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Cylindrical Triple Integrals · Tutorial 625 of 999

The Cylindrical Volume Element r dz dr dtheta

See why cylindrical volume is weighted by the distance \(r\) from the axis, then use the resulting volume element to set up and evaluate triple integrals.

Intermediate 9 min read

What You'll Learn

  • Write the cylindrical coordinate map from \((r,\theta,z)\) to Cartesian coordinates.
  • Derive the Jacobian determinant and identify the cylindrical volume factor.
  • Explain the same factor using the dimensions of a small wedge-shaped volume element.
  • Distinguish the volume element from the order of integration.
  • Set up and evaluate volumes using the cylindrical volume element.
  • Recognize why omitting the factor \(r\) gives an incorrect volume.

Why Cylindrical Volume Needs a Factor

A small change in Cartesian coordinates describes a little rectangular box. In cylindrical coordinates, a small change in angle does not have the same physical width everywhere: at radius \(r\), an angular change \(d\theta\) spans an arc of length approximately \(r\,d\theta\). Farther from the axis, the same angular change covers more distance, so it sweeps out more volume.

This is why a cylindrical triple integral includes a factor of \(r\). The factor is not an extra part of the integrand chosen by convention. It accounts for how cylindrical coordinates stretch space. We will derive it using the Jacobian determinant and interpret it using the dimensions of a small wedge-shaped cell.

Cylindrical Coordinates and the Volume Element

Cylindrical coordinates describe a point by its distance \(r\) from the \(z\)-axis, its angle \(\theta\) in the \(xy\)-plane, and its height \(z\). The corresponding Cartesian coordinates are given by

$$ x=r\cos\theta,\qquad y=r\sin\theta,\qquad z=z. $$

We take \(r\geq 0\). An angular interval should be chosen so that points in the region are represented once, apart from boundary overlaps. The axis \(r=0\) has no unique angle, but this coordinate degeneracy occurs on a set of zero volume.

Definition: In cylindrical coordinates, the volume element is $$ dV=r\,dz\,dr\,d\theta. $$ The differentials can be listed in another order to match an iterated integral, but the factor \(r\) remains. For bounds written in the order \(z\), then \(r\), then \(\theta\), the notation \(r\,dz\,dr\,d\theta\) makes that order explicit.

For a suitable function \(f\) over a solid \(D\), the change-of-variables formula takes the form

$$ \iiint_D f(x,y,z)\,dV = \int_{\alpha}^{\beta} \int_{a(\theta)}^{b(\theta)} \int_{g(r,\theta)}^{h(r,\theta)} f(r\cos\theta,r\sin\theta,z)\,r\,dz\,dr\,d\theta. $$

Here, the bounds describe the solid in cylindrical coordinates: \(\alpha\leq\theta\leq\beta\), \(a(\theta)\leq r\leq b(\theta)\), and \(g(r,\theta)\leq z\leq h(r,\theta)\). The formula applies when these bounds cover the solid without overlapping interiors and the integrand is integrable there. For volume, set \(f=1\). The factor \(r\) is the Jacobian factor; the bounds and the transformed integrand still have to match the particular solid and quantity being integrated.

Deriving the Factor from the Jacobian

The change-of-variables theorem says that a differentiable coordinate map scales small volumes by the absolute value of its Jacobian determinant. For the map from \((r,\theta,z)\) to \((x,y,z)\), its derivative matrix is

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

The entries come from differentiating each coordinate with respect to \(r\), \(\theta\), and \(z\). Expanding the determinant along the third row gives

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

Because \(r\geq 0\), the absolute value of this determinant is \(r\). Thus a small volume element in \((r,\theta,z)\)-coordinates becomes a Cartesian volume element multiplied by \(r\). This proves the volume factor in the formula above.

Why the Jacobian appears: A differentiable map is approximately linear on a sufficiently small region. A linear map scales volume by the absolute value of its determinant. Therefore the determinant of the coordinate map gives the local volume scaling; in cylindrical coordinates that scaling is \(r\).

The Wedge-Shaped Cell Interpretation

The same result can be understood geometrically. Consider a small cell near a point with radius \(r>0\). Its radial thickness is approximately \(dr\). Its height is \(dz\). Its width in the direction around the \(z\)-axis is the arc length \(r\,d\theta\), since arc length equals radius times angle when the angle is measured in radians. The cell is therefore approximately a rectangular box with volume

$$ (dr)(r\,d\theta)(dz)=r\,dz\,dr\,d\theta. $$

The cell is actually slightly wedge-shaped: its outer side is a little longer than its inner side. The rectangular approximation becomes accurate as the cell dimensions shrink, and the Jacobian calculation gives the exact limiting volume factor. This picture also explains why the factor cannot be constant. At a larger radius, the same \(d\theta\) gives a larger arc width and therefore a larger cell.

Worked Examples

Worked Example: A Full Circular Cylinder

Find the volume of the solid with \(0\leq r\leq 2\), \(0\leq\theta\leq 2\pi\), and \(0\leq z\leq 3\). The solid is specified directly in cylindrical coordinates, so we integrate \(1\) using the volume element \(r\,dz\,dr\,d\theta\):

$$ \begin{aligned} V &=\int_0^{2\pi}\int_0^2\int_0^3 r\,dz\,dr\,d\theta\\ &=\int_0^{2\pi}\int_0^2 \left[rz\right]_0^3\,dr\,d\theta\\ &=\int_0^{2\pi}\int_0^2 3r\,dr\,d\theta\\ &=\int_0^{2\pi}\left[\frac{3r^2}{2}\right]_0^2\,d\theta\\ &=\int_0^{2\pi}6\,d\theta\\ &=\left[6\theta\right]_0^{2\pi}\\ &=12\pi. \end{aligned} $$

As a check, the cylinder has base area \(\pi(2)^2=4\pi\) and height \(3\), so its volume is \((4\pi)(3)=12\pi\), agreeing with the integral.

Worked Example: An Annular Sector with Fixed Height

Find the volume of the solid with \(1\leq r\leq 3\), \(\frac{\pi}{6}\leq\theta\leq\frac{\pi}{2}\), and \(0\leq z\leq 4\). The angle range has width \(\frac{\pi}{2}-\frac{\pi}{6}=\frac{\pi}{3}\). Integrating in the order \(z,r,\theta\) gives

$$ \begin{aligned} V &=\int_{\pi/6}^{\pi/2}\int_1^3\int_0^4 r\,dz\,dr\,d\theta\\ &=\int_{\pi/6}^{\pi/2}\int_1^3 4r\,dr\,d\theta\\ &=\int_{\pi/6}^{\pi/2}\left[2r^2\right]_1^3\,d\theta\\ &=\int_{\pi/6}^{\pi/2}(18-2)\,d\theta\\ &=\int_{\pi/6}^{\pi/2}16\,d\theta\\ &=16\left(\frac{\pi}{2}-\frac{\pi}{6}\right)\\ &=\frac{16\pi}{3}. \end{aligned} $$

To check, the sector of the annulus has area \(\frac12(\frac{\pi}{3})(3^2-1^2)=\frac12(\frac{\pi}{3})(8)=\frac{4\pi}{3}\). Multiplying by height \(4\) gives \(\frac{16\pi}{3}\), as required.

Worked Example: A Solid with Height Increasing with Radius

Find the volume of the solid described by \(0\leq r\leq 1\), \(0\leq\theta\leq2\pi\), and \(0\leq z\leq2r\). At each fixed \(r\), the vertical height is \(2r\). Use that height as the inner bound and include the Jacobian factor:

$$ \begin{aligned} V &=\int_0^{2\pi}\int_0^1\int_0^{2r} r\,dz\,dr\,d\theta\\ &=\int_0^{2\pi}\int_0^1 r[z]_0^{2r}\,dr\,d\theta\\ &=\int_0^{2\pi}\int_0^1 2r^2\,dr\,d\theta\\ &=\int_0^{2\pi}\left[\frac{2r^3}{3}\right]_0^1\,d\theta\\ &=\int_0^{2\pi}\frac23\,d\theta\\ &=\frac23(2\pi)\\ &=\frac{4\pi}{3}. \end{aligned} $$

As a check on the radial contribution, a thin ring at radius \(r\) has circumference \(2\pi r\), thickness \(dr\), and height \(2r\). Its approximate volume is \((2\pi r)(dr)(2r)=4\pi r^2\,dr\). Integrating from \(0\) to \(1\) gives \(\int_0^1 4\pi r^2\,dr=\frac{4\pi}{3}\), consistent with the triple integral.

A Common Mistake: Omitting the Factor \(r\)

The most common error is to use \(dz\,dr\,d\theta\) as if a small change \(d\theta\) represented the same physical width for every radius. For the full cylinder in the first example, this incorrect calculation would give

$$ \int_0^{2\pi}\int_0^2\int_0^3 1\,dz\,dr\,d\theta =(2\pi)(2)(3)=12\pi. $$

In this particular arithmetic, the result happens to equal the correct volume by coincidence: the missing radial factor would have contributed \(\int_0^2 r\,dr=2\), while the incorrect expression contributes \(\int_0^2 1\,dr=2\). That accidental agreement does not make the setup valid.

For example, take a cylinder of radius \(1\) and height \(3\). The correct calculation is \(\int_0^{2\pi}\int_0^1\int_0^3 r\,dz\,dr\,d\theta=3\pi\). Omitting \(r\) instead gives \((2\pi)(1)(3)=6\pi\), which is twice the correct volume. The correction is always to include the Jacobian factor, even when a particular calculation might conceal its omission.

Takeaway: The bounds describe which points are included. The factor \(r\) describes how cylindrical coordinate intervals measure volume. A correct setup needs both.

Using the Element Reliably

When setting up a cylindrical triple integral, first identify the coordinate bounds for the solid and the order in which they are to be integrated. Then write the volume factor \(r\) alongside the differentials. For a volume, the integrand is \(1\); for another triple integral, transform the given integrand by substituting \(x=r\cos\theta\) and \(y=r\sin\theta\), and multiply by \(r\). The factor is not a replacement for transforming the function or determining valid bounds.

At \(r=0\), the Jacobian determinant vanishes because all angles describe the same point on the axis. This does not mean that the axis has positive or negative volume; it is a boundary of the coordinate description and has zero volume. The usual integral formula remains valid for ordinary solids that meet the axis, as in the third example. Likewise, a duplicated angular boundary such as \(\theta=0\) and \(\theta=2\pi\) does not duplicate a positive volume.

The essential geometric idea is simple: an angular interval has physical width proportional to its distance from the axis. The Jacobian records that proportionality exactly, and the wedge-shaped cell picture makes its meaning visible. Whenever cylindrical coordinates are used for a triple integral, the volume element must therefore carry the factor \(r\).

Check Your Understanding

Use the volume element and the reasoning in this tutorial to answer the following questions.

  1. What is the Jacobian determinant of the map \(x=r\cos\theta,\ y=r\sin\theta,\ z=z\), and why is its absolute value \(r\) for \(r\geq0\)?
  2. At a fixed radius \(r\), what is the approximate physical width corresponding to a small angular change \(d\theta\)?
  3. Write an iterated integral for the volume of the solid \(0\leq r\leq 2\), \(0\leq\theta\leq\pi\), \(1\leq z\leq 5\), using the order \(dz\,dr\,d\theta\).
  4. Why can omitting \(r\) accidentally produce the right answer for some bounds without being a valid method?
  5. What does the zero Jacobian at \(r=0\) indicate about the cylindrical coordinate description?

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