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Mass, Moments, and Centroids · Tutorial 478 of 998

Uniform Rectangular Laminae

Use surface density and first moments to calculate a rectangle’s mass and centroid, and prove that a uniform rectangle’s centroid is its geometric center.

Intermediate 9 min read

What You'll Learn

  • Define a coordinate-aligned rectangular lamina and its constant surface density
  • Calculate mass from density and area, with consistent physical units
  • Set up and evaluate first moments about the coordinate axes
  • Derive centroid coordinates for a rectangle with arbitrary bounds
  • Verify that the centroid agrees with the rectangle’s geometric center
  • Identify and correct a common error when assigning first moments

Why Start With a Uniform Rectangle?

A lamina is a thin, flat object whose mass is distributed across a plane. To describe its mass and location, we use its surface density and its first moments. A rectangle with uniform density is a useful case because each part of the sheet has the same mass per unit area, so its centroid should be at its geometric center.

The calculation makes that expectation precise. It also clarifies how the coordinate axes affect the numerical values of the first moments: a rectangle’s centroid stays at its geometric center, but its moments can be positive, negative, or zero depending on where the axes lie.

Definitions and Conventions

Consider a rectangular lamina occupying the coordinate-aligned region \(R=[a,b]\times[c,d]\), where \(a<b\) and \(c<d\). Its width is \(b-a\), and its height is \(d-c\). Let its constant surface density be \(\sigma>0\), measured as mass per unit area.

Definition: The mass of a lamina with surface density \(\rho(x,y)\) over a region \(R\) is \(m=\iint_R \rho(x,y)\,dA\). For a uniform lamina, \(\rho(x,y)=\sigma\) throughout \(R\), so \(m=\sigma\operatorname{Area}(R)\).

The first moment about the \(x\)-axis weights each small mass by its vertical distance \(y\) from that axis. The first moment about the \(y\)-axis weights each small mass by its horizontal distance \(x\). With these standard conventions, the centroid coordinates are found by dividing the corresponding first moments by total mass.

Definition: For a lamina of mass \(m>0\), its first moments and centroid coordinates are $$ M_x=\iint_R y\rho(x,y)\,dA,\qquad M_y=\iint_R x\rho(x,y)\,dA, \qquad \bar{x}=\frac{M_y}{m},\qquad \bar{y}=\frac{M_x}{m}. $$

The subscripts on the moments name the axes about which they are taken. In particular, \(M_x\) uses the \(y\)-coordinate, and \(M_y\) uses the \(x\)-coordinate. If the coordinates and lengths are measured in metres and mass in kilograms, then \(\sigma\) has units \(\mathrm{kg/m^2}\), \(m\) has units \(\mathrm{kg}\), and each first moment has units \(\mathrm{kg\cdot m}\). Dividing a moment by mass therefore gives a length, as a coordinate should.

General Formulas for a Uniform Rectangle

For the rectangle \(R=[a,b]\times[c,d]\), its area is the product of its width and height. Since the density is constant, the mass is that area multiplied by \(\sigma\). The first moments follow by integrating the appropriate coordinate over the rectangle.

$$ m=\sigma(b-a)(d-c). $$
$$ M_x=\sigma\int_a^b\int_c^d y\,dy\,dx, \qquad M_y=\sigma\int_a^b\int_c^d x\,dy\,dx. $$

For \(M_x\), the inner integral is \(\int_c^d y\,dy=(d^2-c^2)/2\), which does not depend on \(x\). The outer integral therefore multiplies it by \(b-a\). For \(M_y\), the inner integral of \(x\) with respect to \(y\) is \(x(d-c)\); integrating that expression with respect to \(x\) gives the result below.

$$ \begin{aligned} M_x &=\sigma\int_a^b\left(\frac{d^2-c^2}{2}\right)dx =\frac{\sigma(b-a)(d^2-c^2)}{2},\\ M_y &=\sigma\int_a^b x(d-c)\,dx =\frac{\sigma(d-c)(b^2-a^2)}{2}. \end{aligned} $$

These expressions are valid for any \(a<b\) and \(c<d\), including rectangles that lie partly or entirely on either side of an axis. The signed coordinates in the integrals are important: a portion below the \(x\)-axis contributes negatively to \(M_x\), and a portion left of the \(y\)-axis contributes negatively to \(M_y\).

Derivation: The Centroid Is the Geometric Center

Now divide each first moment by the mass. The factor \(\sigma\) cancels, as does the rectangle’s dimension perpendicular to the moment’s coordinate. Since \(b^2-a^2=(b-a)(a+b)\) and \(d^2-c^2=(d-c)(c+d)\), the simplification gives the center coordinates.

Theorem: A uniform lamina occupying \(R=[a,b]\times[c,d]\) has centroid $$ \left(\bar{x},\bar{y}\right) = \left(\frac{a+b}{2},\frac{c+d}{2}\right). $$ This is the geometric center of the rectangle.
1
Find the horizontal centroid coordinate.
Using \(\bar{x}=M_y/m\), substitute the formulas for \(M_y\) and \(m\): $$ \bar{x} =\frac{\sigma(d-c)(b^2-a^2)/2}{\sigma(b-a)(d-c)} =\frac{b^2-a^2}{2(b-a)} =\frac{(b-a)(a+b)}{2(b-a)} =\frac{a+b}{2}. $$
2
Find the vertical centroid coordinate.
Using \(\bar{y}=M_x/m\), substitute the formulas for \(M_x\) and \(m\): $$ \bar{y} =\frac{\sigma(b-a)(d^2-c^2)/2}{\sigma(b-a)(d-c)} =\frac{d^2-c^2}{2(d-c)} =\frac{(d-c)(c+d)}{2(d-c)} =\frac{c+d}{2}. $$
3
Interpret the result.
The midpoint of the interval from \(a\) to \(b\) is \((a+b)/2\), and the midpoint of the interval from \(c\) to \(d\) is \((c+d)/2\). Thus the computed centroid is precisely the rectangle’s geometric center.

This derivation also explains why the constant density does not affect the centroid: it multiplies both first moments and the mass by the same factor. The location is determined by the geometry of the rectangle.

Worked Calculations

Worked Example: A Rectangle in the First Quadrant

A uniform rectangular sheet occupies \(R=[0,4]\times[0,3]\), with distances in metres and surface density \(\sigma=2.5\ \mathrm{kg/m^2}\). Find its mass, first moments, and centroid.

The width is \(4-0=4\ \mathrm{m}\), and the height is \(3-0=3\ \mathrm{m}\). Therefore its area and mass are

$$ \operatorname{Area}(R)=4\cdot 3=12\ \mathrm{m^2}, \qquad m=2.5\cdot12=30\ \mathrm{kg}. $$

For the first moment about the \(x\)-axis, integrate \(y\):

$$ \begin{aligned} M_x &=2.5\int_0^4\int_0^3 y\,dy\,dx\\ &=2.5\int_0^4\left[\frac{y^2}{2}\right]_0^3dx\\ &=2.5\int_0^4\frac{9}{2}\,dx\\ &=2.5\cdot\frac{9}{2}\cdot4 =45\ \mathrm{kg\cdot m}. \end{aligned} $$

For the first moment about the \(y\)-axis, integrate \(x\):

$$ \begin{aligned} M_y &=2.5\int_0^4\int_0^3 x\,dy\,dx\\ &=2.5\int_0^4 3x\,dx\\ &=2.5\cdot3\left[\frac{x^2}{2}\right]_0^4\\ &=2.5\cdot3\cdot8 =60\ \mathrm{kg\cdot m}. \end{aligned} $$

Thus

$$ \bar{x}=\frac{M_y}{m}=\frac{60}{30}=2\ \mathrm{m}, \qquad \bar{y}=\frac{M_x}{m}=\frac{45}{30}=1.5\ \mathrm{m}. $$

Check: The geometric center is \(((0+4)/2,(0+3)/2)=(2,1.5)\ \mathrm{m}\), matching the moment calculation. The moments also have the expected units: \(\mathrm{kg\cdot m}\).

Worked Example: A Rectangle Offset From Both Axes

A sheet occupies \(R=[-2,4]\times[1,5]\), with coordinates in centimetres and uniform density \(\sigma=3\ \mathrm{g/cm^2}\). Find its mass, first moments, and centroid.

The width is \(4-(-2)=6\ \mathrm{cm}\), and the height is \(5-1=4\ \mathrm{cm}\). Hence

$$ \operatorname{Area}(R)=6\cdot4=24\ \mathrm{cm^2}, \qquad m=3\cdot24=72\ \mathrm{g}. $$

Using the general moment formulas,

$$ \begin{aligned} M_x &=\frac{3(6)(5^2-1^2)}{2} =\frac{18(25-1)}{2} =9\cdot24 =216\ \mathrm{g\cdot cm},\\ M_y &=\frac{3(4)(4^2-(-2)^2)}{2} =\frac{12(16-4)}{2} =6\cdot12 =72\ \mathrm{g\cdot cm}. \end{aligned} $$

The centroid is

$$ \bar{x}=\frac{72}{72}=1\ \mathrm{cm}, \qquad \bar{y}=\frac{216}{72}=3\ \mathrm{cm}. $$

Check: The geometric center has coordinates \((-2+4)/2=1\) and \((1+5)/2=3\), as calculated. Here both moments are positive because the rectangle lies above the \(x\)-axis and extends on both sides of the \(y\)-axis, with its horizontal center to the right of that axis.

Worked Example: A Rectangle Below the \(x\)-Axis

A uniform sheet occupies \(R=[2,6]\times[-3,1]\), with distances in metres and density \(\sigma=0.8\ \mathrm{kg/m^2}\). Calculate its mass, moments, and centroid, paying attention to signs.

The width and height are each \(4\ \mathrm{m}\), so

$$ \operatorname{Area}(R)=4\cdot4=16\ \mathrm{m^2}, \qquad m=0.8\cdot16=12.8\ \mathrm{kg}. $$

The first moments are

$$ \begin{aligned} M_x &=\frac{0.8(4)(1^2-(-3)^2)}{2} =1.6(1-9) =1.6(-8) =-12.8\ \mathrm{kg\cdot m},\\ M_y &=\frac{0.8(4)(6^2-2^2)}{2} =1.6(36-4) =1.6(32) =51.2\ \mathrm{kg\cdot m}. \end{aligned} $$

Therefore

$$ \bar{x}=\frac{51.2}{12.8}=4\ \mathrm{m}, \qquad \bar{y}=\frac{-12.8}{12.8}=-1\ \mathrm{m}. $$

Check: The geometric center is \(((2+6)/2,(-3+1)/2)=(4,-1)\ \mathrm{m}\), which agrees. The negative \(M_x\) and negative \(\bar{y}\) are appropriate because the rectangle’s vertical midpoint is below the \(x\)-axis.

A Common Moment-Labeling Error

A frequent mistake is to use the same coordinate in a moment and in the centroid coordinate with the matching subscript. That reverses the standard convention. For example, in the first worked example, the incorrect assignment \(\bar{x}=M_x/m\) and \(\bar{y}=M_y/m\) would give

$$ \bar{x}_{\text{wrong}}=\frac{45}{30}=1.5\ \mathrm{m}, \qquad \bar{y}_{\text{wrong}}=\frac{60}{30}=2\ \mathrm{m}. $$

Those coordinates swap the rectangle’s horizontal and vertical midpoints. The correction is to remember that the moment about the \(y\)-axis contains \(x\), while the moment about the \(x\)-axis contains \(y\): \(\bar{x}=M_y/m\) and \(\bar{y}=M_x/m\). Substituting the correct moments gives \(\bar{x}=60/30=2\ \mathrm{m}\) and \(\bar{y}=45/30=1.5\ \mathrm{m}\).

Why the Axis Location Matters

The centroid is a property of the lamina’s mass distribution, while a first moment is measured relative to a chosen axis. Moving the axes changes the coordinates used to calculate the moments, so the moments change. In the examples, the negative vertical moment in the third calculation records the fact that the rectangle lies mostly below the \(x\)-axis; it does not mean that the sheet has negative mass.

For a uniform rectangle, the final centroid coordinates can also be found by taking the midpoint of each coordinate interval. The integral calculation is still valuable: it establishes the result from the definitions of mass and first moment, checks the role of the density, and makes sign conventions explicit. For a different shape or a nonuniform density, a midpoint shortcut need not apply.

Key takeaway: For \(R=[a,b]\times[c,d]\) with constant surface density \(\sigma>0\), the mass is \(\sigma(b-a)(d-c)\), and the centroid is \(\left((a+b)/2,(c+d)/2\right)\). The first moments are \(M_x=\iint_R y\sigma\,dA\) and \(M_y=\iint_R x\sigma\,dA\).

Check Your Understanding

Use the definitions and formulas in this tutorial to answer the following questions.

  1. A uniform rectangle occupies \([1,7]\times[2,4]\) and has density \(5\ \mathrm{kg/m^2}\). What are its area and mass?
  2. For the rectangle in question 1, calculate \(M_x\) and \(M_y\), including units.
  3. What is the centroid of a uniform rectangle occupying \([-5,3]\times[-2,6]\)? Explain why neither coordinate must be positive.
  4. In the standard convention, which coordinate is integrated to find \(M_x\), and which is integrated to find \(M_y\)?
  5. Why does multiplying the surface density of a uniform rectangle by a positive constant change its mass but not its centroid?

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