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The Missing-r Error in Cylindrical Integrals

Understand the cylindrical volume element and diagnose how forgetting its factor \(r\) alters computed volumes and masses.

Intermediate 9 min read

What You'll Learn

  • Identify the Jacobian factor in the cylindrical volume element
  • Explain geometrically why cylindrical volume scales with distance from the axis
  • Correct triple integrals that omit the factor r
  • Compare correct and incorrect volume calculations on the same solid
  • Apply the factor correctly when integrating a density to find mass
  • Distinguish a missing factor r from a radial-coordinate substitution error

Why the Factor \(r\) Matters

Cylindrical coordinates are useful when a solid or integrand has circular structure. They also introduce a particular source of error: replacing \(x\) and \(y\) by cylindrical expressions but leaving the volume element unchanged. The resulting integral may look plausible, yet it measures the wrong quantity.

The essential rule is that cylindrical volume is not \(dr\,d\theta\,dz\). A small change in angle sweeps out a wider arc when it is farther from the axis, so the volume associated with a fixed angular change grows with \(r\). This geometric scaling is represented by the Jacobian factor \(r\).

Key rule: In cylindrical coordinates, \(dV=r\,dr\,d\theta\,dz\), up to the order in which the differentials are written. Thus a Cartesian triple integral of \(f(x,y,z)\) becomes an integral of \(f(r\cos\theta,r\sin\theta,z)\,r\) over the corresponding cylindrical bounds.

The factor belongs to the volume element, not just to integrals whose integrand visibly contains a radial term. For volume, the Cartesian integrand is \(1\), but the cylindrical integrand is still \(r\). For mass, the transformed density is multiplied by \(r\) as well.

Where \(r\) Comes From

The cylindrical coordinate map is \(x=r\cos\theta\), \(y=r\sin\theta\), and \(z=z\), with \(r\geq 0\). Its Jacobian determinant measures how the coordinate map scales a small volume. In the order \(r,\theta,z\), the derivative matrix is:

$$ \frac{\partial(x,y,z)}{\partial(r,\theta,z)} = \begin{pmatrix} \cos\theta & -r\sin\theta & 0\\ \sin\theta & r\cos\theta & 0\\ 0 & 0 & 1 \end{pmatrix}. $$

Expanding the determinant along the last row leaves the determinant of the upper-left \(2\) by \(2\) block:

$$ \det\!\left(\frac{\partial(x,y,z)}{\partial(r,\theta,z)}\right) = \cos\theta(r\cos\theta)-(-r\sin\theta)(\sin\theta) = r(\cos^2\theta+\sin^2\theta) = r. $$

Because \(r\geq 0\), the absolute value of this determinant is \(r\). The change-of-variables formula therefore gives the volume element \(dV=r\,dr\,d\theta\,dz\). At \(r=0\), the determinant is zero because all angles describe the same axis point; this coordinate singularity does not change the integral over a three-dimensional region.

There is also a direct geometric interpretation. At radius \(r\), a small angle change \(d\theta\) corresponds to an arc length \(r\,d\theta\). A thin wedge-shaped piece of volume consequently has dimensions \(dr\), \(r\,d\theta\), and \(dz\). Its volume is approximately \(r\,dr\,d\theta\,dz\). Without \(r\), the calculation treats angular slices near the axis and far from the axis as though they had equal width.

General formula: If a solid \(E\) is described by cylindrical bounds, then $$ \iiint_E f(x,y,z)\,dV = \iiint_{\text{cylindrical bounds}} f(r\cos\theta,r\sin\theta,z)\,r\,dr\,d\theta\,dz. $$ The differentials may be ordered to match the chosen integration order. The factor \(r\) is included exactly once.

Worked Example: The Volume of a Cylinder

Worked Example: Comparing the Correct and Missing-\(r\) Integrals

Consider a cylinder of radius \(3\) and height \(2\), centered on the \(z\)-axis, with its bottom at \(z=0\). Its cylindrical bounds are \(0\leq r\leq 3\), \(0\leq\theta\leq2\pi\), and \(0\leq z\leq2\). For volume, the Cartesian integrand is \(1\).

The correct integral includes the Jacobian:

$$ V = \int_0^2\int_0^{2\pi}\int_0^3 r\,dr\,d\theta\,dz = \int_0^2\int_0^{2\pi} \left[\frac{r^2}{2}\right]_0^3d\theta\,dz = \int_0^2\int_0^{2\pi}\frac{9}{2}\,d\theta\,dz. $$

Integrating over angle and height gives:

$$ V = \int_0^2 9\pi\,dz = 9\pi[z]_0^2 = 18\pi. $$

If the \(r\) is omitted, the incorrect calculation is:

$$ V_{\text{wrong}} = \int_0^2\int_0^{2\pi}\int_0^3 1\,dr\,d\theta\,dz = \int_0^2\int_0^{2\pi}3\,d\theta\,dz = \int_0^2 6\pi\,dz = 12\pi. $$

As a check, the geometric cylinder formula gives \(\pi(3)^2(2)=18\pi\), agreeing with the correct integral. The omitted-\(r\) result is only \(\frac{2}{3}\) of the true volume. The error is not a rounding issue: it comes from using the wrong volume element.

Worked Example: Volume Under a Paraboloid

Worked Example: A Paraboloid Above a Disk

Let \(E\) lie above the disk \(0\leq r\leq2\) in the plane \(z=0\) and below the paraboloid \(z=4-r^2\). The height is nonnegative throughout the disk, since \(4-r^2\geq0\) for \(0\leq r\leq2\). The angular bounds are \(0\leq\theta\leq2\pi\).

The volume integral with the necessary factor is:

$$ V = \int_0^{2\pi}\int_0^2\int_0^{4-r^2}r\,dz\,dr\,d\theta = \int_0^{2\pi}\int_0^2(4-r^2)r\,dr\,d\theta. $$

Integrate first in \(r\):

$$ \int_0^2(4-r^2)r\,dr = \int_0^2(4r-r^3)\,dr = \left[2r^2-\frac{r^4}{4}\right]_0^2 = (8-4)-0 = 4. $$

Then integrate over the full angle:

$$ V = \int_0^{2\pi}4\,d\theta = [4\theta]_0^{2\pi} = 8\pi. $$

Without the Jacobian, the height would be integrated against \(dr\,d\theta\), producing:

$$ V_{\text{wrong}} = \int_0^{2\pi}\int_0^2(4-r^2)\,dr\,d\theta = \int_0^{2\pi} \left[4r-\frac{r^3}{3}\right]_0^2d\theta = \int_0^{2\pi}\frac{16}{3}\,d\theta = \frac{32\pi}{3}. $$

These answers differ: \(\frac{32\pi}{3}\neq8\pi\). To check the correct calculation, the radial antiderivative \(2r^2-\frac{r^4}{4}\) differentiates to \(4r-r^3=(4-r^2)r\), the required radial integrand. The incorrect calculation differentiates an antiderivative for \(4-r^2\), not for the volume integrand \((4-r^2)r\).

The Same Factor Is Needed for Mass

For a density \(\rho(x,y,z)\), mass is the integral of density over volume. After converting the density to cylindrical coordinates, the mass element is \(\rho(r\cos\theta,r\sin\theta,z)\,r\,dr\,d\theta\,dz\). Density conversion and volume scaling are separate operations: replacing the density does not replace the Jacobian.

Worked Example: Mass with Radially Varying Density

Take a cylinder with \(0\leq r\leq2\), \(0\leq\theta\leq2\pi\), and \(0\leq z\leq3\). Let its density be \(\rho(r,\theta,z)=r\), in consistent units. The correct mass is:

$$ M = \int_0^3\int_0^{2\pi}\int_0^2 \rho(r,\theta,z)\,r\,dr\,d\theta\,dz = \int_0^3\int_0^{2\pi}\int_0^2 r^2\,dr\,d\theta\,dz. $$

Evaluate the radial integral:

$$ \int_0^2r^2\,dr = \left[\frac{r^3}{3}\right]_0^2 = \frac{8}{3}. $$

Then evaluate the angle and height integrals:

$$ M = \int_0^3\int_0^{2\pi}\frac{8}{3}\,d\theta\,dz = \int_0^3\frac{16\pi}{3}\,dz = \frac{16\pi}{3}[z]_0^3 = 16\pi. $$

If the Jacobian is omitted, the density \(r\) is integrated without an additional factor:

$$ M_{\text{wrong}} = \int_0^3\int_0^{2\pi}\int_0^2r\,dr\,d\theta\,dz = \int_0^3\int_0^{2\pi}\left[\frac{r^2}{2}\right]_0^2d\theta\,dz = \int_0^3\int_0^{2\pi}2\,d\theta\,dz = 12\pi. $$

Check the correct radial antiderivative: \(\frac{d}{dr}\left(\frac{r^3}{3}\right)=r^2\), which is the product of density \(r\) and Jacobian \(r\). The wrong calculation instead integrates the density alone. Its result does not represent the mass of this three-dimensional cylinder.

A Reliable Diagnostic

When setting up a cylindrical integral, check each part independently. First, write the original Cartesian integrand or density. Second, express that function in terms of \(r,\theta,z\). Third, attach the Jacobian factor \(r\) to account for the volume element. Finally, apply the bounds. Keeping these steps distinct makes it easier to catch an omission before evaluating the integral.

1
Transform the function.
For example, replace \(x^2+y^2\) by \(r^2\), or replace a density by its expression in cylindrical variables.
2
Write the volume element.
Include \(r\) once, so the integrand for the triple integral contains the transformed function multiplied by \(r\).
3
Check the result.
For a volume, the starting function is \(1\), so the cylindrical integrand should be \(r\). For mass, it should be the transformed density times \(r\).

The factor is not an extra density term, nor is it a radial bound. It is the scale change in volume caused by the coordinate transformation. It is included once regardless of integration order. If the original Cartesian integrand itself contains factors that become \(r\)-dependent, those remain in the transformed function, and the Jacobian contributes its own \(r\) as well.

Common Mistake: Changing Coordinates but Not the Volume Element

A frequent incorrect setup for a solid of revolution is to convert its height or density to a function of \(r\), then integrate that expression with \(dr\,d\theta\,dz\). For instance, the volume under \(z=4-r^2\) over \(r\leq2\) is sometimes written as:

$$ \int_0^{2\pi}\int_0^2(4-r^2)\,dr\,d\theta. $$

This expression uses the correct height but the wrong volume element. Its value, \(\frac{32\pi}{3}\), is not the volume. The correction is to multiply by the Jacobian before integrating:

$$ \int_0^{2\pi}\int_0^2(4-r^2)\,r\,dr\,d\theta = 8\pi. $$

There is a related opposite error: including \(r\) twice. If \(r\) has already been included as the Jacobian, do not insert another factor merely because the integral is cylindrical. The reliable test is to begin with the transformed original function, then multiply it by exactly one Jacobian factor.

Why the Error Changes the Answer

Omitting \(r\) changes how strongly different radial locations contribute. In the disk \(0\leq r\leq R\), a thin ring near radius \(r\) has circumference approximately \(2\pi r\), so a ring of thickness \(dr\) has area approximately \(2\pi r\,dr\). The area associated with rings grows with radius. A calculation using just \(dr\) gives each radial increment the same weight and therefore does not add up the correct areas or volumes.

This is why checking only the bounds is insufficient. Bounds determine which points are included; the Jacobian determines how the coordinate increments represent physical volume. A correct cylindrical setup requires both. The three examples show the practical consequence: the missing factor changes volume and mass values, even when the region and its bounds have been described correctly.

Check Your Understanding

Answer these questions by identifying both the transformed function and the volume element.

  1. What is the cylindrical volume element, and what geometric feature accounts for its factor \(r\)?
  2. For volume under \(z=9-r^2\), what factor must multiply the height in the cylindrical integral?
  3. A density becomes \(r^2+z\) in cylindrical coordinates. What expression should be integrated with respect to \(dr\,d\theta\,dz\) to calculate mass?
  4. Why is the Jacobian still needed when the original Cartesian integrand is simply \(1\)?
  5. How would you identify an integral in which the Jacobian has accidentally been included twice?

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