Density, Centers of Mass and Moments of Inertia

Geometry becomes a physical model through density

For nonnegative volume density $\delta(x,y,z)$ , mass is $m=\iiint_E\delta\,dV$ . If $m>0$ , the center coordinates are

\bar x=\frac{\iiint_E x\delta\,dV}{m},\quad\bar y=\frac{\iiint_E y\delta\,dV}{m},\quad\bar z=\frac{\iiint_E z\delta\,dV}{m}.

A centroid uses uniform density. Symmetry of the shape only implies symmetry of mass if the density shares it. The center lies in the convex hull of the mass distribution, but need not lie inside a nonconvex object such as a ring.

Visual guide

VISUAL GUIDEDensity shifts the center toward the heavier side

Slice density 1 + x
Geometric midpoint
Center of mass x = 5/9

In a unit cube with density δ = 1 + x, slices farther right carry more mass. The x-coordinate of the center is 5/9, to the right of the geometric midpoint 1/2. The other coordinates stay at 1/2 by symmetry.

Worked example: density increases across a box

In the unit cube, let $\delta=1+x$ . The mass is $\int_0^1(1+x)dx=3/2$ . The first $x$ moment is $\int_0^1x(1+x)dx=5/6$ , so $\bar x=5/9$ . Symmetry in $y,z$ gives $\bar y=\bar z=1/2$ .

The center shifts toward the denser side. Using the geometric midpoint in every coordinate would ignore the density model. The coordinate numerator has mass-times-length units, so division by mass gives a length.

Worked example: rotational inertia

Moment of inertia about the $z$ -axis is $I_z=\iiint_E(x^2+y^2)\delta\,dV$ . For a uniform cylinder of radius $R$ , height $H$ , and density $\delta_0$ , cylindrical coordinates give

I_z=\delta_0\int_0^H\int_0^{2\pi}\int_0^R r^2r\,dr\,d\theta\,dz=\frac{\delta_0\pi HR^4}{2}=\frac12mR^2.

This is a second moment weighted by squared distance to the axis, not a first moment used to find a center coordinate. Inertia has mass-times-length-squared units.

For a thin lamina, replace volume density and $dV$ with area density and $dA$ . The same weighted-average logic applies. Always define the reference axis and origin, since moments depend on them.

Practice

Find the centroid of a uniform box $[0,2]\times[0,4]\times[0,6]$ .
Does a density symmetric in $x$ on a symmetric region imply $\bar x=0$ ?
Which factor replaces $x^2+y^2$ for inertia about the $x$ -axis?

Show worked solutions

$(1,2,3)$ by symmetry.
Yes, provided positive finite mass exists: the first-moment integrand is odd in $x$ .
Squared perpendicular distance $y^2+z^2$ .

Further study

MIT OpenCourseWare: Multivariable Calculus provides a full university course with additional lectures and exercises.