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A Worked Workshop on Wire Mass and Mean Density

Learn to combine segment and circular-arc contributions to find a wire’s total mass and average density.

Intermediate 10 min read

What You'll Learn

  • Set up mass integrals for joined wire pieces with coordinate-dependent density
  • Parameterize straight segments and circular arcs with bounds matching the wire
  • Include the correct speed or arc-length factor in each integral
  • Add piecewise masses and lengths to find total mass and mean density
  • Check units and verify that the calculated average is consistent with mass

From a Density Field to a Wire’s Mass

A wire may be made from several connected pieces, and its density may vary with position. To find the total mass, calculate the mass contributed by each piece and add those contributions. To find the mean density, divide the total mass by the wire’s total length.

The calculations are straightforward once each piece has the right parameterization and bounds. The main things to keep track of are the density evaluated along the curve, the length represented by a parameter interval, and the units of the result.

Definition: If a wire occupies a curve \(C\), has linear density \(\delta\) along that curve, and has finite length \(L\), its mass is \(M=\int_C \delta\,ds\). Its average density with respect to arc length is \(\overline{\delta}=M/L\).

Here, \(ds\) denotes an element of arc length. If a piece of the wire is parameterized by \(\mathbf{r}(t)\), for \(a\le t\le b\), then \(ds=\|\mathbf{r}'(t)\|\,dt\). The corresponding mass contribution is therefore the integral of the density evaluated at \(\mathbf{r}(t)\), multiplied by the speed \(\|\mathbf{r}'(t)\|\).

$$ M_C=\int_a^b \delta(\mathbf{r}(t))\|\mathbf{r}'(t)\|\,dt, \qquad L_C=\int_a^b \|\mathbf{r}'(t)\|\,dt. $$

For the examples below, coordinates are given by their numerical values in meters, and density formulas give numerical values in kilograms per meter. Thus, for example, \(\delta(x,y)=2+x\) means that the density is \(2+x\) kg/m when \(x\) is entered as its numerical coordinate in meters. Parameter values and bounds are chosen to trace each physical piece once.

Why Mass Adds Across Joined Pieces

Suppose a wire consists of finitely many smooth pieces \(C_1,\ldots,C_n\), joined at endpoints. The endpoints contribute no length, so they contribute no mass to a line integral. The mass of the whole wire is the sum of the masses of its pieces. Its length is likewise the sum of the piece lengths.

$$ M=\sum_{j=1}^{n} M_j, \qquad L=\sum_{j=1}^{n} L_j, \qquad \overline{\delta}=\frac{\sum_{j=1}^{n}M_j}{\sum_{j=1}^{n}L_j}. $$

This additivity explains why each piece can be parameterized and integrated separately, even when the pieces have different shapes or require different parameters. It also explains why the average density is not usually the ordinary average of the separate piece averages: each piece’s average must be weighted by its length.

Derivation: On a short arc-length element of length \(\Delta s\), the mass is approximately the local density times that length, \(\delta\,\Delta s\). Summing these contributions over a partition gives approximations to total mass. As the partition becomes finer, these sums approach \(\int_C\delta\,ds\). If the curve is parameterized by \(\mathbf{r}(t)\), a small parameter change \(\Delta t\) corresponds to length approximately \(\|\mathbf{r}'(t)\|\Delta t\). Substituting this length into the sums gives the parameter integral \(\int_a^b\delta(\mathbf{r}(t))\|\mathbf{r}'(t)\|\,dt\). For adjacent pieces, the sums of the contributions become sums of their integrals, which gives the additivity formulas above.

Worked Example: An L-Shaped Wire with Varying Density

Worked Example: An L-Shaped Wire with Varying Density

An L-shaped wire consists of the segment from \((0,0)\) to \((2,0)\), followed by the segment from \((2,0)\) to \((2,1)\). Its density is \(\delta(x,y)=2+x\). Find its total mass and average density.

Parameterize the horizontal segment by \(\mathbf{r}_1(t)=(t,0)\), with \(0\le t\le2\). Its speed is \(\|\mathbf{r}'_1(t)\|=1\), and its density along the segment is \(2+t\). Therefore,

$$ M_1=\int_0^2(2+t)(1)\,dt =\left[2t+\frac{t^2}{2}\right]_0^2 =(4+2)-0 =6\text{ kg}. $$

The horizontal segment has length

$$ L_1=\int_0^2 1\,dt=[t]_0^2=2\text{ m}. $$

Parameterize the vertical segment by \(\mathbf{r}_2(t)=(2,t)\), with \(0\le t\le1\). Its speed is also \(1\). Since \(x=2\) throughout this piece, its density is constant at \(2+2=4\) kg/m. Thus,

$$ M_2=\int_0^1 4(1)\,dt=[4t]_0^1=4\text{ kg}, \qquad L_2=\int_0^1 1\,dt=1\text{ m}. $$

Add the two masses and the two lengths:

$$ M=M_1+M_2=6+4=10\text{ kg}, \qquad L=L_1+L_2=2+1=3\text{ m}. $$

The average density is

$$ \overline{\delta}=\frac{M}{L} =\frac{10\text{ kg}}{3\text{ m}} =\frac{10}{3}\text{ kg/m}. $$

Check: Multiplying the average density by the total length recovers the mass: \((10/3\text{ kg/m})(3\text{ m})=10\text{ kg}\). The mass units and average-density units are therefore consistent.

Worked Example: A Coordinate-Dependent Density on a Circular Arc

Worked Example: A Coordinate-Dependent Density on a Circular Arc

A wire is the upper semicircle of radius \(2\) m centered at the origin. Its density is \(\delta(x,y)=3+y\). Find its mass and average density.

Use \(\mathbf{r}(\theta)=(2\cos\theta,2\sin\theta)\), with \(0\le\theta\le\pi\). This interval traces the upper semicircle once. The speed is

$$ \|\mathbf{r}'(\theta)\| =\|(-2\sin\theta,2\cos\theta)\| =\sqrt{4\sin^2\theta+4\cos^2\theta} =2. $$

Along the arc, \(y=2\sin\theta\), so the density becomes \(3+2\sin\theta\). The mass is

$$ \begin{aligned} M &=\int_0^\pi(3+2\sin\theta)(2)\,d\theta\\ &=\int_0^\pi(6+4\sin\theta)\,d\theta\\ &=[6\theta-4\cos\theta]_0^\pi\\ &=(6\pi-4\cos\pi)-(0-4\cos 0)\\ &=(6\pi+4)-(-4)\\ &=6\pi+8\text{ kg}. \end{aligned} $$

The arc length is

$$ L=\int_0^\pi 2\,d\theta=[2\theta]_0^\pi=2\pi\text{ m}. $$

Consequently,

$$ \overline{\delta} =\frac{6\pi+8\text{ kg}}{2\pi\text{ m}} =\left(3+\frac{4}{\pi}\right)\text{ kg/m}. $$

Check: Multiplying the average density by the length gives \(\left(3+\frac{4}{\pi}\right)(2\pi)=6\pi+8\), so \(\overline{\delta}L=M\). The density on the arc ranges from \(3\) kg/m at the endpoints to \(5\) kg/m at the top. The calculated average \(3+4/\pi\) kg/m is between those values, as expected.

Worked Example: A Joined Segment and Circular Arc

Worked Example: A Joined Segment and Circular Arc

A wire consists of a straight segment from \((0,0)\) to \((1,0)\), followed by a quarter-circle arc from \((1,0)\) to \((2,1)\). The arc has radius \(1\) m and center \((1,1)\). The density on both pieces is \(\delta(x,y)=2+x\). Find the total mass and mean density.

For the straight segment, take \(\mathbf{r}_1(t)=(t,0)\), \(0\le t\le1\). The speed is \(1\), and the density along it is \(2+t\). Hence

$$ M_1=\int_0^1(2+t)\,dt =\left[2t+\frac{t^2}{2}\right]_0^1 =2+\frac12 =\frac52\text{ kg}, \qquad L_1=\int_0^1 1\,dt=1\text{ m}. $$

For the arc, use \(\mathbf{r}_2(\theta)=(1+\cos\theta,1+\sin\theta)\), with \(-\pi/2\le\theta\le0\). At \(\theta=-\pi/2\), the point is \((1,0)\); at \(\theta=0\), it is \((2,1)\). The parameter therefore follows the required quarter-circle. Its speed is

$$ \|\mathbf{r}'_2(\theta)\| =\|(-\sin\theta,\cos\theta)\| =\sqrt{\sin^2\theta+\cos^2\theta} =1. $$

Along the arc, \(x=1+\cos\theta\), so the density is \(2+x=3+\cos\theta\). Its mass and length are

$$ \begin{aligned} M_2 &=\int_{-\pi/2}^{0}(3+\cos\theta)(1)\,d\theta\\ &=[3\theta+\sin\theta]_{-\pi/2}^{0}\\ &=0-\left(-\frac{3\pi}{2}-1\right)\\ &=\frac{3\pi}{2}+1\text{ kg}, \end{aligned} \qquad L_2=\int_{-\pi/2}^{0}1\,d\theta=\frac{\pi}{2}\text{ m}. $$

Combining the piece contributions gives

$$ M=M_1+M_2 =\frac52+\frac{3\pi}{2}+1 =\frac72+\frac{3\pi}{2}\text{ kg}, \qquad L=L_1+L_2=1+\frac{\pi}{2}\text{ m}. $$

Therefore, the mean density is

$$ \overline{\delta} =\frac{\frac72+\frac{3\pi}{2}}{1+\frac{\pi}{2}}\text{ kg/m} =\frac{7+3\pi}{2+\pi}\text{ kg/m}. $$

Check: Multiplying the result by the total length gives \(\frac{7+3\pi}{2+\pi}(1+\frac{\pi}{2})=(7+3\pi)/2=\frac72+\frac{3\pi}{2}\) kg, exactly the calculated mass. The segment contributes kilograms through \((5/2\text{ kg/m})(1\text{ m})\), and the arc contributes kilograms through its density integral; all mass terms have matching units.

Why the Mean Must Be Length-Weighted

Each piece can have a different average density. A longer piece contributes more length over which that density is present, so the overall mean is a length-weighted average of the piece means. If piece \(j\) has length \(L_j\) and mean density \(\overline{\delta}_j\), then its mass is \(M_j=L_j\overline{\delta}_j\). Hence

$$ \overline{\delta} =\frac{\sum_{j=1}^{n}L_j\overline{\delta}_j}{\sum_{j=1}^{n}L_j}. $$

For instance, if a \(2\)-m piece has mean density \(3\) kg/m and a \(1\)-m piece has mean density \(6\) kg/m, their combined mean is \((2\cdot3+1\cdot6)/(2+1)=4\) kg/m, not \((3+6)/2\) kg/m.

A Common Mistake: Averaging the Piece Means Directly

In the L-shaped example, the horizontal segment’s mean density is \(6/2=3\) kg/m and the vertical segment’s mean density is \(4/1=4\) kg/m. A tempting but incorrect calculation is to average those two numbers equally:

$$ \frac{3+4}{2}=\frac72\text{ kg/m}. $$

This gives the two segments equal influence despite their different lengths. The horizontal piece is twice as long as the vertical one. The correction is to divide total mass by total length:

$$ \frac{6+4}{2+1}=\frac{10}{3}\text{ kg/m}. $$

This correction works because mass accumulates over length. More generally, use the piece lengths as weights, or first add the masses and lengths and then form their ratio.

A Reliable Calculation Checklist

1
Separate the pieces.
List each segment or arc and identify its endpoints and shape.
2
Parameterize and set bounds.
Check that the parameter interval traces exactly the required piece, in the intended direction and exactly once.
3
Evaluate density and speed.
Substitute the parameterization into the density, then multiply by \(\|\mathbf{r}'(t)\|\) in the mass integral.
4
Find mass and length piece by piece.
Keep separate totals for mass and length, with their units.
5
Combine and verify.
Compute \(\overline{\delta}=M/L\), check that its units are mass per length, and confirm that \(\overline{\delta}L=M\).

For a physical wire, density should be nonnegative along the wire. A finite-length piecewise smooth wire with an integrable density has a well-defined mass integral. If the wire has positive total length, its average density is defined by dividing that mass by the total length.

Check Your Understanding

For each question, identify the mass and length contributions needed before calculating the final average.

  1. A segment is parameterized by \(\mathbf{r}(t)=(t,0)\), \(0\le t\le3\), with density \(\delta(x,y)=1+x\). What integrand gives its mass?
  2. Why must the speed factor \(\|\mathbf{r}'(t)\|\) appear in a parameterized mass integral?
  3. A wire has two pieces with masses \(5\) kg and \(9\) kg and lengths \(2\) m and \(6\) m. What is its total mass and average density?
  4. For a circle of radius \(r\) parameterized by \((r\cos\theta,r\sin\theta)\), what is the speed with respect to \(\theta\)?
  5. Two joined pieces have different mean densities. Why is their combined mean generally not the unweighted average of those two means?

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