Centroids of Rotationally Symmetric Solids

Introduction

A centroid describes the balance point of a solid. For a solid with constant density, it is the geometric center of mass. If density varies from place to place, the centroid shifts toward regions containing more mass.

For a solid symmetric about the $z$-axis, rotational symmetry gives an immediate simplification: the centroid lies on that axis, provided the density shares the same symmetry. The remaining task is to calculate its vertical coordinate by comparing the solid's mass with its moment about the $xy$-plane.

Mass and Centroid in Cylindrical Coordinates

Let a solid $E$ have nonnegative density $\rho$, and suppose its total mass $M$ is positive and finite. The mass-weighted average of each coordinate gives the centroid $(\bar{x},\bar{y},\bar{z})$: each coordinate's first moment is divided by $M$.

Definition: The mass and centroid of a solid $E$ with density $\rho$ are $$ M=\iiint_E \rho\,dV,\qquad \bar{x}=\frac{1}{M}\iiint_E x\rho\,dV,\qquad \bar{y}=\frac{1}{M}\iiint_E y\rho\,dV,\qquad \bar{z}=\frac{1}{M}\iiint_E z\rho\,dV. $$ The integrals defining the first moments must exist, and $M>0$.

In cylindrical coordinates, $x=r\cos\theta$, $y=r\sin\theta$, and the volume element is $dV=r\,dz\,dr\,d\theta$. The factor $r$ is the cylindrical-coordinate Jacobian. For a solid of revolution about the $z$-axis, let $D$ describe its radial and vertical bounds in the $rz$-plane. At each $(r,z)$ in $D$, the solid includes the full range $0\leq\theta\leq2\pi$. If the density is also rotationally symmetric, write it as $\rho(r,z)$, with no dependence on $\theta$.

Symmetry result: For a solid that includes every angle from $0$ to $2\pi$ at each point of its $rz$-description, with density independent of $\theta$, the centroid satisfies $\bar{x}=\bar{y}=0$. Its remaining coordinate is $$ \bar{z}= \frac{\displaystyle\iiint_E z\rho\,dV} {\displaystyle\iiint_E \rho\,dV}. $$

The result requires full rotational symmetry of both the solid and the density. A solid that is rotationally symmetric but has a density concentrated on one side need not have $\bar{x}=\bar{y}=0$.

Why the Horizontal Coordinates Vanish

The first moment about the $yz$-plane is $\iiint_E x\rho\,dV$. In cylindrical coordinates, its angular dependence is $\cos\theta$. For each fixed $r$ and $z$, the contribution from one side of the axis is canceled by the contribution from the opposite side:

\int_0^{2\pi}\cos\theta\,d\theta = [\sin\theta]_0^{2\pi} =0.

Likewise, the first moment about the $xz$-plane has angular dependence $\sin\theta$, and its full-angle integral is zero. In both cases, the density and the $rz$-bounds are constant as $\theta$ varies, so the angular integral can be evaluated separately. Dividing either zero first moment by the positive mass gives a zero centroid coordinate.

\int_0^{2\pi}\sin\theta\,d\theta = [-\cos\theta]_0^{2\pi} =0.

For the vertical coordinate, $z$ does not vary with $\theta$. The mass and vertical first moment therefore take the forms

M=\int\!\!\int_D\int_0^{2\pi}\rho(r,z)\,r\,d\theta\,dz\,dr, \qquad M_z=\int\!\!\int_D\int_0^{2\pi}z\rho(r,z)\,r\,d\theta\,dz\,dr, \qquad \bar{z}=\frac{M_z}{M}.

Here $M_z$ denotes the first moment used to find the vertical centroid coordinate. The bounds describing $D$ must match the particular solid; the symmetry result does not replace the work of specifying those bounds.

Worked Examples

Worked Example: A Uniform Solid Cylinder

Find the centroid of a cylinder of radius $2$ and height $3$, with $0\leq z\leq3$ and constant density $c>0$.

The region has $0\leq r\leq2$ and $0\leq\theta\leq2\pi$. Symmetry gives $\bar{x}=\bar{y}=0$. Its mass is

\begin{aligned} M &=\int_0^3\int_0^2\int_0^{2\pi}c\,r\,d\theta\,dr\,dz\\ &=c(3)\left(\int_0^2r\,dr\right)(2\pi)\\ &=3c\left[\frac{r^2}{2}\right]_0^2(2\pi)\\ &=3c(2)(2\pi)=12\pi c. \end{aligned}

The vertical first moment is

\begin{aligned} M_z &=\int_0^3\int_0^2\int_0^{2\pi}zc\,r\,d\theta\,dr\,dz\\ &=c\left(\int_0^3z\,dz\right)\left(\int_0^2r\,dr\right)(2\pi)\\ &=c\left[\frac{z^2}{2}\right]_0^3(2)(2\pi)\\ &=c\left(\frac{9}{2}\right)(2)(2\pi)=18\pi c. \end{aligned}

Thus $\bar{z}=M_z/M=(18\pi c)/(12\pi c)=3/2$, and the centroid is $(0,0,3/2)$. Check: $M\bar{z}=(12\pi c)(3/2)=18\pi c=M_z$, as required.

Worked Example: A Cylinder With Radially Varying Density

Find the centroid of the cylinder $0\leq r\leq1$, $0\leq z\leq2$, with density $\rho(r,z)=1+r^2$.

The density depends on radius but not angle, so the full rotational symmetry gives $\bar{x}=\bar{y}=0$. Compute the mass:

\begin{aligned} M &=\int_0^2\int_0^1\int_0^{2\pi}(1+r^2)r\,d\theta\,dr\,dz\\ &=2(2\pi)\int_0^1(r+r^3)\,dr\\ &=4\pi\left[\frac{r^2}{2}+\frac{r^4}{4}\right]_0^1\\ &=4\pi\left(\frac12+\frac14\right)=3\pi. \end{aligned}

For the vertical first moment, the $z$-integral is $\int_0^2z\,dz=2$:

\begin{aligned} M_z &=\int_0^2\int_0^1\int_0^{2\pi}z(1+r^2)r\,d\theta\,dr\,dz\\ &=(2\pi)\left(\int_0^2z\,dz\right)\left(\int_0^1(r+r^3)\,dr\right)\\ &=(2\pi)(2)\left(\frac12+\frac14\right)=3\pi. \end{aligned}

Therefore $\bar{z}=3\pi/(3\pi)=1$, so the centroid is $(0,0,1)$. Check: $M\bar{z}=(3\pi)(1)=3\pi=M_z$. The radial variation changes the mass, but because it is the same at every height, it does not shift the vertical centroid away from the cylinder's midpoint.

Worked Example: A Paraboloid With Height-Dependent Density

Find the centroid of the solid $0\leq r\leq1$, $0\leq z\leq1-r^2$, with density $\rho(r,z)=1+z$.

This region and density are rotationally symmetric, so $\bar{x}=\bar{y}=0$. For each fixed $r$, set $h=1-r^2$, the height of the solid above that radius. Integrating the density vertically gives

\int_0^h(1+z)\,dz=h+\frac{h^2}{2}.

Using $h=1-r^2$, the mass is

\begin{aligned} M &=2\pi\int_0^1r\left((1-r^2)+\frac{(1-r^2)^2}{2}\right)dr\\ &=\pi\int_0^1\left(t+\frac{t^2}{2}\right)dt \quad\text{where }t=1-r^2,\;r\,dr=-\frac12dt\\ &=\pi\left[\frac{t^2}{2}+\frac{t^3}{6}\right]_0^1\\ &=\pi\left(\frac12+\frac16\right)=\frac{2\pi}{3}. \end{aligned}

For the vertical first moment, integrate $z(1+z)$ from $0$ to $h$:

\int_0^h z(1+z)\,dz=\frac{h^2}{2}+\frac{h^3}{3}.

Consequently,

\begin{aligned} M_z &=2\pi\int_0^1r\left(\frac{(1-r^2)^2}{2}+\frac{(1-r^2)^3}{3}\right)dr\\ &=\pi\int_0^1\left(\frac{t^2}{2}+\frac{t^3}{3}\right)dt\\ &=\pi\left[\frac{t^3}{6}+\frac{t^4}{12}\right]_0^1\\ &=\pi\left(\frac16+\frac1{12}\right)=\frac{\pi}{4}. \end{aligned}

It follows that $\bar{z}=(\pi/4)/(2\pi/3)=3/8$, and the centroid is $(0,0,3/8)$. Check: $M\bar{z}=(2\pi/3)(3/8)=\pi/4=M_z$. The height-dependent density gives extra weight to higher parts of the solid, and the computed centroid reflects that weighting.

A Common Pitfall: Using the Unweighted Midpoint

A centroid is a mass-weighted average, not necessarily the midpoint of a solid's vertical interval. For example, take a cylinder of radius $1$ and height $1$ with density $\rho=1+z$. Its mass and vertical moment are

\begin{aligned} M&=\pi\int_0^1(1+z)\,dz =\pi\left(1+\frac12\right)=\frac{3\pi}{2},\\ M_z&=\pi\int_0^1z(1+z)\,dz =\pi\left(\frac12+\frac13\right)=\frac{5\pi}{6}. \end{aligned}

A common incorrect calculation treats the density as uniform and assigns the centroid the interval's midpoint, $\bar{z}=1/2$. That would imply $M_z=M\bar{z}=(3\pi/2)(1/2)=3\pi/4$, which is not the actual moment $5\pi/6$. The correction is to include the density in both mass and first-moment integrals: $\bar{z}=M_z/M=(5\pi/6)/(3\pi/2)=5/9$.

How to Set Up a Centroid Calculation

Describe the solid and density.
Identify the radial and vertical bounds, and check whether both the solid and density are independent of angle.

Use the symmetry conclusion when its conditions hold.
For full rotational symmetry, set $\bar{x}=\bar{y}=0$; do not assume this for a partial sector or an angle-dependent density.

Compute mass and vertical moment.
Include the density and the cylindrical volume factor $r$ in both integrals.

Divide and check.
Calculate $\bar{z}=M_z/M$, then verify $M\bar{z}=M_z$ and state all three centroid coordinates.

Why the Symmetry Test Matters

Symmetry can eliminate two first-moment calculations entirely, but only when the density respects the same rotation as the solid. When both are rotationally symmetric, every horizontal direction has a corresponding opposite direction with equal mass, so horizontal contributions cancel. The remaining vertical coordinate is a weighted average: if density is positive, it lies between the lowest and highest $z$-values occupied by the solid.

These centroid calculations use mass and first moments. They are distinct from calculations involving how mass is distributed relative to an axis, which require different integrals.

Check Your Understanding

Answer these questions using the definitions and symmetry conditions above.

What assumptions about the solid and density allow you to conclude that $\bar{x}=\bar{y}=0$?
Why does the cylindrical volume element contribute a factor of $r$ to the mass integral?
For a rotationally symmetric solid, which two quantities are needed to calculate $\bar{z}$?
How would an angle-dependent density affect the symmetry argument for the horizontal centroid coordinates?
Why can a density that increases with height move the centroid above the midpoint of the solid's vertical interval?