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Area, Mass, and Moments · Tutorial 581 of 999

Deriving Centroid Formulas From First Moments

Learn how first moments determine the center of mass, and why constant density reduces those formulas to the geometric centroid.

Intermediate 9 min read

What You'll Learn

  • Identify the first moment about each coordinate axis and the coordinate it determines.
  • State conditions ensuring that mass and both first moments are finite.
  • Derive the center-of-mass formulas by matching first moments.
  • Reduce the formulas to geometric centroid coordinates for uniform density.
  • Compute and check centers of mass for uniform and variable-density laminae.

Why First Moments Determine a Center

A planar lamina may spread its mass over a region rather than concentrate it at one point. The first moments summarize how that distributed mass is positioned relative to the coordinate axes. The center of mass is the point that has the same total mass and the same first moments as the whole lamina.

This description leads to the coordinate formulas, but the axis labels require care: the first moment about the \(x\)-axis determines the \(y\)-coordinate, while the first moment about the \(y\)-axis determines the \(x\)-coordinate. We will derive this rather than rely on memorizing it.

Mass and First Moments

Let \(R\) be a measurable region in the plane, and let \(\rho(x,y)\) be its nonnegative mass density per unit area. Assume that the total mass is finite and positive, and that the absolute first moments are finite. In particular, the following conditions suffice:

$$ 0<m=\iint_R \rho(x,y)\,dA<\infty, \qquad \iint_R (|x|+|y|)\rho(x,y)\,dA<\infty. $$

The absolute-integrability condition matters: finite mass alone does not ensure that either first moment is finite. The mass element at \((x,y)\) is \(dm=\rho(x,y)\,dA\), so the total mass is \(m=\iint_R \rho\,dA\).

Definition: The first moment about the \(x\)-axis is \(M_x=\iint_R y\rho(x,y)\,dA\). The first moment about the \(y\)-axis is \(M_y=\iint_R x\rho(x,y)\,dA\). The coordinate in each integrand is the perpendicular distance from the corresponding axis, with its sign.

These integrals can be signed. For example, mass below the \(x\)-axis contributes negatively to \(M_x\), because there \(y<0\). The stated absolute-integrability condition ensures both moments are finite real numbers, even if the region extends on both sides of an axis.

Deriving the Center-of-Mass Formulas

Suppose a single point mass \(m\) is placed at \((\bar{x},\bar{y})\). Its first moment about the \(x\)-axis is \(m\bar{y}\), since every unit of that mass is at signed vertical distance \(\bar{y}\). Its first moment about the \(y\)-axis is \(m\bar{x}\), since its signed horizontal distance from that axis is \(\bar{x}\).

The center of mass of the lamina is defined as the point mass with the same total mass and first moments as the distributed lamina. Matching those quantities gives \(m\bar{y}=M_x\) and \(m\bar{x}=M_y\). Because \(m>0\), division by \(m\) is valid.

Theorem: If a planar lamina has finite positive mass and finite absolute first moments, its center-of-mass coordinates are
$$ \bar{x}=\frac{M_y}{m} =\frac{\displaystyle\iint_R x\rho(x,y)\,dA} {\displaystyle\iint_R \rho(x,y)\,dA}, \qquad \bar{y}=\frac{M_x}{m} =\frac{\displaystyle\iint_R y\rho(x,y)\,dA} {\displaystyle\iint_R \rho(x,y)\,dA}. $$

The order of the subscripts can initially seem surprising. But the first moment about the \(y\)-axis uses \(x\), so it determines the horizontal coordinate \(\bar{x}\). The first moment about the \(x\)-axis uses \(y\), so it determines the vertical coordinate \(\bar{y}\).

1
Compute the mass.
Evaluate \(m=\iint_R \rho\,dA\), and confirm that it is finite and positive.
2
Compute both first moments.
Use \(y\rho\) for \(M_x\), and \(x\rho\) for \(M_y\).
3
Divide by the mass.
Calculate \(\bar{x}=M_y/m\) and \(\bar{y}=M_x/m\), then check that multiplying each coordinate by \(m\) recovers its corresponding first moment.

Worked Example: A Rectangle with Variable Density

Worked Example: A Rectangle with Variable Density

Let \(R=[0,2]\times[0,1]\) and \(\rho(x,y)=1+x\). Find the center of mass. Since \(R\) is bounded and \(\rho\) is continuous and nonnegative, the mass and absolute first moments are finite. The density is positive, so the mass is positive.

First compute the mass. Integrating with respect to \(y\) gives

$$ m=\int_0^2\int_0^1(1+x)\,dy\,dx =\int_0^2(1+x)(1-0)\,dx =\left[x+\frac{x^2}{2}\right]_0^2 =2+2=4. $$

For the moment about the \(y\)-axis, use \(x\rho(x,y)\):

$$ M_y=\int_0^2\int_0^1x(1+x)\,dy\,dx =\int_0^2(x+x^2)\,dx =\left[\frac{x^2}{2}+\frac{x^3}{3}\right]_0^2 =2+\frac{8}{3} =\frac{14}{3}. $$

For the moment about the \(x\)-axis, use \(y\rho(x,y)\):

$$ M_x=\int_0^2\int_0^1y(1+x)\,dy\,dx =\int_0^2(1+x)\left[\frac{y^2}{2}\right]_0^1\,dx =\frac12\int_0^2(1+x)\,dx =\frac12(4)=2. $$

Therefore,

$$ \bar{x}=\frac{M_y}{m} =\frac{14/3}{4}=\frac76, \qquad \bar{y}=\frac{M_x}{m} =\frac{2}{4}=\frac12. $$

Check by multiplying each coordinate by the mass: \(4(7/6)=14/3=M_y\) and \(4(1/2)=2=M_x\). The center is \((7/6,1/2)\), within the rectangle.

Uniform Density and the Geometric Centroid

For a region \(R\) with finite positive area \(A=\iint_R 1\,dA\), the geometric centroid is the area-weighted balance point: each area element contributes equally, regardless of its location. To obtain its coordinates from the center-of-mass formulas, assign the region a constant positive density \(\rho(x,y)=c\). Its mass is \(m=cA\); its moments are \(M_y=c\iint_R x\,dA\) and \(M_x=c\iint_R y\,dA\).

Substitution into the center-of-mass formulas cancels the common factor \(c\). Thus the centroid coordinates are

$$ \bar{x}=\frac{1}{A}\iint_R x\,dA, \qquad \bar{y}=\frac{1}{A}\iint_R y\,dA. $$
Derivation: If \(c>0\) is constant, then \(m=cA\), \(M_y=c\iint_R x\,dA\), and \(M_x=c\iint_R y\,dA\). Hence \(M_y/m=(c\iint_R x\,dA)/(cA)=(1/A)\iint_R x\,dA\), and \(M_x/m=(c\iint_R y\,dA)/(cA)=(1/A)\iint_R y\,dA\). These are exactly the geometric centroid formulas.

For these area formulas to define finite coordinates, require \(0<A<\infty\) and \(\iint_R(|x|+|y|)\,dA<\infty\). The same integrability issue that arises for mass also applies to geometric centroids.

Worked Example: The Geometric Centroid of a Triangle

Worked Example: The Geometric Centroid of a Triangle

Consider the triangle \(R\) bounded by \(x=0\), \(y=0\), and \(y=2-\frac{2x}{3}\), where \(0\leq x\leq3\). Find its geometric centroid using area and first moments. The region is bounded, so its area and coordinate integrals are finite.

First find the area:

$$ A=\int_0^3\int_0^{2-2x/3}1\,dy\,dx =\int_0^3\left(2-\frac{2x}{3}\right)dx =\left[2x-\frac{x^2}{3}\right]_0^3 =6-3=3. $$

The area moment used for \(\bar{x}\) is \(\iint_R x\,dA\):

$$ \iint_R x\,dA =\int_0^3 x\left(2-\frac{2x}{3}\right)dx =\int_0^3\left(2x-\frac{2x^2}{3}\right)dx =\left[x^2-\frac{2x^3}{9}\right]_0^3 =9-6=3. $$

The area moment used for \(\bar{y}\) is \(\iint_R y\,dA\):

$$ \iint_R y\,dA =\int_0^3\left[\frac{y^2}{2}\right]_0^{2-2x/3}dx =\frac12\int_0^3\left(2-\frac{2x}{3}\right)^2dx =2\int_0^3\left(1-\frac{x}{3}\right)^2dx. $$

Expanding and integrating gives

$$ 2\int_0^3\left(1-\frac{2x}{3}+\frac{x^2}{9}\right)dx =2\left[x-\frac{x^2}{3}+\frac{x^3}{27}\right]_0^3 =2(3-3+1)=2. $$

Therefore \(\bar{x}=3/3=1\) and \(\bar{y}=2/3\). Check: \(A\bar{x}=3(1)=3=\iint_R x\,dA\), and \(A\bar{y}=3(2/3)=2=\iint_R y\,dA\). The centroid is \((1,2/3)\).

Worked Example: A Lamina with Density Depending on Both Coordinates

Worked Example: A Lamina with Density Depending on Both Coordinates

Let \(R=[0,1]\times[0,2]\) and \(\rho(x,y)=x+y+1\). Find the center of mass. The density is continuous and positive on this bounded region, so all required integrals are finite and the mass is positive.

Integrating the density with respect to \(y\) gives the mass:

$$ m=\int_0^1\int_0^2(x+y+1)\,dy\,dx =\int_0^1(2x+2+2)\,dx =\int_0^1(2x+4)\,dx =\left[x^2+4x\right]_0^1=5. $$

For \(M_y\), integrate \(x\rho\):

$$ M_y=\int_0^1\int_0^2x(x+y+1)\,dy\,dx =\int_0^1(2x^2+4x)\,dx =\left[\frac{2x^3}{3}+2x^2\right]_0^1 =\frac23+2=\frac83. $$

For \(M_x\), integrate \(y\rho\):

$$ M_x=\int_0^1\int_0^2y(x+y+1)\,dy\,dx =\int_0^1\left(2x+\frac83+2\right)dx =\int_0^1\left(2x+\frac{14}{3}\right)dx =\left[x^2+\frac{14x}{3}\right]_0^1 =\frac{17}{3}. $$

It follows that

$$ \bar{x}=\frac{M_y}{m} =\frac{8/3}{5}=\frac{8}{15}, \qquad \bar{y}=\frac{M_x}{m} =\frac{17/3}{5}=\frac{17}{15}. $$

Check: \(5(8/15)=8/3=M_y\) and \(5(17/15)=17/3=M_x\). Both coordinates lie within the rectangle, as expected for this example.

A Common Mix-Up: Assigning the Wrong Moment

A frequent error is to set \(\bar{x}=M_x/m\) and \(\bar{y}=M_y/m\), reversing the coordinates. For the triangle above, this incorrect assignment would give \(\bar{x}=2/3\) and \(\bar{y}=1\), because its moments are \(M_x=2\), \(M_y=3\), and its area is \(3\).

That assignment fails the defining moment check: \(m\bar{y}=3(1)=3\), which does not equal \(M_x=2\). The correct assignment is \(\bar{x}=M_y/m=3/3=1\) and \(\bar{y}=M_x/m=2/3\). Now \(m\bar{x}=3=M_y\) and \(m\bar{y}=2=M_x\). Remember that a moment about an axis uses distance from that axis.

Why the Integrability Condition Matters

It is not enough to know that a lamina has finite positive mass. A region can extend far enough that its mass is finite while its first moment is infinite. For instance, take \(R=\{(x,y):x\geq1,\ 0<y<1/x^2\}\) and constant density \(\rho=1\). Its mass is

$$ m=\int_1^\infty \frac{1}{x^2}\,dx=1, $$

but its moment about the \(y\)-axis is

$$ M_y=\int_1^\infty\int_0^{1/x^2}x\,dy\,dx =\int_1^\infty \frac{1}{x}\,dx=\infty. $$

Thus \(M_y/m\) is not a finite coordinate. The finite absolute first-moment hypothesis rules out this failure and ensures that the formulas actually define finite center-of-mass coordinates.

When the Formulas Are Useful

The derivation separates the calculation into quantities with distinct roles: mass measures how much material there is, while each first moment measures its signed distribution relative to an axis. For nonuniform density, the density remains inside all three integrals. For uniform density, its constant value cancels, leaving the area-weighted centroid formulas. In either case, matching the moments provides a direct check on the result.

Check Your Understanding

Use the axis labels and moment formulas carefully in each question.

  1. Which coordinate is obtained from the first moment about the \(x\)-axis, and why?
  2. State conditions on mass and absolute first moments that ensure finite center-of-mass coordinates.
  3. For constant density \(c>0\), explain why \(c\) cancels from both center-of-mass formulas.
  4. A lamina has \(m=6\), \(M_x=9\), and \(M_y=12\). Find its center-of-mass coordinates and check them by multiplication.
  5. Why does finite positive mass by itself not guarantee that the center-of-mass coordinates are finite?

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