Mass of a Wire With Continuous Density

From Local Density to Total Mass

A wire can have different amounts of mass in different regions. Its linear density describes how much mass is concentrated per unit of length at each point. To find the wire's total mass, we add the contributions from all its small pieces of length.

When density is constant, this amounts to multiplying density by length. When density varies continuously along the wire, the corresponding sum becomes a scalar line integral. The key is to measure each small piece by its arc length, not merely by a change in the parameter used to describe the curve.

Definition: Let \(C\) be a wire and let \(\lambda\) assign a nonnegative linear density to each point of \(C\). The total mass of the wire is

m=\int_C \lambda\,ds.

Here \(ds\) denotes an element of arc length. If length is measured in meters and mass in kilograms, then \(\lambda\) has units of kilograms per meter and \(m\) has units of kilograms.

The density may be given directly as a function of position, such as \(\lambda(x,y)\), or as a function of a parameter along the wire. In either case, its value at a point describes mass per unit length at that point.

Deriving the Parameter Formula

Suppose a wire is traced by a parametrization \(\mathbf r(t)\), \(a\le t\le b\). To use the usual arc-length element \(ds=\|\mathbf r'(t)\|\,dt\), we require enough regularity for this derivative and the length formula to make sense. In this tutorial, assume that \(\mathbf r\) is piecewise continuously differentiable and has finite length, and that the density \(\lambda\) is continuous along the curve. These conditions ensure that the mass integral is well-defined.

Here is why the parameter formula follows. Divide the parameter interval into small subintervals. On a subinterval near \(t_i\), the density is approximately \(\lambda(\mathbf r(t_i))\), while the corresponding small piece of wire has length approximately \(\|\mathbf r'(t_i)\|\Delta t_i\). The mass of that piece is therefore approximately the product of those two quantities. Adding the approximations and taking the limit gives the integral below.

Describe the wire.
Use a piecewise continuously differentiable parametrization \(\mathbf r(t)\) on \([a,b]\) with finite length.

Compute the speed.
On each smooth portion, the length of a short piece is approximately \(\|\mathbf r'(t)\|\Delta t\).

Add the mass contributions.
Each short piece contributes approximately its density times its length. The limiting sum is the mass integral.

More explicitly, for a partition with sample points \(t_i^*\), the approximate sum is \(\sum_i \lambda(\mathbf r(t_i^*))\|\mathbf r'(t_i^*)\|\Delta t_i\). On each smooth portion, the limit of these sums is the ordinary integral of the continuous function \(\lambda(\mathbf r(t))\|\mathbf r'(t)\|\). Adding across the finitely many smooth portions yields the parameter formula:

Parameter formula: If \(C\) is parametrized as above and \(\lambda\) is continuous along it, then

m=\int_C\lambda\,ds =\int_a^b \lambda(\mathbf r(t))\|\mathbf r'(t)\|\,dt,

where the integral is taken piece by piece if the parametrization is only piecewise continuously differentiable.

The factor \(\|\mathbf r'(t)\|\) converts a change in parameter into a change in physical length. The density must also be evaluated at the point \(\mathbf r(t)\) on the wire. In a coordinate description, this means substituting the parametrization into the density function before integrating.

Worked Examples

Worked Example: A Straight Wire with Varying Density

A straight wire lies on the \(x\)-axis from \(x=0\) meters to \(x=2\) meters. At position \(x\), its density is \(\lambda(x)=4+x\) kilograms per meter, with \(x\) expressed numerically in meters. Find its mass.

Parametrize the wire by \(\mathbf r(t)=(t,0)\), where \(0\le t\le 2\). Its derivative and speed are

\mathbf r'(t)=(1,0), \qquad \|\mathbf r'(t)\|=\sqrt{1^2+0^2}=1.

The density along the wire is \(\lambda(\mathbf r(t))=4+t\). Therefore,

\begin{aligned} m &=\int_0^2 (4+t)(1)\,dt\\ &=\left[4t+\frac{t^2}{2}\right]_0^2\\ &=\left(8+\frac{4}{2}\right)-0\\ &=10\text{ kg}. \end{aligned}

Check: Differentiating \(4t+\frac{t^2}{2}\) gives \(4+t\), the integrand. The density is positive throughout the wire, and the resulting mass is positive. The units are \((\text{kg}/\text{m})(\text{m})=\text{kg}\).

Worked Example: A Curved Wire Given by a Graph

A wire follows the parabola \(y=x^2\) from \((0,0)\) to \((1,1)\). Its density at \((x,y)\) is \(\lambda(x,y)=1/\sqrt{1+4x^2}\) kilograms per meter, with coordinates expressed numerically in meters. Find its mass.

Use the graph parameterization \(\mathbf r(t)=(t,t^2)\), \(0\le t\le1\). Then

\mathbf r'(t)=(1,2t), \qquad \|\mathbf r'(t)\|=\sqrt{1+4t^2}.

Along the curve, \(x=t\), so the density becomes \(\lambda(\mathbf r(t))=1/\sqrt{1+4t^2}\). Substituting both density and speed into the mass formula gives

\begin{aligned} m &=\int_0^1 \frac{1}{\sqrt{1+4t^2}}\sqrt{1+4t^2}\,dt\\ &=\int_0^1 1\,dt\\ &=[t]_0^1\\ &=1\text{ kg}. \end{aligned}

Check: The two square-root factors cancel because the density was chosen to be the reciprocal of the speed along this parametrization. The antiderivative \(t\) differentiates to \(1\), and evaluating it at the endpoints gives \(1-0=1\). The mass has units \((\text{kg}/\text{m})(\text{m})=\text{kg}\).

Worked Example: A Spatial Wire with Density Depending on Height

A wire traces the helix \(\mathbf r(t)=(\cos t,\sin t,t)\) for \(0\le t\le\pi\). Its density is \(\lambda(x,y,z)=(2+z)\) kilograms per meter, where the coordinates are expressed numerically in meters. Find its mass.

Differentiate the parametrization and compute its speed:

\mathbf r'(t)=(-\sin t,\cos t,1), \qquad \|\mathbf r'(t)\| =\sqrt{\sin^2t+\cos^2t+1} =\sqrt{2}.

Since \(z=t\) along the helix, its density there is \(\lambda(\mathbf r(t))=2+t\). Thus,

\begin{aligned} m &=\int_0^\pi (2+t)\sqrt{2}\,dt\\ &=\sqrt{2}\left[2t+\frac{t^2}{2}\right]_0^\pi\\ &=\sqrt{2}\left(2\pi+\frac{\pi^2}{2}\right)\text{ kg}. \end{aligned}

Check: The derivative of \(2t+\frac{t^2}{2}\) is \(2+t\), so the integration is correct. The density is positive for \(0\le t\le\pi\), and the mass is positive. The speed factor accounts for the length of the helix per unit of parameter.

Why the Speed Factor Matters

A parameter measures progress through a description of the curve; it is not automatically a length. For a straight wire parametrized at unit speed, \(\|\mathbf r'(t)\|=1\), so the distinction can be easy to miss. On a general curve, the speed may vary, or may be a constant different from one. In either case, omitting it changes the amount of wire represented by each parameter interval.

For instance, in the helix example, integrating the density alone would give the incorrect calculation

\int_0^\pi(2+t)\,dt =2\pi+\frac{\pi^2}{2},

which is missing the factor \(\sqrt{2}\). This wrong result treats each increase \(dt\) as if it represented a length \(dt\). But the helix has speed \(\sqrt{2}\), so its arc-length element is \(ds=\sqrt{2}\,dt\). The correct mass is therefore \(\sqrt{2}\left(2\pi+\frac{\pi^2}{2}\right)\) kilograms, not \(2\pi+\frac{\pi^2}{2}\) kilograms.

Common pitfall: Do not replace \(ds\) by \(dt\) unless the parametrization has unit speed. For a parametrized wire, first find \(\|\mathbf r'(t)\|\), then integrate \(\lambda(\mathbf r(t))\|\mathbf r'(t)\|\).

Interpreting the Result

The mass formula is a continuous version of adding the masses of many short wire segments. On a small segment, density times length gives an approximate mass; the line integral adds these contributions over the entire curve. Because a physical linear density is nonnegative, the total mass is nonnegative as well. A continuous density on a finite-length wire gives a finite mass.

The formula also distinguishes the role of density from the role of geometry. The function \(\lambda\) describes how mass is distributed, while \(ds\) measures the length of each piece. Both are needed: a density value alone is not a piece's mass, and a length alone does not account for how heavy the material is at that location.

Check Your Understanding

Use the mass definition and parameter formula to answer these questions.

What does linear density measure, and what are its units if mass is measured in kilograms and length in meters?
A wire has parametrization \(\mathbf r(t)\). What factor converts a small parameter interval into an approximate arc length?
For \(\mathbf r(t)=(t,3t)\), \(0\le t\le1\), find the speed \(\|\mathbf r'(t)\|\).
Why must a density given as \(\lambda(x,y)\) be evaluated along the parametrized curve before integration?
What error occurs if one computes \(\int_a^b\lambda(\mathbf r(t))\,dt\) for a parametrization whose speed is not one?