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Mass, Moments, and Centroids · Tutorial 496 of 998

Checking Density Units and Physical Plausibility

Use units, bounds, and scaling behavior to check whether mass and centroid results are physically and mathematically plausible.

Intermediate 9 min read

What You'll Learn

  • Check density, mass, first-moment, and centroid units
  • Use nonnegative density to verify mass and centroid bounds
  • Identify how multiplying density changes mass and centroid
  • Predict how uniform geometric scaling changes dimensions and mass
  • Audit worked lamina and solid calculations with independent checks

Why Plausibility Checks Matter

A mass or centroid integral can be evaluated correctly while the model or setup is wrong. A missing factor, an incorrect density unit, or a centroid outside the object’s coordinate range may not be obvious from the final arithmetic alone. Physical plausibility checks provide independent ways to catch these problems.

The checks in this tutorial concern four related questions: are the units consistent, is the density physically nonnegative, does the centroid lie within the object’s coordinate bounds, and does the result change in the expected way when density or size changes?

Key idea: A valid mass and centroid calculation must agree with both the integral definitions and the physical dimensions and location of the object.

Units in Mass and Centroid Calculations

For a lamina with areal density \(\sigma\), the density has units of mass per area. If length is measured in centimetres and mass in grams, then \(\sigma\) is measured in \(\mathrm{g/cm^2}\). Since an area element \(dA\) has units \(\mathrm{cm^2}\), the mass integral has units of grams.

$$ m=\iint_R \sigma\,dA. $$

For a solid with volume density \(\rho\), the density has units of mass per volume. Thus, if lengths are in centimetres and mass is in grams, \(\rho\) has units \(\mathrm{g/cm^3}\), and \(\iiint_E \rho\,dV\) has units of grams.

First moments include one additional factor of length. For example, a lamina’s first moment used to find its \(x\)-coordinate is \(\iint_R x\sigma\,dA\). Its units are \(\mathrm{cm}\cdot\mathrm{g/cm^2}\cdot\mathrm{cm^2}=\mathrm{g\,cm}\). Dividing this moment by mass gives a centroid coordinate in centimetres.

$$ \bar{x}=\frac{\iint_R x\sigma\,dA}{m}, \qquad \bar{y}=\frac{\iint_R y\sigma\,dA}{m}. $$

For a solid, the same check applies with volume density and volume elements. The numerator for \(\bar{x}\), for instance, has units of mass times length, so the quotient has units of length. A result for \(\bar{x}\) reported in square centimetres or grams is a warning that something has gone wrong.

Unit check: Density times the appropriate area or volume element must have units of mass. A first moment must have units of mass times length, and a centroid coordinate must have units of length.

Nonnegative Density and Centroid Bounds

A physical mass density is nonnegative throughout the material. If the density is nonnegative and the region has positive mass, the centroid is a weighted average of the coordinates occupied by the object. It cannot lie beyond the coordinate extremes of that material.

Theorem: Suppose a lamina lies entirely between the vertical lines \(x=a\) and \(x=b\), where \(a\le b\), and has nonnegative areal density and positive total mass. Then \(a\le\bar{x}\le b\). The corresponding statement holds for each coordinate of a solid or lamina.

Here is a proof for the \(x\)-coordinate. At every point of the lamina, \(a\le x\le b\). Multiplying this inequality by the nonnegative density preserves its direction. Integrating preserves the inequalities as well:

$$ a\iint_R \sigma\,dA \le \iint_R x\sigma\,dA \le b\iint_R \sigma\,dA. $$

The mass \(m=\iint_R\sigma\,dA\) is positive, so division by \(m\) gives \(a\le\bar{x}\le b\). The proof for the other coordinates uses their corresponding coordinate bounds in the same way. These bounds are useful even when the centroid integral is complicated: if a computed coordinate falls outside the object’s range, check the setup, signs, and arithmetic.

Worked Example: Checking a Uniform Solid

Worked Example: Checking a Uniform Solid

A rectangular solid occupies \(0\le x\le2\), \(0\le y\le3\), and \(0\le z\le4\), with lengths in centimetres. Its constant density is \(5\ \mathrm{g/cm^3}\). Find its mass and centroid, and check their units and location.

The volume is \(2\cdot3\cdot4=24\ \mathrm{cm^3}\), so

$$ m=\iiint_E 5\,dV =5(24) =120\ \mathrm{g}. $$

For the \(x\)-coordinate, integrate \(x\rho\). The \(y\)- and \(z\)-integrals contribute factors \(3\) and \(4\):

$$ \iiint_E x\rho\,dV =5\left(\int_0^2 x\,dx\right) \left(\int_0^3dy\right) \left(\int_0^4dz\right) =5(2)(3)(4) =120\ \mathrm{g\,cm}. $$

Therefore, \(\bar{x}=120/120=1\ \mathrm{cm}\). Similarly, the first moments for \(y\) and \(z\) are

$$ 5(2)\left(\int_0^3y\,dy\right)(4) =5(2)\left(\frac{9}{2}\right)(4) =180\ \mathrm{g\,cm}, $$

and

$$ 5(2)(3)\left(\int_0^4z\,dz\right) =5(2)(3)(8) =240\ \mathrm{g\,cm}. $$

Thus \(\bar{y}=180/120=1.5\ \mathrm{cm}\) and \(\bar{z}=240/120=2\ \mathrm{cm}\). The centroid is \((1,1.5,2)\ \mathrm{cm}\). Each coordinate lies within its respective interval, and each first moment divided by mass has units of centimetres. As a further check, the constant-density centroid is at the midpoint of each interval: \(1\), \(1.5\), and \(2\).

What Changes When Density or Size Changes?

If the density is multiplied everywhere by a positive constant \(c\), mass and every first moment are multiplied by \(c\). Their ratio, each centroid coordinate, is unchanged. This follows directly from the integral definitions: the same factor appears in the numerator and denominator.

Uniform geometric scaling has a different effect. Suppose every coordinate of a lamina is multiplied by \(s>0\), and the density at corresponding points keeps the same numerical value in mass per area. Lengths and centroid coordinates are multiplied by \(s\), area and mass by \(s^2\). For a solid with unchanged numerical density per volume, volume and mass are multiplied by \(s^3\), while centroid coordinates are multiplied by \(s\). These mass factors assume the same density value after scaling; a change in density would contribute an additional factor.

Scaling check: Multiplying density alone changes mass but not the centroid. Scaling all lengths by \(s\) scales centroid coordinates by \(s\); with fixed density, it scales lamina mass by \(s^2\) and solid mass by \(s^3\).

Worked Example: A Lamina With Variable Density

Worked Example: A Lamina With Variable Density

Consider the rectangle \(0\le x\le2\), \(0\le y\le1\), measured in centimetres, with areal density \(\sigma(x,y)=2+x\ \mathrm{g/cm^2}\). Find its mass and centroid, then check plausibility.

The mass is

$$ m=\int_0^2\int_0^1(2+x)\,dy\,dx =\int_0^2(2+x)\,dx =\left[2x+\frac{x^2}{2}\right]_0^2 =6\ \mathrm{g}. $$

The first moment that determines \(\bar{x}\) is

$$ \int_0^2\int_0^1x(2+x)\,dy\,dx =\int_0^2(2x+x^2)\,dx =\left[x^2+\frac{x^3}{3}\right]_0^2 =4+\frac{8}{3} =\frac{20}{3}\ \mathrm{g\,cm}. $$

Hence \(\bar{x}=(20/3)/6=10/9\ \mathrm{cm}\).

For the other coordinate,

$$ \int_0^2\int_0^1y(2+x)\,dy\,dx =\left(\int_0^1y\,dy\right)\left(\int_0^2(2+x)\,dx\right) =\frac12(6) =3\ \mathrm{g\,cm}. $$

Thus \(\bar{y}=3/6=1/2\ \mathrm{cm}\), and the centroid is \((10/9,1/2)\ \mathrm{cm}\). Both coordinates lie in the rectangle’s bounds. Also, the density is larger at larger \(x\), so a centroid with \(\bar{x}>1\), the rectangle’s midpoint coordinate, is plausible. The calculated value \(10/9\) passes that check.

Worked Example: Testing Density and Scaling

Worked Example: Testing Density and Scaling

A rod occupies \(0\le x\le3\ \mathrm{m}\), with linear density \(\lambda(x)=2+x\ \mathrm{kg/m}\). Check its mass and centroid, then predict what happens if its density is tripled everywhere.

The mass is

$$ m=\int_0^3(2+x)\,dx =\left[2x+\frac{x^2}{2}\right]_0^3 =6+\frac{9}{2} =\frac{21}{2}\ \mathrm{kg}. $$

The first moment is

$$ \int_0^3x(2+x)\,dx =\left[x^2+\frac{x^3}{3}\right]_0^3 =9+9 =18\ \mathrm{kg\,m}. $$

Therefore, \(\bar{x}=18/(21/2)=12/7\ \mathrm{m}\), which lies between \(0\) and \(3\ \mathrm{m}\). If the density is tripled, the new mass is \(3(21/2)=63/2\ \mathrm{kg}\), and the new first moment is \(3(18)=54\ \mathrm{kg\,m}\). Their ratio is \(54/(63/2)=12/7\ \mathrm{m}\), so the centroid is unchanged. This agrees with the general density-multiplication check.

A Common Plausibility Mistake

Suppose a positive-density lamina lies in \(0\le x\le2\), but a calculation reports \(\bar{x}=2.4\). One might accept the result because the integrals were evaluated mechanically. That is not justified: the centroid bound requires \(0\le\bar{x}\le2\). The result signals a problem to investigate, such as a reversed sign, incorrect moment integrand, or arithmetic error. It does not, by itself, identify which step failed.

Another common error is to suppose that doubling density doubles the centroid coordinates. It doubles mass and first moments, but the centroid is their ratio. If both numerator and denominator are multiplied by \(2\), their quotient is unchanged.

1
Check dimensions.
Confirm that the density and differential element produce mass, and that moments produce mass times length.
2
Check the density and mass.
For physical density, look for nonnegative values and a positive total mass when material is present.
3
Check coordinate bounds.
Compare each centroid coordinate with the minimum and maximum coordinate of the material.
4
Check how the answer responds.
Test whether density multiplication or uniform geometric scaling changes mass and centroid as expected.

Check Your Understanding

Use units, bounds, and scaling behavior to assess each situation.

  1. A solid has density in \(\mathrm{kg/m^3}\) and volume in \(\mathrm{m^3}\). What are the units of its mass?
  2. A lamina lies entirely within \(1\le x\le5\) and has nonnegative density and positive mass. What bounds must its \(\bar{x}\) satisfy?
  3. If the density of an object is multiplied everywhere by \(4\), what happens to its mass and centroid?
  4. For a solid with unchanged density, what factor multiplies its mass when every length is scaled by \(3\)?
  5. A centroid coordinate has units of \(\mathrm{g\,cm}\). What dimensional check should be made to locate the likely error?

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