Mass of Wires With Piecewise Density

Why Density Changes Require Care

A wire may be made from sections of different materials, or its mass per unit length may vary from one part to another. In either case, one density value cannot describe the whole wire. To find the total mass, identify where the density changes, calculate the mass of each portion using its own density, and add the results.

The integration is with respect to arc length, not merely the parameter used to describe the curve. A parameter interval of length \(1\), for instance, need not represent one unit of wire length.

Definition: The linear density \(\lambda\) of a wire is its mass per unit length. If a wire follows a curve \(C\) and has linear density \(\lambda\) at each point, its mass is the scalar line integral \(m=\int_C \lambda\,ds\). For a parametrization \(\mathbf r(t)\), \(a\le t\le b\), this becomes \(m=\int_a^b \lambda(\mathbf r(t))\lVert\mathbf r'(t)\rVert\,dt\), provided the density and parametrization make the integral well-defined.

Here \(ds=\lVert\mathbf r'(t)\rVert\,dt\) is the arc-length element. The density has units of mass per length, so multiplying it by a length element gives mass. For a wire with piecewise continuous density and finitely many change points, the integral can be evaluated one piece at a time.

Splitting the Wire at Density Changes

Suppose a wire is parametrized by \(\mathbf r(t)\) on \([a,b]\), and its density changes at parameter values \(t_1,\ldots,t_{n-1}\), where \(a=t_0<t_1<\cdots<t_n=b\). On each interval \([t_{i-1},t_i]\), use the density formula that applies there. The total mass is the sum of the masses of those pieces.

m = \sum_{i=1}^{n} \int_{t_{i-1}}^{t_i} \lambda_i(\mathbf r(t))\lVert\mathbf r'(t)\rVert\,dt.

The formulas \(\lambda_i\) are the density formulas for the respective portions of the wire. If density is specified directly as a function of the parameter, write that formula in place of \(\lambda_i(\mathbf r(t))\). If density is specified by regions in the plane or space, first find where the curve enters and leaves each region, then split the parameter interval at those values.

Why the Piecewise Formula Works

Mass adds: the mass of two non-overlapping portions of a wire is the sum of their masses. The scalar line integral has the same additivity property. To see how that gives the piecewise formula, take the ordinary parametrized mass integral and divide its interval at the density-change points.

Write the mass integral.
For a parametrization \(\mathbf r(t)\), the integrand is the density at \(\mathbf r(t)\), multiplied by the speed \(\lVert\mathbf r'(t)\rVert\).

Partition the parameter interval.
Choose each \(t_i\) where the density formula changes along the curve. On each open subinterval, one density formula applies.

Use additivity of integration.
The integral over \([a,b]\) is the sum of the integrals over \([t_{i-1},t_i]\). Each term is the mass of one portion.

In equation form, if the density on each portion is \(\lambda_i\), additivity gives

\int_a^b \lambda(\mathbf r(t))\lVert\mathbf r'(t)\rVert\,dt = \sum_{i=1}^{n} \int_{t_{i-1}}^{t_i} \lambda_i(\mathbf r(t))\lVert\mathbf r'(t)\rVert\,dt.

This argument relies on the density being integrable on each piece and on the curve having a well-defined arc-length element there. A finite jump in density does not prevent the integral from existing. The value assigned to the density at a single change point does not affect the mass: one point has zero arc length. This is different from adding a separate point mass, which is not represented by a linear-density integral.

Worked Examples

Worked Example: A Straight Wire with One Change Point

A straight wire runs from \((0,0)\) to \((3,0)\), with coordinates measured in meters. Its density is \(2\ \text{kg/m}\) for \(0\le x<1\), and \(5\ \text{kg/m}\) for \(1\le x\le3\). Find its mass.

Parametrize the wire by \(\mathbf r(t)=(t,0)\), \(0\le t\le3\). Then \(\mathbf r'(t)=(1,0)\), so the speed is \(\sqrt{1^2+0^2}=1\ \text{m per unit }t\). The change point \(x=1\) corresponds to \(t=1\). Therefore,

\begin{aligned} m &=\int_0^1 2(1)\,dt+\int_1^3 5(1)\,dt\\ &=2(1-0)+5(3-1)\\ &=2+10\\ &=12\ \text{kg}. \end{aligned}

The first portion has length \(1\ \text{m}\) and mass \(2\ \text{kg}\); the second has length \(2\ \text{m}\) and mass \(10\ \text{kg}\). Their sum is \(12\ \text{kg}\), as calculated by the integrals.

Worked Example: A Circular Arc with Two Density Values

A wire follows the circular arc \(\mathbf r(\theta)=(2\cos\theta,2\sin\theta)\), \(0\le\theta\le\pi/2\), with lengths measured in meters. Its density is \(3\ \text{kg/m}\) for \(0\le\theta<\pi/3\), and \(1\ \text{kg/m}\) for \(\pi/3\le\theta\le\pi/2\). Find its mass.

Differentiate the parametrization: \(\mathbf r'(\theta)=(-2\sin\theta,2\cos\theta)\). Its speed is

\lVert\mathbf r'(\theta)\rVert = \sqrt{4\sin^2\theta+4\cos^2\theta} = 2.

Split the integral at the change point \(\theta=\pi/3\). The mass on the first portion is

\int_0^{\pi/3}3(2)\,d\theta = 6\left(\frac{\pi}{3}-0\right) = 2\pi\ \text{kg}.

The mass on the second portion is

\int_{\pi/3}^{\pi/2}1(2)\,d\theta = 2\left(\frac{\pi}{2}-\frac{\pi}{3}\right) = 2\left(\frac{\pi}{6}\right) = \frac{\pi}{3}\ \text{kg}.

Thus the total mass is

m=2\pi+\frac{\pi}{3}=\frac{7\pi}{3}\ \text{kg}.

As a check, the first arc has length \(2(\pi/3)=2\pi/3\ \text{m}\), giving mass \(3(2\pi/3)=2\pi\ \text{kg}\). The second has length \(2(\pi/6)=\pi/3\ \text{m}\), giving mass \(\pi/3\ \text{kg}\). These agree with the integral contributions.

Worked Example: A Two-Segment Wire with Varying Densities

A wire consists of a horizontal segment from \((0,0)\) to \((3,0)\), followed by a vertical segment from \((3,0)\) to \((3,4)\). Coordinates are measured in meters. On the horizontal segment, the density at parameter \(t=x\) is \(1+t\ \text{kg/m}\), for \(0\le t\le3\). On the vertical segment, the density at parameter \(t=y\) is \(2+t\ \text{kg/m}\), for \(0\le t\le4\). Find the total mass.

For the horizontal segment, use \(\mathbf r_1(t)=(t,0)\), \(0\le t\le3\). Its speed is \(1\). The mass is

\begin{aligned} m_1 &=\int_0^3(1+t)(1)\,dt\\ &=\left[t+\frac{t^2}{2}\right]_0^3\\ &=3+\frac{9}{2}\\ &=\frac{15}{2}\ \text{kg}. \end{aligned}

For the vertical segment, use \(\mathbf r_2(t)=(3,t)\), \(0\le t\le4\). Its derivative is \((0,1)\), so its speed is also \(1\). Its mass is

\begin{aligned} m_2 &=\int_0^4(2+t)(1)\,dt\\ &=\left[2t+\frac{t^2}{2}\right]_0^4\\ &=8+8\\ &=16\ \text{kg}. \end{aligned}

Adding the segment masses gives

m=m_1+m_2=\frac{15}{2}+16=\frac{47}{2}=23.5\ \text{kg}.

The segment lengths are \(3\ \text{m}\) and \(4\ \text{m}\), consistent with the parameter intervals because both speeds equal \(1\). The total \(23.5\ \text{kg}\) is the sum of the two computed contributions.

A Common Mistake: Using One Density for the Whole Wire

In the first example, a tempting but incorrect calculation is to use the first density, \(2\ \text{kg/m}\), over the entire \(3\ \text{m}\) wire. That gives \(2(3)=6\ \text{kg}\), which ignores the section whose density is \(5\ \text{kg/m}\). The correction is to split at \(x=1\): the first portion contributes \(2(1)=2\ \text{kg}\), and the second contributes \(5(2)=10\ \text{kg}\), for a total of \(12\ \text{kg}\).

A related setup error is to multiply density by \(dt\) without accounting for speed. The correct element is \(ds=\lVert\mathbf r'(t)\rVert\,dt\). In the examples above, the speeds happened to be \(1\) or constant, but for a general parametrization the speed factor must remain in the integral.

Practical Checks Before Adding the Masses

Check the breakpoints: Every place where the density formula changes along the wire must be a boundary between pieces.
Check the intervals: The pieces should cover the entire wire, with no gaps or unintended overlaps.
Check the arc-length factor: Include \(\lVert\mathbf r'(t)\rVert\) unless the parameter itself is arc length.
Check the units: Density times length must have units of mass.
Check the total: Add the mass contributions, not the density values alone.

If a curve crosses the same density boundary more than once, each crossing creates a split in the parameter interval. A density rule defined by spatial regions may therefore produce several separate parameter intervals with the same density formula. Each interval contributes its own integral, and all contributions are included in the total mass.

Check Your Understanding

For each question, identify the relevant density pieces and arc-length factors before evaluating any integrals.

A straight wire is parametrized by \(\mathbf r(t)=(t,0)\), \(0\le t\le4\). Its density is \(2\ \text{kg/m}\) for \(0\le t<2\) and \(3\ \text{kg/m}\) for \(2\le t\le4\). What is its mass?
Why does a jump in density at a single point not change the value of the mass integral?
For a parametrized wire, what factor must multiply the density in an integral with respect to the parameter \(t\)?
A wire crosses a boundary between two density regions twice. How should the parameter interval be divided?
What are the units of \(\lambda\,ds\) if \(\lambda\) is measured in kilograms per meter and \(ds\) in meters?