(\mathrm{kg/m}\) | \(dm\approx\lambda\,dx\) | \(m=\int_a^b\lambda(x)\,dx\) |
| Thin sheet or lamina | Area density \(\sigma\), in \(\mathrm{kg/m^2}\) | \(dm\approx\sigma\,dA\) | \(m=\iint_R\sigma(x,y)\,dA\) |
| Three-dimensional solid | Volume density \(\rho\), in \(\mathrm{kg/m^3}\) | \(dm\approx\rho\,dV\) | \(m=\iiint_B\rho(x,y,z)\,dV\) |
Here \(R\) denotes the region occupied by a lamina, and \(B\) denotes the region occupied by a solid. The symbols \(dA\) and \(dV\) represent infinitesimal area and volume elements. For a rod along the \(x\)-axis, \(dx\) is its length element. In each case, density may vary with position.
Why the Integral Gives Mass
Consider a rod occupying the interval from \(x=a\) to \(x=b\), with a continuous linear density \(\lambda(x)\). Divide the interval into small subintervals. In a subinterval of length \(\Delta x_i\), the density is approximately constant if the subinterval is sufficiently short. If \(x_i^*\) is a point in that subinterval, its mass is approximately \(\lambda(x_i^*)\Delta x_i\). Adding these approximations gives a Riemann sum.
As the largest subinterval length tends to zero, the Riemann sums converge to the integral. Therefore, for a rod with integrable linear density, its mass is
The same reasoning applies to small pieces of a sheet or solid. For a lamina, each piece contributes approximately its area density times its area; for a solid, each piece contributes approximately its volume density times its volume.
Divide it into small pieces whose lengths, areas, or volumes can be measured.
Use density at a representative point multiplied by the measure of that piece.
The sum of the piece masses approximates the whole mass.
As the pieces become arbitrarily small, the sum becomes the appropriate integral.
For a continuous, nonnegative density on a bounded body, this limiting argument can be justified directly. On each small piece, the actual density lies between its minimum and maximum. Thus the actual mass on that piece lies between the minimum density times the piece's measure and the maximum density times that measure. The difference between these bounds is at most the density's oscillation on the piece multiplied by its measure. Continuity on a bounded closed body makes the maximum oscillation tend to zero as the pieces shrink. Since the body's total length, area, or volume is finite, the total difference between the upper and lower estimates tends to zero. The limiting sum therefore gives the total mass. More generally, the corresponding formula applies whenever the density is integrable over the body.
Worked Example: Mass of a Rod with Variable Linear Density
Worked Example: Mass of a Rod with Variable Linear Density
A rod extends from \(x=0\) to \(x=2\,\mathrm{m}\). Its linear density is \(\lambda(x)=(2\,\mathrm{kg/m})+(3\,\mathrm{kg/m^2})x\). Find its mass.
The density has units of mass per length: the first term is in \(\mathrm{kg/m}\), and the second term has units \((\mathrm{kg/m^2})(\mathrm{m})=\mathrm{kg/m}\). The mass is the integral of linear density with respect to length:
Check by differentiating the antiderivative: \(\frac{d}{dx}\left((2\,\mathrm{kg/m})x+(1.5\,\mathrm{kg/m^2})x^2\right)=(2\,\mathrm{kg/m})+(3\,\mathrm{kg/m^2})x=\lambda(x)\). The result also agrees with the average density: the endpoint densities are \(2\,\mathrm{kg/m}\) and \(8\,\mathrm{kg/m}\), so their average is \(5\,\mathrm{kg/m}\); multiplying by the \(2\,\mathrm{m}\) length gives \(10\,\mathrm{kg}\).
Worked Example: Mass of a Uniform Lamina
Worked Example: Mass of a Uniform Lamina
A rectangular sheet covers \(0\le x\le3\,\mathrm{m}\) and \(0\le y\le2\,\mathrm{m}\). Its constant area density is \(\sigma=4\,\mathrm{kg/m^2}\). Find its mass.
For a lamina, multiply area density by an area element and integrate over the sheet:
Check using the rectangle's area: \((3\,\mathrm{m})(2\,\mathrm{m})=6\,\mathrm{m^2}\), and \((4\,\mathrm{kg/m^2})(6\,\mathrm{m^2})=24\,\mathrm{kg}\). The units cancel to kilograms, as required.
Worked Example: Mass of a Solid with Variable Volume Density
Worked Example: Mass of a Solid with Variable Volume Density
A solid occupies \(0\le x\le2\,\mathrm{m}\), \(0\le y\le1\,\mathrm{m}\), and \(0\le z\le1\,\mathrm{m}\). Its volume density is \(\rho(x,y,z)=\left(1+\frac{x}{1\,\mathrm{m}}\right)\mathrm{kg/m^3}\). Find its mass.
The density is \(1\,\mathrm{kg/m^3}\) at \(x=0\) and \(3\,\mathrm{kg/m^3}\) at \(x=2\,\mathrm{m}\). Integrate it over the solid:
Check with average density. Because the density varies linearly from \(1\,\mathrm{kg/m^3}\) to \(3\,\mathrm{kg/m^3}\) along the \(x\)-direction, its average is \(2\,\mathrm{kg/m^3}\). The solid's volume is \((2\,\mathrm{m})(1\,\mathrm{m})(1\,\mathrm{m})=2\,\mathrm{m^3}\), so average density times volume gives \((2\,\mathrm{kg/m^3})(2\,\mathrm{m^3})=4\,\mathrm{kg}\), agreeing with the integral.
Units Prevent a Common Density Error
The measure in the integral must match the dimension used in the density. Multiplying area density by a length does not calculate mass: the units would be \((\mathrm{kg/m^2})(\mathrm{m})=\mathrm{kg/m}\), not kilograms.
For example, for the rectangular sheet above, a mistaken calculation might use only its \(3\,\mathrm{m}\) width and write \((4\,\mathrm{kg/m^2})(3\,\mathrm{m})=12\,\mathrm{kg/m}\). This is not a mass, and it ignores the other \(2\,\mathrm{m}\) dimension. The correction is to use the full area, \(6\,\mathrm{m^2}\), giving \((4\,\mathrm{kg/m^2})(6\,\mathrm{m^2})=24\,\mathrm{kg}\).
Choosing and Interpreting the Mass Integral
To model a mass correctly, first identify whether the material is treated as a rod, a sheet, or a solid. Then use the corresponding density and integrate over the entire occupied interval or region. If the density is constant, the integral reduces to density times length, area, or volume. If density varies, it must remain inside the integral as a function of position.
Density should normally be nonnegative because mass cannot be negative. A density equal to zero in part of a region contributes no mass there. A density formula that is given only on part of a body must be used with care: the integral must still cover the body's full extent, using the appropriate density on each part. The basic principle remains the same—add local mass contributions across the whole body.
Check Your Understanding
Use the density's units and the body's dimensions to decide which mass integral applies.
- What are the SI units of linear, area, and volume density?
- A wire has constant linear density \(6\,\mathrm{kg/m}\) and length \(4\,\mathrm{m}\). What integral gives its mass, and what is the mass?
- A sheet has area density \(2\,\mathrm{kg/m^2}\). Why is multiplying this density by a length alone not enough to obtain its mass?
- For a solid with volume density \(\rho(x,y,z)\), what measure must accompany the density in the mass integral?
- Explain in one or two sentences why adding the masses of increasingly small pieces leads to an integral.