First Moments and Centroids of Solids

From Mass to a Solid’s Centroid

A solid occupies three-dimensional space, so its center of mass requires three coordinates. Each coordinate is a mass-weighted average of the corresponding coordinate of every small part of the solid. The integrals that produce these averages are called first moments.

When a solid has uniform density and its cross-sections are simple shapes, slicing can make these integrals manageable. The mass of each thin slice depends on its cross-sectional area, while its contribution to a first moment also depends on where that slice lies or where its cross-section’s centroid lies.

We will use \(x,y,z\) as the coordinate variables and assume a constant, positive density \(\rho\). For such a solid, mass is density times volume. The centroid of the solid is its center of mass under this uniform-density assumption.

First Moments and Center-of-Mass Coordinates

For a small mass element \(dm\) at position \((x,y,z)\), its contributions to the three first moments are \(x\,dm\), \(y\,dm\), and \(z\,dm\). Adding these contributions throughout the solid gives the first moments with respect to the coordinate planes.

Definition: If a solid has total mass \(m>0\), its first moments about the \(yz\)-, \(xz\)-, and \(xy\)-planes are, respectively,

M_{yz}=\int x\,dm,\qquad M_{xz}=\int y\,dm,\qquad M_{xy}=\int z\,dm.

The center of mass, or centroid for uniform density, is \((\bar{x},\bar{y},\bar{z})\), where

\bar{x}=\frac{M_{yz}}{m},\qquad \bar{y}=\frac{M_{xz}}{m},\qquad \bar{z}=\frac{M_{xy}}{m}.

The subscripts identify the plane: the moment about the \(yz\)-plane uses the perpendicular coordinate \(x\); the moment about the \(xz\)-plane uses \(y\); and the moment about the \(xy\)-plane uses \(z\). This distinction is important when setting up the integrals.

For a uniform-density solid occupying a region \(E\), \(dm=\rho\,dV\), where \(dV\) is a small volume element. Thus the same definitions can be written as volume integrals:

m=\rho\int_E dV,\qquad M_{yz}=\rho\int_E x\,dV,\qquad M_{xz}=\rho\int_E y\,dV,\qquad M_{xy}=\rho\int_E z\,dV.

All three first moments and the mass contain the same density factor. Consequently, when density is uniform, it cancels in each coordinate ratio. The location of the centroid therefore depends on the shape of the solid, not on the numerical value of its uniform density.

Why the Moment Ratios Give the Coordinates

The formulas follow from the same weighted-average idea used for a finite collection of point masses. Suppose masses \(m_1,\ldots,m_n\) lie at \(x\)-coordinates \(x_1,\ldots,x_n\). Their center-of-mass \(x\)-coordinate is \(\sum_i x_i m_i/\sum_i m_i\): mass farther in the positive \(x\)-direction contributes more to the average.

A solid can be divided into small pieces with masses \(\Delta m_i\), located approximately at \((x_i,y_i,z_i)\). The total mass is approximated by \(\sum_i\Delta m_i\), and the \(x\)-weighted total mass by \(\sum_i x_i\Delta m_i\). In the limit as the pieces become arbitrarily small, these sums become integrals. The same reasoning applies to the other coordinates.

Derivation: The center-of-mass coordinate in the \(x\)-direction is the total \(x\)-weighted mass divided by total mass. For a continuous solid, these two quantities are \(M_{yz}=\int x\,dm\) and \(m=\int dm\). Therefore \(\bar{x}=M_{yz}/m\). Replacing \(x\) by \(y\) or \(z\) gives \(\bar{y}=M_{xz}/m\) and \(\bar{z}=M_{xy}/m\).

This derivation also explains why the total mass must be positive: the centroid coordinates are defined by dividing each first moment by \(m\).

Writing the Integrals with Cross-Sectional Slices

Consider slicing the solid with planes perpendicular to the \(x\)-axis. At position \(x\), let the cross-section have area \(A(x)\) and centroid coordinates \((\bar{y}_A(x),\bar{z}_A(x))\) within the \(yz\)-plane. A slice of thickness \(dx\) has volume approximately \(A(x)\,dx\), so its mass is \(\rho A(x)\,dx\).

The slice contributes \(x\) times its mass to \(M_{yz}\). For the other moments, the slice’s mass is distributed across its cross-section: its average \(y\)-coordinate is \(\bar{y}_A(x)\), and its average \(z\)-coordinate is \(\bar{z}_A(x)\). Thus its contributions to \(M_{xz}\) and \(M_{xy}\) use those cross-sectional centroid coordinates.

Slicing formulas: If the solid extends from \(x=a\) to \(x=b\), and each cross-section perpendicular to the \(x\)-axis has area \(A(x)\) and centroid \((\bar{y}_A(x),\bar{z}_A(x))\), then for uniform density \(\rho\),

m=\rho\int_a^b A(x)\,dx,

M_{yz}=\rho\int_a^b xA(x)\,dx,\qquad M_{xz}=\rho\int_a^b \bar{y}_A(x)A(x)\,dx,\qquad M_{xy}=\rho\int_a^b \bar{z}_A(x)A(x)\,dx.

These formulas require cross-sections perpendicular to the chosen axis and correctly specified area and centroid for each section.

The centroid of each cross-section is needed because knowing its area alone does not generally reveal where its mass lies in the \(y\)- and \(z\)-directions. For rectangles, triangles, and disks, the cross-sectional centroid can be found using the standard centroid of that uniform lamina.

Worked Examples

Worked Example: A Solid with Rectangular Cross-Sections

A solid extends over \(0\le x\le2\). At each \(x\), its cross-section is the rectangle \(0\le y\le x\), \(0\le z\le1\). Find its centroid for constant density \(\rho\).

The rectangle has side lengths \(x\) and \(1\), so its area is \(A(x)=x\). Its centroid is at the midpoint of each side: \(\bar{y}_A(x)=x/2\) and \(\bar{z}_A(x)=1/2\). The mass is

m=\rho\int_0^2 x\,dx =\rho\left[\frac{x^2}{2}\right]_0^2 =2\rho.

For the \(x\)-coordinate,

M_{yz}=\rho\int_0^2 xA(x)\,dx =\rho\int_0^2 x^2\,dx =\rho\left[\frac{x^3}{3}\right]_0^2 =\frac{8\rho}{3}, \qquad \bar{x}=\frac{M_{yz}}{m} =\frac{8\rho/3}{2\rho} =\frac{4}{3}.

For the \(y\)-coordinate,

M_{xz}=\rho\int_0^2 \frac{x}{2}\,x\,dx =\frac{\rho}{2}\int_0^2 x^2\,dx =\frac{4\rho}{3}, \qquad \bar{y}=\frac{M_{xz}}{m} =\frac{4\rho/3}{2\rho} =\frac{2}{3}.

For the \(z\)-coordinate,

M_{xy}=\rho\int_0^2 \frac{1}{2}\,x\,dx =\frac{\rho}{2}\left[\frac{x^2}{2}\right]_0^2 =\rho, \qquad \bar{z}=\frac{M_{xy}}{m} =\frac{\rho}{2\rho} =\frac{1}{2}.

The centroid is \((4/3,2/3,1/2)\). To check, the computed moments must equal mass times the corresponding centroid coordinate: \(2\rho(4/3)=8\rho/3=M_{yz}\), \(2\rho(2/3)=4\rho/3=M_{xz}\), and \(2\rho(1/2)=\rho=M_{xy}\).

Worked Example: The Centroid of a Solid Hemisphere

Consider the upper hemisphere of radius \(R\), described by \(0\le z\le R\) and circular cross-sections perpendicular to the \(z\)-axis. Find its centroid for constant density \(\rho\).

At height \(z\), the disk radius is \(\sqrt{R^2-z^2}\), so its area is \(A(z)=\pi(R^2-z^2)\). Each disk is centered at the origin in the \(xy\)-plane, so its centroid coordinates are \(\bar{x}_A(z)=0\) and \(\bar{y}_A(z)=0\). The mass is

m=\rho\pi\int_0^R(R^2-z^2)\,dz =\rho\pi\left[R^2z-\frac{z^3}{3}\right]_0^R =\frac{2\pi\rho R^3}{3}.

The \(x\)- and \(y\)-moments are zero because the corresponding cross-sectional centroid coordinates are zero. Hence \(\bar{x}=0\) and \(\bar{y}=0\). The moment about the \(xy\)-plane is

M_{xy}=\rho\pi\int_0^R z(R^2-z^2)\,dz =\rho\pi\left[\frac{R^2z^2}{2}-\frac{z^4}{4}\right]_0^R =\frac{\pi\rho R^4}{4}.

Therefore,

\bar{z}=\frac{M_{xy}}{m} =\frac{\pi\rho R^4/4}{2\pi\rho R^3/3} =\frac{3R}{8}.

The centroid is \((0,0,3R/8)\). As a check, \(m\bar{z}=(2\pi\rho R^3/3)(3R/8)=\pi\rho R^4/4=M_{xy}\), and the two zero coordinates agree with the zero first moments.

Worked Example: Triangular Cross-Sections That Grow with \(x\)

For \(0\le x\le1\), let each cross-section perpendicular to the \(x\)-axis be a right triangle in the \(yz\)-plane with vertices \((0,0)\), \((1+x,0)\), and \((0,2)\). Find the centroid for constant density \(\rho\).

The triangle’s legs have lengths \(1+x\) and \(2\), so its area is \(A(x)=(1+x)\cdot2/2=1+x\). A right triangle’s centroid lies one-third of the way along each leg from the right-angle vertex. Thus \(\bar{y}_A(x)=(1+x)/3\) and \(\bar{z}_A(x)=2/3\).

m=\rho\int_0^1(1+x)\,dx =\rho\left[x+\frac{x^2}{2}\right]_0^1 =\frac{3\rho}{2}.

The moment and coordinate in the \(x\)-direction are

M_{yz}=\rho\int_0^1x(1+x)\,dx =\rho\left[\frac{x^2}{2}+\frac{x^3}{3}\right]_0^1 =\frac{5\rho}{6}, \qquad \bar{x}=\frac{5\rho/6}{3\rho/2} =\frac{5}{9}.

For the \(y\)-coordinate,

M_{xz}=\rho\int_0^1\frac{1+x}{3}(1+x)\,dx =\frac{\rho}{3}\left[x+x^2+\frac{x^3}{3}\right]_0^1 =\frac{7\rho}{9}, \qquad \bar{y}=\frac{7\rho/9}{3\rho/2} =\frac{14}{27}.

For the \(z\)-coordinate,

M_{xy}=\rho\int_0^1\frac{2}{3}(1+x)\,dx =\frac{2\rho}{3}\left[x+\frac{x^2}{2}\right]_0^1 =\rho, \qquad \bar{z}=\frac{\rho}{3\rho/2} =\frac{2}{3}.

The centroid is \((5/9,14/27,2/3)\). Checking gives \(m\bar{x}=5\rho/6\), \(m\bar{y}=7\rho/9\), and \(m\bar{z}=\rho\), which reproduce the three first moments.

A Common Error: Using the Wrong Coordinate in a Moment

A first moment uses the coordinate perpendicular to its named plane. For the rectangular-section solid above, \(M_{xz}\) is the moment about the \(xz\)-plane, so it weights mass by \(y\), not \(x\). A mistaken setup might use \(M_{xz}=\rho\int_0^2xA(x)\,dx=8\rho/3\), which would give the incorrect result \(\bar{y}=(8\rho/3)/(2\rho)=4/3\).

The correction is to use the cross-section’s average \(y\)-coordinate, \(x/2\): \(M_{xz}=\rho\int_0^2(x/2)A(x)\,dx=4\rho/3\), and hence \(\bar{y}=2/3\). A reliable check is to write all three moment definitions before integrating and match each moment to its coordinate.

Interpreting and Checking the Result

First moments have units of mass times length, while mass has units of mass. Their ratios therefore have units of length, as a coordinate should. In a uniform-density calculation, the density cancels, but including it while setting up the mass and moments helps keep the physical meaning clear.

The centroid is a weighted average of the solid’s coordinates. For a solid with positive mass, it lies within the coordinate range occupied by the solid in each direction. If a computed coordinate lies outside that range, the setup or arithmetic should be checked. For sliced solids, also verify that each cross-sectional area and centroid belongs to the actual slice at the stated coordinate.

Check Your Understanding

Use the definitions and slicing formulas to answer these questions.

Which coordinate is used in the first moment about the \(xz\)-plane?
A solid has cross-sectional area \(A(x)\) and cross-sectional centroid coordinate \(\bar{y}_A(x)\). Write the integrals for its mass and its first moment \(M_{xz}\), assuming constant density \(\rho\).
Why does the density cancel when calculating a centroid for a uniform-density solid?
For the rectangular-section solid in the first worked example, what would be wrong with using \(xA(x)\) in the integral for \(M_{xz}\)?
What are the units of a first moment, and what are the units of a centroid coordinate?