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Curl and Divergence · Tutorial 915 of 999

Curl in Cylindrical Coordinates

Learn to compute cylindrical curl correctly by relating its components to circulation on small oriented patches.

Advanced 9 min read

What You'll Learn

  • Identify the radial, angular, and axial components of curl in cylindrical coordinates
  • Explain why the angular scale factor contributes factors of the radius
  • Derive the curl components using oriented infinitesimal patches
  • Compute curl for axisymmetric fields and fields with angular dependence
  • Check cylindrical answers by converting the field and curl to Cartesian components
  • Recognize common errors involving signs, scale factors, and basis directions

Why Curl Needs Care in Cylindrical Coordinates

In Cartesian coordinates, the three coordinate directions are fixed throughout space. Cylindrical coordinates use a different basis: the radial and angular unit vectors turn as the angle changes. The physical distance represented by an angular change also depends on the radius. Both facts matter when computing curl.

A reliable calculation therefore requires more than replacing \(x\) and \(y\) by cylindrical variables in a Cartesian formula. We must keep track of the field’s cylindrical components, the directions of the unit vectors, the scale factor \(r\), and the orientation of each small loop used to measure circulation.

Key idea: The curl component in a given direction is the limiting circulation per unit area of a small oriented surface perpendicular to that direction. In cylindrical coordinates, angular sides have physical length \(r\,d\theta\), so the radius appears in the component formulas.

Coordinates, Basis, and Field Components

For \(r>0\), cylindrical coordinates \((r,\theta,z)\) are related to Cartesian coordinates by \(x=r\cos\theta\), \(y=r\sin\theta\), and \(z=z\). The associated unit vectors are

$$ \mathbf{e}_r=(\cos\theta,\sin\theta,0),\qquad \mathbf{e}_\theta=(-\sin\theta,\cos\theta,0),\qquad \mathbf{e}_z=(0,0,1). $$

The first two basis vectors depend on \(\theta\); \(\mathbf{e}_z\) does not. A vector field is written

$$ \mathbf{F} = F_r(r,\theta,z)\mathbf{e}_r + F_\theta(r,\theta,z)\mathbf{e}_\theta + F_z(r,\theta,z)\mathbf{e}_z. $$

Here \(F_r,F_\theta,F_z\) are scalar component functions. They may depend on all three coordinates, including \(\theta\). The physical length of a small angular displacement at radius \(r\) is \(r\,d\theta\); the radial and vertical lengths are \(dr\) and \(dz\). We compute the following formula where the component functions are differentiable and \(r>0\).

Formula: If \(\mathbf{F}=F_r\mathbf{e}_r+F_\theta\mathbf{e}_\theta+F_z\mathbf{e}_z\), then
$$ \nabla\times\mathbf{F} = \frac{1}{r}\left( \frac{\partial F_z}{\partial\theta} - \frac{\partial(rF_\theta)}{\partial z} \right)\mathbf{e}_r + \left( \frac{\partial F_r}{\partial z} - \frac{\partial F_z}{\partial r} \right)\mathbf{e}_\theta + \frac{1}{r}\left( \frac{\partial(rF_\theta)}{\partial r} - \frac{\partial F_r}{\partial\theta} \right)\mathbf{e}_z. $$

Why the Formula Has These Terms

Each component follows from circulation around a small coordinate patch, oriented so its normal points in the direction of the component being measured. Divide the circulation by the patch’s area and take the limit as the patch shrinks. The following derivation also makes the signs and scale factors explicit.

1
Patch normal to \(\mathbf{e}_z\).
Take a rectangle with radial width \(dr\) and angular width \(d\theta\), at fixed \(z\), and orient its boundary counterclockwise when viewed from \(+\mathbf{e}_z\). Its area is \(r\,dr\,d\theta\). The radial sides contribute \(F_r(r,\theta,z)\,dr-F_r(r,\theta+d\theta,z)\,dr\). The angular sides contribute \((r+dr)F_\theta(r+dr,\theta,z)\,d\theta-rF_\theta(r,\theta,z)\,d\theta\). To first order, the total circulation is \(\left(\frac{\partial(rF_\theta)}{\partial r}-\frac{\partial F_r}{\partial\theta}\right)dr\,d\theta\). Dividing by the area gives the \(\mathbf{e}_z\) component \(\frac{1}{r}\left(\frac{\partial(rF_\theta)}{\partial r}-\frac{\partial F_r}{\partial\theta}\right)\).
2
Patch normal to \(\mathbf{e}_r\).
Use a patch at fixed \(r\), with dimensions \(r\,d\theta\) and \(dz\), oriented so \(\mathbf{e}_\theta\) followed by \(\mathbf{e}_z\) gives normal \(\mathbf{e}_r\). The circulation to first order is \(\left(\frac{\partial F_z}{\partial\theta}-\frac{\partial(rF_\theta)}{\partial z}\right)d\theta\,dz\). Its area is \(r\,d\theta\,dz\). Dividing gives the \(\mathbf{e}_r\) component.
3
Patch normal to \(\mathbf{e}_\theta\).
Use a rectangle in the \(r,z\) directions, oriented so \(\mathbf{e}_z\) followed by \(\mathbf{e}_r\) gives normal \(\mathbf{e}_\theta\). Its area is \(dr\,dz\). The circulation to first order is \(\left(\frac{\partial F_r}{\partial z}-\frac{\partial F_z}{\partial r}\right)dr\,dz\). Dividing by this area gives the \(\mathbf{e}_\theta\) component.

The first step is especially useful for checking orientation: with normal \(+\mathbf{e}_z\), the radial side at angle \(\theta\) is traversed outward, while the radial side at \(\theta+d\theta\) is traversed inward. That is why their net contribution is \(F_r(r,\theta,z)\,dr-F_r(r,\theta+d\theta,z)\,dr\), not the reverse.

Worked Example: An Axisymmetric Field

Worked Example: An Axisymmetric Field

Let \(\mathbf{F}=rz\,\mathbf{e}_\theta\), so \(F_r=0\), \(F_\theta=rz\), and \(F_z=0\). The components have no angular dependence. Compute the curl.

For the radial component,

$$ (\nabla\times\mathbf{F})_r = \frac{1}{r}\left( \frac{\partial 0}{\partial\theta} - \frac{\partial(r(rz))}{\partial z} \right) = \frac{1}{r}\left(0-\frac{\partial(r^2z)}{\partial z}\right) = \frac{-r^2}{r} = -r. $$

For the angular component,

$$ (\nabla\times\mathbf{F})_\theta = \frac{\partial 0}{\partial z} - \frac{\partial 0}{\partial r} = 0. $$

For the axial component,

$$ (\nabla\times\mathbf{F})_z = \frac{1}{r}\left( \frac{\partial(r(rz))}{\partial r} - \frac{\partial 0}{\partial\theta} \right) = \frac{1}{r}\frac{\partial(r^2z)}{\partial r} = \frac{2rz}{r} = 2z. $$

Thus \(\nabla\times\mathbf{F}=-r\mathbf{e}_r+2z\mathbf{e}_z\). Check by converting the field to Cartesian components: \(rz\mathbf{e}_\theta=(-yz,xz,0)\). The Cartesian curl is \[ \left( \frac{\partial 0}{\partial y}-\frac{\partial(xz)}{\partial z}, \frac{\partial(-yz)}{\partial z}-\frac{\partial 0}{\partial x}, \frac{\partial(xz)}{\partial x}-\frac{\partial(-yz)}{\partial y} \right) = (-x,-y,2z). \] Since \(-x\mathbf{i}-y\mathbf{j}=-r\mathbf{e}_r\), this agrees with the cylindrical result.

Worked Example: A Field with Angular Dependence

Worked Example: A Field with Angular Dependence

Let \(\mathbf{F}=r\cos\theta\,\mathbf{e}_z\). Then \(F_r=F_\theta=0\) and \(F_z=r\cos\theta\), which depends explicitly on \(\theta\). The components are

$$ (\nabla\times\mathbf{F})_r = \frac{1}{r}\frac{\partial(r\cos\theta)}{\partial\theta} = -\sin\theta, \qquad (\nabla\times\mathbf{F})_\theta = -\frac{\partial(r\cos\theta)}{\partial r} = -\cos\theta, \qquad (\nabla\times\mathbf{F})_z=0. $$

Therefore \(\nabla\times\mathbf{F}=-\sin\theta\,\mathbf{e}_r-\cos\theta\,\mathbf{e}_\theta\). For a check, \(r\cos\theta=x\), so \(\mathbf{F}=(0,0,x)\) in Cartesian coordinates. Its curl is \((0,-1,0)\). Because \(\mathbf{j}=\sin\theta\,\mathbf{e}_r+\cos\theta\,\mathbf{e}_\theta\), the cylindrical answer is \(-\mathbf{j}\), as required.

Worked Example: A Radial Component That Varies with Height

Worked Example: A Radial Component That Varies with Height

Let \(\mathbf{F}=z\mathbf{e}_r\), so \(F_r=z\), \(F_\theta=0\), and \(F_z=0\). Then

$$ (\nabla\times\mathbf{F})_r=0,\qquad (\nabla\times\mathbf{F})_\theta = \frac{\partial z}{\partial z} - \frac{\partial 0}{\partial r} = 1,\qquad (\nabla\times\mathbf{F})_z=0. $$

Hence \(\nabla\times\mathbf{F}=\mathbf{e}_\theta\). To check, \(\mathbf{F}=(zx/r,zy/r,0)\) for \(r>0\). Its Cartesian curl is \[ \left( -\frac{\partial(zy/r)}{\partial z}, \frac{\partial(zx/r)}{\partial z}, \frac{\partial(zy/r)}{\partial x}-\frac{\partial(zx/r)}{\partial y} \right) = \left(-\frac{y}{r},\frac{x}{r},0\right) = \mathbf{e}_\theta. \] The cylindrical and Cartesian calculations agree.

Axisymmetry and What It Does Not Remove

A field is axisymmetric when its cylindrical component functions do not depend on \(\theta\). In that case, the angular derivatives in the formula vanish:

$$ \nabla\times\mathbf{F} = -\frac{1}{r}\frac{\partial(rF_\theta)}{\partial z}\mathbf{e}_r + \left( \frac{\partial F_r}{\partial z} - \frac{\partial F_z}{\partial r} \right)\mathbf{e}_\theta + \frac{1}{r}\frac{\partial(rF_\theta)}{\partial r}\mathbf{e}_z. $$

The factors involving \(rF_\theta\) remain. Axisymmetry removes dependence on the angle from the scalar components; it does not make the angular direction a fixed Cartesian direction or eliminate the angular scale factor. Example 1 illustrates this: even with no angular dependence, the curl has both radial and axial components.

A Common Scale-Factor Error

A frequent mistake is to use \(\frac{\partial F_\theta}{\partial r}\) in the axial component instead of differentiating \(rF_\theta\). For \(\mathbf{F}=r^2\mathbf{e}_\theta\), that incorrect calculation gives

$$ (\nabla\times\mathbf{F})_z \ \text{(incorrect)} = \frac{\partial(r^2)}{\partial r} = 2r. $$

The correct calculation includes the angular scale factor:

$$ (\nabla\times\mathbf{F})_z = \frac{1}{r}\frac{\partial(r\cdot r^2)}{\partial r} = \frac{1}{r}\frac{\partial(r^3)}{\partial r} = 3r. $$

A Cartesian check confirms the correction. Since \(r^2\mathbf{e}_\theta=(-ry,rx,0)\), the axial curl component is \(\frac{\partial(rx)}{\partial x}-\frac{\partial(-ry)}{\partial y}=3r\). The missing contribution in the incorrect calculation comes from the fact that the angular direction and its physical scale change with position.

Using the Formula Reliably

Before differentiating, write all three component functions and mark which ones depend on \(r\), \(\theta\), and \(z\). Then keep the output directions attached to the results: the first expression multiplies \(\mathbf{e}_r\), the second \(\mathbf{e}_\theta\), and the third \(\mathbf{e}_z\). In particular, the angular derivative of \(F_r\) appears with a minus sign in the axial component, while the angular derivative of \(F_z\) appears with a plus sign in the radial component.

The formula is stated for \(r>0\), where cylindrical coordinates and their unit vectors are defined. If a field is being considered on the axis, the cylindrical expression alone is not a coordinate description there; use another coordinate description or examine a limit when appropriate.

The main practical lesson is simple: use the cylindrical components as given, preserve the \(rF_\theta\) terms, and orient the small loops consistently. These checks make it much harder to lose a scale factor, reverse a sign, or attach a correct scalar component to the wrong direction.

Check Your Understanding

Try these questions using the formula and the circulation interpretation.

  1. For \(\mathbf{F}=F_r\mathbf{e}_r+F_\theta\mathbf{e}_\theta+F_z\mathbf{e}_z\), which curl component contains \(\frac{\partial(rF_\theta)}{\partial r}\)?
  2. Why does a small angular side at radius \(r\) have length \(r\,d\theta\), and where does this appear in the curl formula?
  3. If the components are axisymmetric, which terms disappear, and which factors involving \(r\) must remain?
  4. For \(\mathbf{F}=r^2\mathbf{e}_\theta\), compute the axial curl component and explain why differentiating \(F_\theta\) alone is incorrect.
  5. For the patch normal to \(+\mathbf{e}_z\), in which direction is the radial side at angle \(\theta+d\theta\) traversed?

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