Why Coordinate Changes Affect Moments of Inertia
A moment of inertia measures how mass is distributed relative to an axis. Mass farther from the axis contributes more because its contribution is weighted by the square of its distance from that axis. For a solid with a complicated boundary, however, setting up the integral in the original coordinates may be awkward. A substitution can simplify the domain, but it also changes the coordinates in the distance factor, the density, and the volume element.
The essential discipline is to transform all three parts consistently. A simple parameter region does not by itself make the integral simple: the physical distance to the axis and the physical density must still be evaluated at the point represented by the new coordinates.
Density is mass per unit volume, so the integrand is density multiplied by squared distance. If length is measured in units of length and mass in units of mass, a moment of inertia has units of mass times length squared.
The Formula After a General Substitution
Let \(F\) map a parameter region \(D'\) in \((u,v,w)\)-space to the physical region \(D\) in \((x,y,z)\)-space: \(F(u,v,w)=(x(u,v,w),y(u,v,w),z(u,v,w))\). Assume the map is one-to-one on the region of integration, is continuously differentiable there, and has nonzero Jacobian determinant in its interior. The change-of-variables formula replaces the physical volume element by \(\left|\det DF(u,v,w)\right|\,du\,dv\,dw\). The absolute value is required because volume is nonnegative, regardless of the orientation of the map.
This formula distinguishes three separate transformations. The distance factor uses the physical coordinates supplied by \(F\); the density is the physical density evaluated at those coordinates; and the Jacobian converts parameter volume to physical volume. In particular, the transformed density is \(\rho\circ F\), not automatically the same formula written in new variables.
For an arbitrary axis, let \(\mathbf{a}\) be a point on the axis and let \(\mathbf{n}\) be a unit vector in its direction. For a physical point \(\mathbf{r}=(x,y,z)\), its squared distance to the axis is \(\lVert\mathbf{r}-\mathbf{a}\rVert^2-((\mathbf{r}-\mathbf{a})\cdot\mathbf{n})^2\). Thus the moment about that axis is the integral of this squared distance times density over the solid. Under a substitution, replace \(\mathbf{r}\) by \(F(u,v,w)\), compose the density with \(F\), and include the absolute Jacobian.
Why the Squared-Distance Factor Has This Form
Here is a short justification of the arbitrary-axis formula. Write \(\mathbf{q}=\mathbf{r}-\mathbf{a}\), and decompose it into its component parallel to the axis and its component perpendicular to the axis. Because \(\mathbf{n}\) is a unit vector, the parallel component is \((\mathbf{q}\cdot\mathbf{n})\mathbf{n}\), whose squared length is \((\mathbf{q}\cdot\mathbf{n})^2\). The two components are perpendicular, so their squared lengths add to \(\lVert\mathbf{q}\rVert^2\). Consequently, the squared length of the perpendicular component—and hence the squared distance to the axis—is
For example, take the \(x\)-axis through the origin. Its direction vector is \(\mathbf{n}=(1,0,0)\), and \(\mathbf{r}\cdot\mathbf{n}=x\). The general squared-distance expression is then \(x^2+y^2+z^2-x^2=y^2+z^2\), giving the coordinate-axis formula in the definition.
Worked Example: A Uniform Ellipsoid
Worked Example: A Uniform Ellipsoid
Find the moment of inertia about the \(x\)-axis of the ellipsoid \(\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}\leq 1\), where \(a,b,c>0\) and the density is the constant \(\rho_0\).
Use the map \(x=au,\ y=bv,\ z=cw\), which sends the unit ball \(u^2+v^2+w^2\leq 1\) to the ellipsoid. Its derivative matrix is diagonal, so \(\det DF=abc\). Since \(a,b,c>0\), the volume factor is \(|\det DF|=abc\). The physical squared distance to the \(x\)-axis becomes \(b^2v^2+c^2w^2\), and the density remains \(\rho_0\). Therefore
By symmetry of the unit ball, the integrals of \(u^2,v^2,w^2\) over it are equal. Their sum is the integral of \(u^2+v^2+w^2\). In spherical coordinates, that sum is \(\int_0^1 r^2r^2\,dr\int_0^\pi\sin\phi\,d\phi\int_0^{2\pi}d\theta =\frac15\cdot2\cdot2\pi=\frac{4\pi}{5}\). Thus each of the three equal integrals is \(\frac{4\pi}{15}\). Substituting the two needed values gives
As a check, when \(a=b=c=R\), the ellipsoid is a ball and the result reduces to \(\frac{8\pi\rho_0R^5}{15}\). The mass of this ball is \(\rho_0\frac{4\pi R^3}{3}\), so the moment is \(\frac{2}{5}MR^2\), consistent with the same expression. The units are also correct: \(\rho_0abc(b^2+c^2)\) has units of mass times length squared.
Worked Example: A Linear Map with a Negative Determinant
Worked Example: A Linear Map with a Negative Determinant
Let \(D'\) be the cube \(-1\leq u,v,w\leq1\), and map it to physical coordinates by \(x=u+v,\ y=u-v,\ z=2w\). For constant density \(\rho_0\), find the moment about the \(z\)-axis.
The derivative matrix and determinant are
Hence the volume factor is \(|\det DF|=4\). The squared distance to the physical \(z\)-axis is \(x^2+y^2=(u+v)^2+(u-v)^2=2u^2+2v^2\). Therefore
The integrand does not depend on \(w\), so its integral over \(w\) contributes a factor of \(2\). Also, \(\int_{-1}^1u^2\,du=\left[\frac{u^3}{3}\right]_{-1}^1=\frac23\), while \(\int_{-1}^1du=2\). Thus the remaining two-dimensional integral is
Multiplying by the density and Jacobian factor yields \(I_z=\frac{128}{3}\rho_0\). As a check, the transformed integral has nonnegative integrand and positive volume factor, so this positive result has the required sign. The negative determinant records orientation reversal; it cannot make the physical volume or the moment negative.
Worked Example: A Nonlinear Map and a Variable Density
Worked Example: A Nonlinear Map and a Variable Density
Consider the map \(x=u,\ y=v,\ z=w+uv\) from the unit cube \(0\leq u,v,w\leq1\). The physical density is \(\rho(x,y,z)=1+x\). Find the moment about the physical \(x\)-axis.
The derivative matrix is
The transformed density is \(\rho(F(u,v,w))=1+u\). The squared distance to the physical \(x\)-axis is \(y^2+z^2=v^2+(w+uv)^2\). The integral to evaluate is therefore
First integrate with respect to \(w\). Expanding the square gives \(v^2+w^2+2uvw+u^2v^2\). Since \(\int_0^1w^2\,dw=\frac13\) and \(\int_0^1 2uvw\,dw=uv\), the result after the \(w\)-integration is
Now integrate each term over \(v\). Using \(\int_0^1v^2\,dv=\frac13\) and \(\int_0^1v\,dv=\frac12\), this becomes
Integrating the resulting terms in \(u\) gives \(\frac23\int_0^1(1+u)\,du+\frac12\int_0^1(u+u^2)\,du+\frac13\int_0^1(u^2+u^3)\,du\). The three contributions are \(\frac23\cdot\frac32=1\), \(\frac12(\frac12+\frac13)=\frac{5}{12}\), and \(\frac13(\frac13+\frac14)=\frac{7}{36}\). Hence
For a check of the integration, the expanded integrand has contributions \(1/2\) from \((1+u)v^2\), \(1/2\) from \((1+u)w^2\), \(5/12\) from the cross term \(2uvw\), and \(7/36\) from \(u^2v^2\). Their sum is again \(29/18\). The density is positive throughout the region, and the squared distance is nonnegative, so a positive moment is reasonable.
A Common Pitfall: Forgetting What Must Be Transformed
In the nonlinear example, it would be incorrect to use the original density formula as \(1+w\), or to leave the distance factor as \(y^2+z^2\) without substituting \(y=v\) and \(z=w+uv\). The density depends on the physical coordinate \(x\), so its transformed value is \(1+u\). The distance factor also uses physical coordinates, so it becomes \(v^2+(w+uv)^2\).
For a concrete wrong calculation, omitting the density factor \(1+u\) would produce the unweighted integral \(\iiint_{[0,1]^3}[v^2+(w+uv)^2]\,du\,dv\,dw\). Its four expanded contributions are \(\frac13\) from \(v^2\), \(\frac13\) from \(w^2\), \(\frac14\) from \(2uvw\), and \(\frac19\) from \(u^2v^2\). This gives \(\frac{37}{36}\), not the correct \(\frac{29}{18}\). The wrong value corresponds to replacing the specified variable density with unit density.
Why the Transformation Is Useful
A suitable substitution can turn an ellipsoid into a ball, a slanted region into a box, or a curved domain into a familiar parameter region. The moment integral may then have symmetric bounds or separable factors that make evaluation manageable. The examples also show why simplification of the domain does not mean that the transformed integral is automatically a constant multiple of a basic volume: the squared-distance factor and density may acquire nonconstant expressions.
When checking a setup, inspect each factor separately: does the squared distance describe distance to the intended physical axis, is the density evaluated at the transformed physical point, and is the volume factor nonnegative? Checking dimensions and the sign of the final result provides additional diagnostics, but neither replaces checking the formula itself.
Check Your Understanding
For each question, identify all factors that must appear in the transformed moment integral before attempting to evaluate it.
- Under \(x=au,\ y=bv,\ z=cw\), what squared-distance factor and volume factor are used for the moment about the \(y\)-axis?
- Why must a Jacobian determinant of \(-6\) contribute a volume factor of \(6\), not \(-6\)?
- If \(\rho(x,y,z)=x^2+z\) and \(F(u,v,w)=(u,v,w+uv)\), what is the transformed density?
- State the squared distance from \(\mathbf{r}\) to an axis through \(\mathbf{a}\) with unit direction vector \(\mathbf{n}\).
- For the map in the nonlinear example, which factor in the integral would be lost by incorrectly treating the density as constant, and what physical assumption would that error make?