Tutorials › Calculus III › Flux and Circulation in a Rotating Vortex Field

Home  /  Tutorials  /  Flux and Circulation Applications

Flux and Circulation Applications · Tutorial 996 of 999

Flux and Circulation in a Rotating Vortex Field

Learn to calculate circulation on both boundary circles of an annulus and verify how their orientations relate to the field’s curl.

Advanced 10 min read

What You'll Learn

  • Express a planar rotating vortex as the sum of solid-body rotation and a point vortex.
  • Compute circulation around a circle of any radius in the field.
  • Assign the correct induced orientation to the inner and outer annular boundaries.
  • Relate boundary circulation to the curl integral over an annulus.
  • Explain why excluding the vortex singularity is essential.

Why an Annulus Reveals the Vortex

A vortex field can circulate around a point even when its curl is zero everywhere the field is defined. The apparent puzzle comes from the point at the center: the field is singular there, so a disk containing that point is not a valid surface on which to apply Stokes’ theorem.

An annulus avoids the singularity. Its boundary has two circles, and the circulation on one must be counted in the direction opposite to the other. For a rotating vortex field, this distinction lets us see both the persistent circulation due to the vortex and the contribution from ordinary rotation.

The Rotating Vortex Field

Write \(r=\sqrt{x^2+y^2}\), and let \(\omega\) and \(\kappa\) be constants. On the region \(r>0\), consider the planar vector field

$$ \mathbf{F}(x,y) = \omega(-y,x) + \kappa\left(-\frac{y}{r^2},\frac{x}{r^2}\right). $$

The first term is solid-body rotation: its tangential speed grows in proportion to distance from the origin. The second term is a point-vortex field: its tangential speed decreases in proportion to \(1/r\). In polar coordinates, with \(\mathbf{e}_{\theta}=(-\sin\theta,\cos\theta)\), this becomes

$$ \mathbf{F}(r,\theta) = \left(\omega r+\frac{\kappa}{r}\right)\mathbf{e}_{\theta}. $$

Take an annulus with inner radius \(a\) and outer radius \(b\), where \(0<a<b\). The field is smooth throughout this annulus because it does not include the origin.

Definition: The circulation of a vector field \(\mathbf{F}\) along an oriented curve \(C\) is the line integral \(\int_C\mathbf{F}\cdot d\mathbf{r}\). For a planar field \(\mathbf{F}=(P,Q)\), its scalar curl is \(\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}\).

For a positively oriented planar region, the outer boundary is traversed counterclockwise and any inner boundary is traversed clockwise. Thus the annulus’s boundary consists of the outer circle traversed counterclockwise and the inner circle traversed clockwise.

Circulation Around a Circle

First consider a circle of radius \(r\), traversed counterclockwise. Parametrize it by \(\mathbf{r}(\theta)=(r\cos\theta,r\sin\theta)\), for \(0\leq\theta\leq2\pi\). Its tangent displacement is \(d\mathbf{r}=r\mathbf{e}_{\theta}\,d\theta\). The field is tangential, so its dot product with this displacement is \((\omega r+\kappa/r)r\,d\theta\). Therefore,

$$ \begin{aligned} \oint_{C_r}\mathbf{F}\cdot d\mathbf{r} &=\int_0^{2\pi}\left(\omega r+\frac{\kappa}{r}\right)r\,d\theta\\ &=\int_0^{2\pi}(\omega r^2+\kappa)\,d\theta\\ &=2\pi(\omega r^2+\kappa). \end{aligned} $$

The solid-body rotation contributes \(2\pi\omega r^2\), which depends on the circle’s radius. The vortex contributes \(2\pi\kappa\), which is the same for every circle around the origin.

Key formula: For a counterclockwise circle of radius \(r\), the circulation is \(2\pi(\omega r^2+\kappa)\). Reversing the direction of traversal changes the sign of the circulation.

Worked Example: A Pure Vortex on an Annulus

Worked Example: A Pure Vortex on an Annulus

Let \(\omega=0\), \(\kappa=3\), and take the annulus \(1\leq r\leq4\). Compute the circulation around each circle counterclockwise, then compute the circulation along the oriented boundary of the annulus.

For either circle, the counterclockwise formula gives \(2\pi(0\cdot r^2+3)=6\pi\). Thus the outer circle has counterclockwise circulation \(6\pi\), and the inner circle, if traversed counterclockwise on its own, also has circulation \(6\pi\).

For the annulus boundary, however, the inner circle is traversed clockwise. Its contribution is therefore \(-6\pi\), while the outer contribution remains \(6\pi\). The total boundary circulation is \(6\pi-6\pi=0\).

As a check, the field has zero curl for \(r>0\), so its curl integral over this annulus is also zero. This agrees with the oriented boundary calculation.

Curl on the Annulus

For a tangential field of the form \(\mathbf{F}=g(r)\mathbf{e}_{\theta}\), its scalar curl at \(r>0\) is \(\frac{1}{r}\frac{d}{dr}(rg(r))\). In the rotating vortex field, \(g(r)=\omega r+\kappa/r\). Consequently,

$$ \begin{aligned} \operatorname{curl}\mathbf{F} &=\frac{1}{r}\frac{d}{dr} \left(r\left(\omega r+\frac{\kappa}{r}\right)\right)\\ &=\frac{1}{r}\frac{d}{dr}(\omega r^2+\kappa)\\ &=2\omega. \end{aligned} $$

The vortex part has zero curl at every point with \(r>0\). That statement is local: it does not say the field is defined at the origin. The annulus excludes the origin, so the curl is defined everywhere on the region where we integrate.

Theorem: If a planar vector field is continuously differentiable on an annulus, its circulation along the positively oriented boundary equals the double integral of its scalar curl over the annulus. The outer boundary is counterclockwise and the inner boundary is clockwise.

Why the Boundary Difference Equals the Curl Integral

For this radial tangential field, the relationship can be verified directly, without leaving any sign implicit. Let \(C_b\) and \(C_a\) denote the outer and inner circles, each initially traversed counterclockwise. Their circulations are \(2\pi b\,g(b)\) and \(2\pi a\,g(a)\), respectively. Since the inner circle is reversed in the annulus boundary, the total is

$$ \oint_{\partial A}\mathbf{F}\cdot d\mathbf{r} = 2\pi b\,g(b)-2\pi a\,g(a). $$

Meanwhile, polar coordinates give the curl integral over the annulus \(A\):

$$ \begin{aligned} \iint_A \operatorname{curl}\mathbf{F}\,dA &=\int_0^{2\pi}\int_a^b \frac{1}{r}\frac{d}{dr}(rg(r))\,r\,dr\,d\theta\\ &=\int_0^{2\pi}\int_a^b\frac{d}{dr}(rg(r))\,dr\,d\theta\\ &=2\pi\bigl(bg(b)-ag(a)\bigr). \end{aligned} $$

The final equality follows by evaluating the antiderivative \(rg(r)\) at \(b\) and \(a\). This is exactly the oriented boundary circulation. The subtraction at the inner radius is not optional: it is the contribution of the clockwise inner boundary.

1
Compute each circle’s circulation.
Use the counterclockwise formula \(2\pi r\,g(r)\) at the inner and outer radii.
2
Apply the annulus orientations.
Keep the outer circulation and subtract the inner counterclockwise circulation.
3
Integrate the curl over the region.
In polar coordinates, the radial factor in the area element cancels the \(1/r\) in the curl formula.
4
Compare the results.
Both calculations give \(2\pi(bg(b)-ag(a))\).

Worked Example: Solid-Body Rotation Plus a Vortex

Worked Example: Solid-Body Rotation Plus a Vortex

Let \(\omega=2\), \(\kappa=3\), and take the annulus \(1\leq r\leq2\). Find the circulation around each boundary circle with its induced orientation, then verify the curl integral.

On the outer circle, \(r=2\), so its counterclockwise circulation is \(2\pi(2\cdot2^2+3)=2\pi(8+3)=22\pi\).

On the inner circle, \(r=1\), the counterclockwise circulation would be \(2\pi(2\cdot1^2+3)=2\pi(2+3)=10\pi\). The annulus orientation reverses this circle, so its actual boundary contribution is \(-10\pi\). The total is \(22\pi-10\pi=12\pi\).

The curl is \(2\omega=4\). The annulus has area \(\pi(2^2-1^2)=3\pi\), so its curl integral is \(4(3\pi)=12\pi\). It matches the boundary circulation.

Worked Example: A Vortex with Radius-Dependent Swirl

Worked Example: A Vortex with Radius-Dependent Swirl

Consider \(\mathbf{F}=r^2\mathbf{e}_{\theta}\) on \(1\leq r\leq2\). Compute the boundary circulation and the curl integral.

Here \(g(r)=r^2\). A counterclockwise circle of radius \(r\) has circulation \(2\pi r\,g(r)=2\pi r^3\). The outer circle contributes \(2\pi(2^3)=16\pi\). The inner circle, counterclockwise on its own, contributes \(2\pi(1^3)=2\pi\); as the annulus’s inner boundary it contributes \(-2\pi\). The total is \(16\pi-2\pi=14\pi\).

The curl is \(\frac{1}{r}\frac{d}{dr}(r^3)=\frac{1}{r}(3r^2)=3r\). Thus

$$ \begin{aligned} \iint_A 3r\,dA &=\int_0^{2\pi}\int_1^2 3r\cdot r\,dr\,d\theta\\ &=2\pi\left[r^3\right]_1^2\\ &=2\pi(8-1)=14\pi. \end{aligned} $$

The curl integral and the oriented boundary circulation both equal \(14\pi\).

The Central Singularity and a Common Mistake

For a pure vortex, the field \(\kappa\mathbf{e}_{\theta}/r\) is undefined at \(r=0\). Its curl is zero on any annulus that excludes the origin, yet a counterclockwise loop around the origin has circulation \(2\pi\kappa\). There is no contradiction: Stokes’ theorem applies on the annulus, whose two boundary circulations cancel, but it cannot be applied to the disk containing the singular point as though the field were smooth there.

A frequent error is to compute both boundary circles counterclockwise and add their circulations. For the example with \(\omega=2\), \(\kappa=3\), and radii \(1\) and \(2\), that incorrect calculation gives \(22\pi+10\pi=32\pi\). It treats the inner circle as counterclockwise, even though the annulus’s induced orientation there is clockwise. Reversing its sign gives \(22\pi-10\pi=12\pi\), which agrees with the curl integral.

Takeaway: On an annulus, the circulation along the boundary is the outer counterclockwise circulation minus the inner counterclockwise circulation. For a point vortex, the equal vortex contributions cancel; the curl integral records the net contribution from the field’s variation across the annulus.

This calculation makes the role of the hole precise. Each circle can carry nonzero circulation from the vortex, while the oriented total around the annulus reflects only the change between the two radii. The curl integral measures that same change across the region.

Check Your Understanding

Use the orientation and formulas developed above to answer these questions.

  1. For a pure vortex with \(\kappa=5\), what is the counterclockwise circulation on a circle of radius \(3\)? What is its contribution when it is the inner boundary of an annulus?
  2. For \(\omega=1\), \(\kappa=2\), and radii \(a=1\), \(b=3\), compute the total circulation along the oriented annular boundary.
  3. For a tangential field \(g(r)\mathbf{e}_{\theta}\), explain why the inner-circle term is subtracted in the annular boundary circulation.
  4. For \(g(r)=r^2\) on \(2\leq r\leq3\), compute the outer and inner boundary contributions and their total.
  5. Why does zero curl on the punctured plane not justify applying Stokes’ theorem to a disk that contains the vortex center?

Previous

Next

Back to Tutorials