Workshop: Mass of a Variable-Density Paraboloid Bowl

Mass and the Center of Mass of a Bowl

A bowl-shaped solid can have a nonuniform density: material near the bottom may differ from material near the rim, and density may also vary with distance from the vertical axis. To find its mass, each small piece of volume must be weighted by its local density. To find its centroid, we also need the moments of that mass distribution.

This workshop considers a solid bounded by a circular paraboloid and a horizontal plane. Its geometry is simple enough to describe with cylindrical coordinates, while its density can depend on both radius and height. The calculation illustrates two distinct roles of cylindrical coordinates: describing the region and incorporating the volume factor in the integral.

For clarity, all numerical coordinates below are measured in one fixed length unit \(L\). Thus \(z=r^2\) means \(z/L=(r/L)^2\), and a density formula gives its numerical value in \(\mathrm{kg}/L^3\). Masses are in kilograms, and centroid coordinates are in \(L\).

Definition: Let \(S\) be a solid with nonnegative density \(\rho\). Its mass is \(M=\iiint_S \rho\,dV\). Its first moment about the \(xy\)-plane is \(M_{xy}=\iiint_S z\rho\,dV\). If \(M>0\), the vertical coordinate of its center of mass is \(\bar z=M_{xy}/M\).

Describe the Paraboloid Bowl

Take the bowl to be the part inside the upward-opening paraboloid \(z=r^2\), above \(z=0\), and below the horizontal plane \(z=h\), where \(h>0\). At a fixed height \(z\), the paraboloid condition gives \(0\le r\le\sqrt z\). At a fixed radius, the bounds are \(r^2\le z\le h\), and the largest radius is \(\sqrt h\).

In cylindrical coordinates, \(0\le\theta\le2\pi\). The volume element is \(dV=r\,dr\,d\theta\,dz\); the factor \(r\) is the cylindrical Jacobian. If \(\rho\) depends only on \(r\) and \(z\), the integrand does not depend on \(\theta\), so integration over the full revolution contributes a factor \(2\pi\).

M=2\pi\int_0^h\int_0^{\sqrt z}\rho(r,z)\,r\,dr\,dz, \qquad M_{xy}=2\pi\int_0^h\int_0^{\sqrt z}z\rho(r,z)\,r\,dr\,dz.

The vertical moment has an extra factor \(z\) because every mass element contributes its mass multiplied by its height above the \(xy\)-plane. Once both integrals are evaluated, their ratio gives \(\bar z\). The bounds can also be written in the other order, as \(0\le r\le\sqrt h\) and \(r^2\le z\le h\); either description covers the same bowl.

Why the Horizontal Centroid Coordinates Vanish

At every height, the bowl contains a complete disk centered on the \(z\)-axis. Since the density depends on \(r\) and \(z\), rotating the solid about that axis does not change either the region or the density. For example, the \(x\)-coordinate of the center of mass is \(\bar x=M^{-1}\iiint_S x\rho\,dV\). In cylindrical coordinates, \(x=r\cos\theta\), so the angular part of this moment is \(\int_0^{2\pi}\cos\theta\,d\theta=0\). Likewise, \(y=r\sin\theta\) and \(\int_0^{2\pi}\sin\theta\,d\theta=0\). Therefore \(\bar x=\bar y=0\).

Consequence of rotational symmetry: For a full paraboloid bowl with density depending only on \(r\) and \(z\), the center of mass lies on the \(z\)-axis. The only centroid coordinate that needs to be calculated is \(\bar z\).

A Useful Formula for a Polynomial Density

For the density \(\rho(r,z)=a+br^2+cz\), with constants chosen so that \(\rho\ge0\) throughout the bowl, the radial integration can be done explicitly. This gives a check on direct calculations and a convenient route to several examples.

\int_0^{\sqrt z}(a+br^2+cz)r\,dr =\frac{(a+cz)z}{2}+\frac{bz^2}{4} =\frac{az}{2}+\frac{(b+2c)z^2}{4}.

Integrating this expression over \(z\) gives \(M\). Multiplying it first by \(z\) gives the first moment. Thus:

M=\frac{\pi a h^2}{2}+\frac{\pi(b+2c)h^3}{6}, \qquad M_{xy}=\frac{\pi a h^3}{3}+\frac{\pi(b+2c)h^4}{8}.

These formulas follow from \(\int_0^h z\,dz=h^2/2\), \(\int_0^h z^2\,dz=h^3/3\), \(\int_0^h z^3\,dz=h^4/4\), and the factor \(2\pi\). They apply to this bowl and this density form; other shapes or density functions require setting up and evaluating the corresponding integrals.

Worked Examples

Worked Example: Density Increasing with Radius and Height

Let \(h=4\) and \(\rho(r,z)=2+r^2+z\). First integrate over \(r\), using \(0\le r\le\sqrt z\):

\begin{aligned} \int_0^{\sqrt z}(2+r^2+z)r\,dr &=(2+z)\frac{(\sqrt z)^2}{2}+\frac{(\sqrt z)^4}{4}\\ &=z+\frac{z^2}{2}+\frac{z^2}{4}\\ &=z+\frac{3z^2}{4}. \end{aligned}

Therefore the mass is

\begin{aligned} M &=2\pi\int_0^4\left(z+\frac{3z^2}{4}\right)dz\\ &=2\pi\left[\frac{z^2}{2}+\frac{z^3}{4}\right]_0^4\\ &=2\pi(8+16)=48\pi\ \mathrm{kg}. \end{aligned}

For the first moment, multiply the radial result by \(z\):

\begin{aligned} M_{xy} &=2\pi\int_0^4\left(z^2+\frac{3z^3}{4}\right)dz\\ &=2\pi\left[\frac{z^3}{3}+\frac{3z^4}{16}\right]_0^4\\ &=2\pi\left(\frac{64}{3}+48\right) =\frac{416\pi}{3}\ \mathrm{kg}\,L. \end{aligned}

Hence \(\bar z=(416\pi/3)/(48\pi)=26/9\ L\). The check \(M\bar z=(48\pi)(26/9)=416\pi/3\ \mathrm{kg}\,L\) recovers the calculated first moment. The antiderivatives also check: differentiating \(z^2/2+z^3/4\) gives \(z+3z^2/4\), and differentiating \(z^3/3+3z^4/16\) gives \(z^2+3z^3/4\).

Worked Example: Density Depending on Height

Use the same bowl with \(h=4\), but let \(\rho(r,z)=1+2z\). This density is independent of \(r\), but the cylindrical volume factor \(r\) remains in the integral. The radial integral is

\int_0^{\sqrt z}(1+2z)r\,dr =(1+2z)\frac{z}{2} =\frac{z}{2}+z^2.

It follows that

\begin{aligned} M &=2\pi\int_0^4\left(\frac z2+z^2\right)dz\\ &=2\pi\left[\frac{z^2}{4}+\frac{z^3}{3}\right]_0^4\\ &=2\pi\left(4+\frac{64}{3}\right) =\frac{152\pi}{3}\ \mathrm{kg}. \end{aligned}

For the vertical moment, the radial integrand becomes \(z^2/2+z^3\):

\begin{aligned} M_{xy} &=2\pi\int_0^4\left(\frac{z^2}{2}+z^3\right)dz\\ &=2\pi\left[\frac{z^3}{6}+\frac{z^4}{4}\right]_0^4\\ &=2\pi\left(\frac{32}{3}+64\right) =\frac{448\pi}{3}\ \mathrm{kg}\,L. \end{aligned}

Thus \(\bar z=(448\pi/3)/(152\pi/3)=56/19\ L\). Check: \((152\pi/3)(56/19)=448\pi/3\ \mathrm{kg}\,L\). Differentiating the mass antiderivative gives \(z/2+z^2\), and differentiating the moment antiderivative gives \(z^2/2+z^3\), as required.

Worked Example: A Shallower Bowl with Radial and Height Dependence

Let \(h=1\) and \(\rho(r,z)=2+r^2+z\). Integrating first in \(r\) gives

\int_0^{\sqrt z}(2+r^2+z)r\,dr=z+\frac{3z^2}{4}.

Therefore

\begin{aligned} M &=2\pi\int_0^1\left(z+\frac{3z^2}{4}\right)dz\\ &=2\pi\left[\frac{z^2}{2}+\frac{z^3}{4}\right]_0^1\\ &=2\pi\left(\frac12+\frac14\right) =\frac{3\pi}{2}\ \mathrm{kg}. \end{aligned}

The first moment is

\begin{aligned} M_{xy} &=2\pi\int_0^1\left(z^2+\frac{3z^3}{4}\right)dz\\ &=2\pi\left[\frac{z^3}{3}+\frac{3z^4}{16}\right]_0^1\\ &=2\pi\left(\frac13+\frac{3}{16}\right) =\frac{25\pi}{24}\ \mathrm{kg}\,L. \end{aligned}

Consequently \(\bar z=(25\pi/24)/(3\pi/2)=25/36\ L\). Multiplying back gives \((3\pi/2)(25/36)=25\pi/24\ \mathrm{kg}\,L\), confirming the centroid ratio. The derivatives of the displayed antiderivatives are respectively \(z+3z^2/4\) and \(z^2+3z^3/4\), so both integrations check.

A Common Error: Omitting the Cylindrical Factor

The factor \(r\) is part of \(dV\), not an optional part of the density. For example, with constant density \(\rho=1\) and \(h=1\), the correct mass is

2\pi\int_0^1\int_0^{\sqrt z}r\,dr\,dz =2\pi\int_0^1\frac z2\,dz =\frac{\pi}{2}.

A common incorrect calculation omits \(r\) and instead writes \(2\pi\int_0^1\int_0^{\sqrt z}1\,dr\,dz\). It gives \(2\pi\int_0^1\sqrt z\,dz=4\pi/3\), which is not the volume of this bowl. The correction is to use \(dV=r\,dr\,d\theta\,dz\), yielding the correct mass \(\pi/2\) for unit density. As a further check, the bowl’s volume is \(\pi h^2/2\), obtained by integrating the cross-sectional disk area \(\pi z\) from \(0\) to \(h\); at \(h=1\), this is indeed \(\pi/2\).

Procedure and Interpretation

Describe the region.
For this bowl, use \(0\le z\le h\), \(0\le r\le\sqrt z\), and \(0\le\theta\le2\pi\).

Write the density and volume element.
Substitute the density \(\rho(r,z)\) and include \(r\) from \(dV=r\,dr\,d\theta\,dz\).

Calculate mass and first moment.
Integrate \(\rho\) for mass and \(z\rho\) for the moment about the \(xy\)-plane.

Use symmetry and take the ratio.
Rotational symmetry gives \(\bar x=\bar y=0\); divide the first moment by the positive mass to obtain \(\bar z\).

For nonnegative density, the computed centroid height must lie between the bottom and top of the bowl: \(0\le\bar z\le h\). Indeed, \(0\le z\le h\) throughout the region, so \(0\le z\rho\le h\rho\); integrating gives \(0\le M_{xy}\le hM\), and dividing by \(M>0\) gives the stated bound. This is a useful plausibility check, though it does not replace checking the integral arithmetic.

Check Your Understanding

Use \(0\le z\le h\), \(0\le r\le\sqrt z\), and the cylindrical volume factor when answering.

For a bowl with \(h=9\), what is the upper radial bound at height \(z\), and what is the largest radius?
Why does the mass integral contain a factor \(r\), even when the density does not depend on \(r\)?
For density \(\rho(r,z)=a+br^2+cz\), which integral defines the first moment about the \(xy\)-plane?
Why are the horizontal coordinates of the center of mass zero when the density depends only on \(r\) and \(z\)?
For a nonnegative density and positive mass, explain why the vertical centroid coordinate cannot exceed the top height \(h\).