Centers of Mass After Coordinate Substitution

From Mass to Center of Mass

A change of variables can turn a complicated solid into a region with simple bounds. But computing a center of mass requires more than transforming the mass integral: each first moment also contains a physical coordinate, and that coordinate must be expressed in the new variables.

The procedure is systematic. Transform the density by composing it with the coordinate map, multiply by the absolute Jacobian, and include the appropriate physical coordinate in each first moment. The center-of-mass coordinates are then the three first moments divided by the mass.

Mass and First Moments

Let $S$ be a solid in physical coordinates $(x,y,z)$, with nonnegative density $\rho(x,y,z)$. Assume the mass and first moments below are finite. The mass $m$ and the first moments about the coordinate planes are

m=\iiint_S \rho(x,y,z)\,dV,\qquad M_{yz}=\iiint_S x\rho(x,y,z)\,dV,

M_{xz}=\iiint_S y\rho(x,y,z)\,dV,\qquad M_{xy}=\iiint_S z\rho(x,y,z)\,dV.

The subscripts indicate the plane used for the moment: for example, $M_{yz}$ weights each mass element by its $x$-coordinate, its distance from the $yz$-plane. If $m>0$, the center of mass 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}.

Definition: The center-of-mass coordinates are the first moments divided by the total mass. Each first moment uses the physical coordinate measuring distance from the corresponding coordinate plane.

The Substitution Formula for Centers of Mass

Suppose a parameter region $U$ has coordinates $(u,v,w)$, and a map $\Phi:U\to S$ is given by

\Phi(u,v,w)=\bigl(x(u,v,w),y(u,v,w),z(u,v,w)\bigr).

For the standard change-of-variables formula, assume that the interiors of the regions are open and that $\Phi$ is a continuously differentiable diffeomorphism between them: it is one-to-one and onto, and its inverse is continuously differentiable. Assume its Jacobian determinant is nonzero in the interiors. Also assume the regions have boundary of volume zero and that the density and coordinate-weighted integrands are measurable and integrable. These conditions let us apply the usual change-of-variables theorem; differentiability alone is not a sufficient hypothesis.

Write $J_\Phi(u,v,w)$ for the determinant of the derivative matrix of $\Phi$. The volume element transforms by the absolute value $\lvert J_\Phi\rvert$, so orientation does not make mass negative. The density in parameter coordinates, including the volume scaling, is $\rho(\Phi(u,v,w))\lvert J_\Phi(u,v,w)\rvert$.

Theorem: Under these hypotheses, the mass and first moments are $$ m=\iiint_U \rho(\Phi(u,v,w))\,\lvert J_\Phi(u,v,w)\rvert\,du\,dv\,dw, $$ $$ M_{yz}=\iiint_U x(u,v,w)\rho(\Phi(u,v,w))\,\lvert J_\Phi(u,v,w)\rvert\,du\,dv\,dw, $$ $$ M_{xz}=\iiint_U y(u,v,w)\rho(\Phi(u,v,w))\,\lvert J_\Phi(u,v,w)\rvert\,du\,dv\,dw,\qquad M_{xy}=\iiint_U z(u,v,w)\rho(\Phi(u,v,w))\,\lvert J_\Phi(u,v,w)\rvert\,du\,dv\,dw. $$

In particular, the factors $x(u,v,w)$, $y(u,v,w)$, and $z(u,v,w)$ are the physical coordinates, composed with the map. They are not automatically $u$, $v$, and $w$.

Why Each Factor Appears

The justification is the ordinary change-of-variables theorem applied separately to the four physical integrands. For mass, take $g(x,y,z)=\rho(x,y,z)$. For the $yz$-moment, take $g(x,y,z)=x\rho(x,y,z)$. The theorem states $\iiint_S g\,dV=\iiint_U g(\Phi(u,v,w))\lvert J_\Phi\rvert\,du\,dv\,dw$ for each integrable $g$. Composing the second integrand gives $x(u,v,w)\rho(\Phi(u,v,w))$; the same argument with $g=y\rho$ and $g=z\rho$ gives the other two moments. Dividing those transformed moments by the transformed mass gives the center.

Compose the density.
Replace $\rho(x,y,z)$ by $\rho(x(u,v,w),y(u,v,w),z(u,v,w))$.

Include the volume factor.
Multiply each integrand by $\lvert J_\Phi(u,v,w)\rvert$.

Include the physical coordinate for each moment.
Use $x(u,v,w)$, $y(u,v,w)$, or $z(u,v,w)$, as appropriate.

Divide by mass.
Calculate $\bar x=M_{yz}/m$, $\bar y=M_{xz}/m$, and $\bar z=M_{xy}/m$.

Worked Example: A Skewed Solid with Density Depending on Position

Worked Example: Transforming All Three First Moments

Let $U=[0,1]^3$, with map $x=u+v,\ y=v,\ z=w$, and density $\rho(x,y,z)=1+x$. Find the mass and center of mass of the image solid.

The derivative matrix and determinant are

D\Phi= \begin{pmatrix} 1&1&0\\ 0&1&0\\ 0&0&1 \end{pmatrix}, \qquad J_\Phi=1.

The transformed density is $1+u+v$. First, the mass is

m=\int_0^1\int_0^1\int_0^1(1+u+v)\,dw\,dv\,du =1+\frac12+\frac12=2.

For the $yz$-moment, the physical $x$-coordinate is $u+v$:

\begin{aligned} M_{yz} &=\int_0^1\int_0^1 (u+v)(1+u+v)\,dv\,du\\ &=\int_0^1\int_0^1\bigl(u+v+u^2+2uv+v^2\bigr)\,dv\,du\\ &=\frac12+\frac12+\frac13+\frac12+\frac13 =\frac{13}{6}. \end{aligned}

For the $xz$-moment, the physical $y$-coordinate is $v$:

\begin{aligned} M_{xz} &=\int_0^1\int_0^1 v(1+u+v)\,dv\,du\\ &=\frac12+\frac14+\frac13 =\frac{13}{12}. \end{aligned}

For the $xy$-moment, the physical $z$-coordinate is $w$. The integrand separates into a $w$-factor and the mass integrand in $u,v$, so

M_{xy} =\left(\int_0^1 w\,dw\right) \left(\int_0^1\int_0^1(1+u+v)\,dv\,du\right) =\frac12\cdot2=1.

Therefore

(\bar x,\bar y,\bar z) =\left(\frac{13/6}{2},\frac{13/12}{2},\frac{1}{2}\right) =\left(\frac{13}{12},\frac{13}{24},\frac12\right).

As a check, multiplying these coordinates by $m=2$ gives $(13/6,13/12,1)$, exactly the three moments computed above.

Worked Example: A Nonlinear Vertical Coordinate

Worked Example: The Jacobian and Physical Height

Let $U=[0,1]\times[0,1]\times[1,2]$, and define $x=u,\ y=v,\ z=w^2$. Let $\rho(x,y,z)=1+z$. Find the mass and center of mass.

The map is one-to-one on the interior, and its Jacobian determinant is $2w>0$. The transformed density, including the absolute Jacobian, is $2w(1+w^2)$. Thus

\begin{aligned} m &=\int_0^1\int_0^1\int_1^2 2w(1+w^2)\,dw\,dv\,du\\ &=\left[w^2+\frac{w^4}{2}\right]_1^2 =12-\frac32=\frac{21}{2}. \end{aligned}

Since $x=u$ and $y=v$, and each has integral $1/2$ over its unit interval,

M_{yz}=\frac12\cdot\frac{21}{2}=\frac{21}{4}, \qquad M_{xz}=\frac12\cdot\frac{21}{2}=\frac{21}{4}.

For the remaining moment, the physical height is $z=w^2$, not $w$:

\begin{aligned} M_{xy} &=\int_1^2 2w^3(1+w^2)\,dw\\ &=\left[\frac{w^4}{2}+\frac{w^6}{3}\right]_1^2\\ &=\left(8+\frac{64}{3}\right)-\left(\frac12+\frac13\right) =\frac{57}{2}. \end{aligned}

The center is therefore

(\bar x,\bar y,\bar z) =\left(\frac{21/4}{21/2},\frac{21/4}{21/2},\frac{57/2}{21/2}\right) =\left(\frac12,\frac12,\frac{19}{7}\right).

As a check, multiplying the coordinates by $21/2$ recovers $M_{yz}=21/4$, $M_{xz}=21/4$, and $M_{xy}=57/2$. The height $19/7$ is between the solid’s physical heights $1$ and $4$.

Worked Example: A Triangular Cross-Section

Worked Example: A Coordinate That Is a Product

Take $U=[1,2]\times[0,1]\times[0,1]$, with $x=u,\ y=uv,\ z=w$, and density $\rho(x,y,z)=1+y$. Find the center of mass of the image.

The derivative matrix is triangular and has determinant $u$, which is positive throughout $U$. Thus the density times the volume factor is $u(1+uv)$. The mass is

\begin{aligned} m &=\int_1^2\int_0^1\int_0^1u(1+uv)\,dw\,dv\,du\\ &=\int_1^2\left(u+\frac{u^2}{2}\right)\,du =\frac32+\frac76=\frac83. \end{aligned}

For $M_{yz}$, include $x=u$:

\begin{aligned} M_{yz} &=\int_1^2\int_0^1 u^2(1+uv)\,dv\,du\\ &=\int_1^2\left(u^2+\frac{u^3}{2}\right)\,du =\frac73+\frac{15}{8} =\frac{101}{24}. \end{aligned}

For $M_{xz}$, the physical coordinate is $y=uv$:

\begin{aligned} M_{xz} &=\int_1^2\int_0^1 u^2v(1+uv)\,dv\,du\\ &=\int_1^2\left(\frac{u^2}{2}+\frac{u^3}{3}\right)\,du =\frac76+\frac54=\frac{29}{12}. \end{aligned}

For $M_{xy}$, the physical coordinate is $z=w$. Its integral over $w\in[0,1]$ is $1/2$, so $M_{xy}=\frac12m=\frac43$. Dividing by the mass gives

(\bar x,\bar y,\bar z) =\left(\frac{101/24}{8/3},\frac{29/12}{8/3},\frac{4/3}{8/3}\right) =\left(\frac{101}{64},\frac{29}{32},\frac12\right).

Multiplying these coordinates by $8/3$ recovers $101/24,\ 29/12,\ 4/3$, respectively. Also $1<101/64<2$, $0<29/32<101/64$, and $0<1/2<1$, consistent with the image solid.

A Common Error: Using a Parameter as a Physical Coordinate

In the last example, the map says $y=uv$, not $y=v$. If $v$ is incorrectly used in the $xz$-moment, the calculation becomes

\begin{aligned} M_{xz}^{\text{wrong}} &=\int_1^2\int_0^1 v\,u(1+uv)\,dv\,du\\ &=\int_1^2\left(\frac{u}{2}+\frac{u^2}{3}\right)\,du =\frac34+\frac79=\frac{55}{36}. \end{aligned}

This gives the incorrect coordinate $\bar y^{\text{wrong}}=(55/36)/(8/3)=55/96$, instead of $29/32=87/96$. The error is not in the integration: it is in substituting the wrong coordinate factor. A reliable check is to write the map’s three physical coordinate formulas beside the three first-moment integrals before evaluating them.

Why Consistency Matters

The center of mass describes the distribution of the original physical mass, not the distribution of parameter values. The Jacobian accounts for how parameter volume changes into physical volume; the composed density accounts for the mass at each physical point; and the physical coordinate in each first moment records that point’s position relative to a coordinate plane. All three ingredients must be present in every transformed integral.

Key takeaway: For a center-of-mass calculation, transform the mass and each first moment separately. In every integral, use the composed density, the absolute Jacobian, and—when calculating a first moment—the corresponding composed physical coordinate.

Check Your Understanding

Answer these questions using the definitions and procedure above.

In a transformed $yz$-moment, which physical coordinate must multiply the density?
Why does the change-of-variables formula use the absolute value of the Jacobian determinant?
If a map has physical coordinate $z=w^2$, what coordinate factor belongs in the $xy$-moment?
For the map $x=u,\ y=uv,\ z=w$, write the transformed $xz$-moment integrand, including the density $\rho=1+y$ and Jacobian factor.
What three conditions or ingredients should you check before dividing a first moment by the mass to obtain a center coordinate?