Centroids and First Moments of Surface Laminae

From Surface Mass to Center of Mass

A surface lamina is a thin sheet whose mass is distributed over a surface. If its density varies from place to place, the center of mass is generally not the geometric center of the sheet. Locations with greater density contribute more strongly to the total mass and to its first moments.

The calculation uses the same scalar surface integrals used to find surface mass: multiply the density by the surface area element and integrate. For the center of mass, weight that mass element by its \(x\)-, \(y\)-, or \(z\)-coordinate before integrating.

Definition: Let \(S\) be a surface in three-dimensional space with nonnegative surface mass density \(\rho\), measured in mass per unit area. Its mass is \(M=\int_S \rho\,dS\). If \(0<M<\infty\) and the coordinate-weighted integrals are finite, its first moments and center of mass are defined by the formulas below.

First Moments and the Center of Mass

The first moment about the \(yz\)-plane weights each mass element by its \(x\)-coordinate. The first moment about the \(xz\)-plane uses the \(y\)-coordinate, and the first moment about the \(xy\)-plane uses the \(z\)-coordinate. The coordinate plane named in each subscript is the plane whose perpendicular coordinate supplies the weight.

M_{yz}=\int_S x\rho\,dS,\qquad M_{xz}=\int_S y\rho\,dS,\qquad M_{xy}=\int_S z\rho\,dS.

When \(M>0\), the center of mass is the point \((\bar{x},\bar{y},\bar{z})\) whose coordinates are the first moments divided by the total mass:

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

The density is part of every integral. In particular, a surface with variable density cannot generally be treated as if each unit of area had equal mass. A constant positive density guarantees that the center of mass equals the geometric centroid; a nonconstant density generally moves the center away from it, although symmetry can make the two points coincide in special cases.

Why the First-Moment Formulas Give the Center

A mass element at position \(\mathbf{r}=(x,y,z)\) contributes \(\rho\,dS\) units of mass. The center of mass is the mass-weighted average of the position vectors. Equivalently, it is the point \(\mathbf{c}=(\bar{x},\bar{y},\bar{z})\) for which the total mass-weighted displacement from \(\mathbf{c}\) is zero.

\int_S(\mathbf{r}-\mathbf{c})\rho\,dS=\mathbf{0}.

Distribute the integral across the vector subtraction and use the fact that \(\mathbf{c}\) is constant over \(S\). The result is \( \int_S\mathbf{r}\rho\,dS-\mathbf{c}\int_S\rho\,dS=\mathbf{0}\). Since \(\int_S\rho\,dS=M>0\), solve for \(\mathbf{c}\):

\mathbf{c} =\frac{1}{M}\int_S\mathbf{r}\rho\,dS =\left( \frac{\int_S x\rho\,dS}{M}, \frac{\int_S y\rho\,dS}{M}, \frac{\int_S z\rho\,dS}{M} \right).

This establishes the coordinate formulas: their numerators are precisely the first moments about the coordinate planes. The condition \(M>0\) matters because division by zero would not define a center of mass. If the density is identically zero, the lamina has no positive mass and these formulas do not apply.

Setting Up the Integrals on a Surface

If a surface is parametrized by \(\mathbf{r}(u,v)=(x(u,v),y(u,v),z(u,v))\) on a parameter domain \(D\), then its area element is \(dS=\lVert\mathbf{r}_u\times\mathbf{r}_v\rVert\,du\,dv\), wherever the parametrization is regular. Thus, for example, the \(x\)-coordinate first moment is

M_{yz} =\iint_D x(u,v)\rho(\mathbf{r}(u,v)) \lVert\mathbf{r}_u\times\mathbf{r}_v\rVert\,du\,dv.

The other two first moments use \(y(u,v)\) and \(z(u,v)\) in place of \(x(u,v)\). For a graph \(z=f(x,y)\), the area element is \(\sqrt{1+f_x^2+f_y^2}\,dx\,dy\); the density and each coordinate in the integrand must still be evaluated at the graph point.

Conditions for finite quantities: A compact piecewise smooth surface has finite area, and a continuous density on it is bounded. These conditions ensure finite mass and finite absolute first moments. More generally, on a bounded surface of finite area, a bounded density also ensures these quantities are finite. Boundedness of the surface alone is not enough: on a noncompact surface a continuous density may be unbounded and nonintegrable.

Worked Examples

Worked Example: A Rectangular Lamina with Variable Density

Let \(S\) be the rectangle \(0\leq x\leq2\), \(0\leq y\leq1\), \(z=0\), with density \(\rho(x,y,0)=1+x\). Find its mass, first moments, and center of mass. Since the lamina lies in the \(xy\)-plane, \(dS=dx\,dy\).

First compute the mass:

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

The moment about the \(yz\)-plane is

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

The moment about the \(xz\)-plane and that about the \(xy\)-plane are

M_{xz} =\int_0^1 y\,dy\int_0^2(1+x)\,dx =\frac12\cdot4=2, \qquad M_{xy}=\int_0^1\int_0^2 0\cdot(1+x)\,dx\,dy=0.

Therefore,

(\bar{x},\bar{y},\bar{z}) =\left(\frac{14/3}{4},\frac{2}{4},\frac{0}{4}\right) =\left(\frac76,\frac12,0\right).

Check by multiplying each coordinate by the mass: \(4(7/6)=14/3=M_{yz}\), \(4(1/2)=2=M_{xz}\), and \(4(0)=0=M_{xy}\). The center lies in the plane \(z=0\), as it must for a lamina contained in that plane.

Worked Example: A Tilted Planar Patch

Parametrize a plane patch by \(\mathbf{r}(u,v)=(u,v,u+2v)\), where \(0\leq u\leq1\) and \(0\leq v\leq1\). Let \(\rho=1+u\). Find its center of mass.

The tangent vectors are \(\mathbf{r}_u=(1,0,1)\) and \(\mathbf{r}_v=(0,1,2)\). Their cross product is \((-1,-2,1)\), with length \(\sqrt{1+4+1}=\sqrt6\). Hence \(dS=\sqrt6\,du\,dv\). The mass is

M=\sqrt6\int_0^1\int_0^1(1+u)\,dv\,du =\sqrt6\int_0^1(1+u)\,du =\frac32\sqrt6.

Using \(x=u\), \(y=v\), and \(z=u+2v\), the three first moments are

\begin{aligned} M_{yz} &=\sqrt6\int_0^1u(1+u)\,du =\sqrt6\left(\frac12+\frac13\right) =\frac56\sqrt6,\\ M_{xz} &=\sqrt6\left(\int_0^1v\,dv\right) \left(\int_0^1(1+u)\,du\right) =\sqrt6\cdot\frac12\cdot\frac32 =\frac34\sqrt6,\\ M_{xy} &=\sqrt6\int_0^1\int_0^1(u+2v)(1+u)\,dv\,du\\ &=\sqrt6\left(\int_0^1u(1+u)\,du +2\int_0^1v\,dv\int_0^1(1+u)\,du\right)\\ &=\sqrt6\left(\frac56+\frac32\right) =\frac73\sqrt6. \end{aligned}

Dividing by \(M\) gives \((\bar{x},\bar{y},\bar{z})=(5/9,1/2,14/9)\). As a check, the plane satisfies \(z=x+2y\); the computed coordinates satisfy \(14/9=5/9+2(1/2)\). Also, multiplying the coordinates by \(M\) recovers \(M_{yz}=5\sqrt6/6\), \(M_{xz}=3\sqrt6/4\), and \(M_{xy}=7\sqrt6/3\), as required.

Worked Example: A Curved Cylindrical Patch

Consider the unit-cylinder patch \(\mathbf{r}(u,v)=(\cos u,\sin u,v)\), with \(0\leq u\leq\pi/2\) and \(0\leq v\leq1\). Let its density be \(\rho=1+v\). Find its center of mass.

Here \(\mathbf{r}_u=(-\sin u,\cos u,0)\) and \(\mathbf{r}_v=(0,0,1)\), so \(\mathbf{r}_u\times\mathbf{r}_v=(\cos u,\sin u,0)\) and \(\lVert\mathbf{r}_u\times\mathbf{r}_v\rVert=1\). The mass is

M=\int_0^{\pi/2}du\int_0^1(1+v)\,dv =\frac{\pi}{2}\left(1+\frac12\right) =\frac{3\pi}{4}.

The coordinate-weighted integrals give

\begin{aligned} M_{yz} &=\int_0^{\pi/2}\cos u\,du\int_0^1(1+v)\,dv =1\cdot\frac32=\frac32,\\ M_{xz} &=\int_0^{\pi/2}\sin u\,du\int_0^1(1+v)\,dv =1\cdot\frac32=\frac32,\\ M_{xy} &=\int_0^{\pi/2}du\int_0^1v(1+v)\,dv =\frac{\pi}{2}\left(\frac12+\frac13\right) =\frac{5\pi}{12}. \end{aligned}

Thus \((\bar{x},\bar{y},\bar{z})=(2/\pi,2/\pi,5/9)\). Check by multiplying by \(3\pi/4\): the resulting moments are \(3/2\), \(3/2\), and \(5\pi/12\), respectively, matching the integrals above.

A Common Mistake: Using Area Instead of Mass

In the rectangular example, the geometric center of the rectangle is \((1,1/2,0)\). Using that point for the center of mass would be wrong: it treats each unit of area as having the same mass, although \(\rho=1+x\) increases with \(x\). The incorrect \(x\)-coordinate would imply \(M_{yz}=M\bar{x}=4(1)=4\), but direct integration gives \(M_{yz}=14/3\), not \(4\). The correct coordinate is \(\bar{x}=M_{yz}/M=(14/3)/4=7/6\). In general, the density must appear in both the mass integral and every first-moment integral.

Why These Quantities Matter

The first moments summarize how the surface’s mass is distributed relative to the coordinate planes. Dividing them by total mass produces a location that balances the mass-weighted position of the lamina. The calculation also provides a useful consistency check: multiplying each center coordinate by the total mass must recover its corresponding first moment.

For physical applications, ensure that the density is nonnegative and that the total mass is finite and positive. A continuous density on a compact piecewise smooth surface meets the needed finiteness conditions. Do not infer finiteness merely from the surface being bounded: a bounded, noncompact surface can support a continuous density that becomes unbounded near its boundary and has infinite integral.

Check Your Understanding

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

Which coordinate is used in the first moment about the \(xz\)-plane?
How does the surface area element enter the first-moment integral for a parametrized patch?
Why must the total mass be positive before the center-of-mass formulas can be used?
For a lamina contained in \(z=0\), what is its first moment about the \(xy\)-plane?
Why does a bounded surface with a continuous density not, by itself, guarantee finite mass?