Mass of Solids With Radial or Height-Dependent Density

Mass Depends on Both Density and Volume

A solid’s volume alone does not determine its mass. If the material is denser in some regions than others, those regions contribute more mass even when their volumes are equal. To calculate mass, we add the contributions from small pieces throughout the solid.

For a cylindrical solid, cylindrical coordinates are often useful when its boundaries or density depend on distance from the vertical axis or on height. The key is to express the density and the bounds in the same coordinates, then integrate density against the cylindrical volume element.

Definition: If a solid \(E\) has a nonnegative, integrable volume density \(\rho\), its mass is the triple integral of density over the solid. In cylindrical coordinates, with bounds \(a\le r\le b\), \(\alpha\le\theta\le\beta\), and \(g(r,\theta)\le z\le h(r,\theta)\), this is

M=\int_{\alpha}^{\beta}\int_a^b\int_{g(r,\theta)}^{h(r,\theta)} \rho(r,\theta,z)\,r\,dz\,dr\,d\theta.

Here \(r\) is the distance from the \(z\)-axis, \(\theta\) is the angular coordinate, and \(z\) is height. The factor \(r\) is part of the volume element, not part of the density.

The density may depend on \(r\), on \(z\), or on both. If it is originally given in Cartesian coordinates as a function of \(x,y,z\), use \(x=r\cos\theta\) and \(y=r\sin\theta\) to rewrite it. In particular, the distance from the \(z\)-axis satisfies \(\sqrt{x^2+y^2}=r\). The bounds must describe the same solid after conversion.

Why the Mass Integral Has This Form

For a small cylindrical-coordinate cell, the volume element is approximately \(r\,\Delta r\,\Delta\theta\,\Delta z\). The \(r\) appears because a small change \(\Delta\theta\) at radius \(r\) spans an arc of length approximately \(r\Delta\theta\). Multiplying that arc length by \(\Delta r\) and \(\Delta z\) gives the cell’s approximate volume.

The cell’s mass is approximately its density times its volume. Adding these contributions over a partition of the solid gives sums of the form \(\sum \rho\,r\,\Delta r\,\Delta\theta\,\Delta z\). In the limit as the cells become small, these sums give the triple integral of \(\rho r\). Thus the integral accounts for both the material’s density and the geometry of cylindrical coordinates.

Describe the solid.
Choose cylindrical bounds that cover the region exactly once, apart from boundary overlaps.

Rewrite the density.
Express every occurrence of \(x\) and \(y\) in terms of \(r\) and \(\theta\), where needed.

Include the volume factor.
Multiply the density by \(r\) in the integrand.

Integrate and check.
Evaluate in the stated order and verify the result by a separate calculation or by checking the integral’s components.

Density That Depends on Radius

A radial density has the form \(\rho(r)\): at a fixed distance from the axis, it has the same value at every angle and height. A solid with greater density farther from the axis, for example, can have a density that increases with \(r\). When setting up the mass integral, multiply this function by the cylindrical volume factor \(r\).

Worked Example: A Cylinder with Radially Increasing Density

Consider the full cylinder \(0\le r\le2\), \(0\le\theta\le2\pi\), \(0\le z\le3\), with density \(\rho(r)=2+r\). Find its mass.

The density is already in cylindrical form. The mass integral includes the additional factor \(r\):

\begin{aligned} M &=\int_0^{2\pi}\int_0^2\int_0^3(2+r)r\,dz\,dr\,d\theta\\ &=\int_0^{2\pi}\int_0^2 3(2r+r^2)\,dr\,d\theta\\ &=2\pi\cdot 3\left[r^2+\frac{r^3}{3}\right]_0^2\\ &=6\pi\left(4+\frac{8}{3}\right)\\ &=6\pi\cdot\frac{20}{3}=40\pi. \end{aligned}

For a check, split the density into its constant part \(2\) and radial part \(r\). Their mass contributions are \(2\cdot 12\pi=24\pi\) and \(12\pi\int_0^2r^2\,dr=16\pi\), respectively. Their sum is \(24\pi+16\pi=40\pi\), agreeing with the integral.

Density That Depends on Height

A height-dependent density has the form \(\rho(z)\). Its value is the same at all points with the same height, though the solid’s vertical bounds may vary with \(r\) and \(\theta\). In that situation, the bounds still need to describe the actual solid: a density depending only on \(z\) does not imply the solid has constant height.

Worked Example: A Paraboloid Cap with Density Proportional to Height

Let \(E\) be the solid \(0\le r\le1\), \(0\le\theta\le2\pi\), between \(z=r^2\) and \(z=1\). Suppose \(\rho(z)=z\). Find the mass.

The lower and upper vertical bounds are \(r^2\) and \(1\). Include \(r\) from the volume element, then integrate first with respect to \(z\):

\begin{aligned} M &=\int_0^{2\pi}\int_0^1\int_{r^2}^1 z\,r\,dz\,dr\,d\theta\\ &=2\pi\int_0^1 r\left[\frac{z^2}{2}\right]_{r^2}^1\,dr\\ &=2\pi\int_0^1\frac{r}{2}(1-r^4)\,dr\\ &=\pi\int_0^1(r-r^5)\,dr\\ &=\pi\left[\frac{r^2}{2}-\frac{r^6}{6}\right]_0^1\\ &=\pi\left(\frac12-\frac16\right)=\frac{\pi}{3}. \end{aligned}

To check the vertical integration, the antiderivative of \(z\) is \(z^2/2\), whose values at the bounds give \(\frac12-\frac{r^4}{2}\). The remaining radial integral evaluates to \(\frac12-\frac16=\frac13\), so the mass is \(\pi/3\).

Density That Depends on Radius and Height

A density such as \(\rho(r,z)=r+z\) varies with both distance from the axis and height. The same procedure applies: substitute the density into the mass integral, multiply by \(r\), and use bounds for the solid. If the solid occupies only part of the full angular range, the angular limits must describe that sector.

Worked Example: An Annular Sector with Combined Density

Consider the annular sector \(1\le r\le2\), \(0\le\theta\le\frac{\pi}{2}\), \(0\le z\le3\), with density \(\rho(r,z)=r+z\). Find its mass.

The density times the cylindrical volume factor is \((r+z)r=r^2+rz\). Integrate first over height:

\begin{aligned} M &=\int_0^{\pi/2}\int_1^2\int_0^3(r+z)r\,dz\,dr\,d\theta\\ &=\int_0^{\pi/2}\int_1^2 \left(3r^2+\frac{9}{2}r\right)\,dr\,d\theta\\ &=\frac{\pi}{2}\left( \left[r^3\right]_1^2+ \frac{9}{4}\left[r^2\right]_1^2 \right)\\ &=\frac{\pi}{2}\left((8-1)+\frac94(4-1)\right)\\ &=\frac{\pi}{2}\left(7+\frac{27}{4}\right) =\frac{55\pi}{8}. \end{aligned}

Check by separating the two density terms. The \(r\) part contributes \(\frac{\pi}{2}\int_1^2 3r^2\,dr=\frac{7\pi}{2}\). The \(z\) part contributes \(\frac{\pi}{2}\int_1^2\frac92r\,dr=\frac{27\pi}{8}\). Adding gives \(\frac{28\pi}{8}+\frac{27\pi}{8}=\frac{55\pi}{8}\), as before.

A Common Error: Treating \(r\) as Only a Coordinate

The coordinate \(r\) also appears in the cylindrical volume element. Omitting that factor changes the integral and generally gives the wrong mass. For the first example, the incorrect setup without the volume factor would give

\int_0^{2\pi}\int_0^2\int_0^3(2+r)\,dz\,dr\,d\theta =2\pi\cdot3\left[2r+\frac{r^2}{2}\right]_0^2 =36\pi.

This is not the mass integral: the cylindrical cell volume includes \(r\), so the integrand must be \((2+r)r\), yielding \(40\pi\). The density and volume element have distinct roles. Write them separately when setting up the integral—\(\rho\) for density and \(r\,dz\,dr\,d\theta\) for volume—before multiplying.

Using the Formula Reliably

Before evaluating, check that each bound describes the intended solid and that the density is expressed in the chosen coordinates. A density given as a function of \(z\) needs no conversion, while a Cartesian radial expression such as \(\sqrt{x^2+y^2}\) becomes \(r\). Do not change the density merely because the volume element contains \(r\): multiply the two factors only when forming the integrand.

Mass is nonnegative when the density is nonnegative. This gives a basic check on the setup and result. For a constant density \(\rho_0\), the formula reduces to \(\rho_0\) times the volume integral of the solid, since \(\rho_0\) is constant and can be factored out. For a varying density, the integral weights each part of the solid by its local density instead.

Key takeaway: For a cylindrical-coordinate description of a solid, integrate \(\rho(r,\theta,z)r\) over the solid’s cylindrical bounds. Convert the density and bounds consistently, and keep the factor \(r\) in the integrand.

Check Your Understanding

Use the mass integral and cylindrical volume element to answer these questions.

What factor must multiply the density in a cylindrical-coordinate mass integral, and why does it appear?
For a radial density \(\rho(r)=3r\), what is the integrand after including the cylindrical volume factor?
For a height-dependent density \(\rho(z)=z^2\), does the density need to be rewritten when the solid is already described in cylindrical coordinates?
In the annular-sector example, which bounds describe the angular extent of the solid?
What incorrect mass results in the first worked example if the cylindrical volume factor is omitted?