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Flux and Circulation Applications · Tutorial 999 of 999

A Mixed Flow Problem with Sources and Vortices

Learn how source and vortex terms affect flux and circulation, and how to choose and verify direct or theorem-based calculations when a singularity may lie inside the region.

Advanced 11 min read

What You'll Learn

  • Write a planar flow as the sum of a source field and a vortex field.
  • Distinguish boundary integrals that can be computed directly from those eligible for an integral theorem.
  • Track outward normals and positive boundary orientations on annuli.
  • Apply the divergence theorem and Green’s theorem only when the field is smooth throughout the region.
  • Compare direct flux and circulation calculations with theorem-based results.

Introduction

A flow can contain more than one recognizable component. A source sends flow outward, while a vortex circulates around a point. Their effects on flux and circulation are different, so a useful calculation begins by keeping the components distinct.

The location of the source and vortex matters just as much as the shape of the boundary. A boundary curve may avoid a singular point even though the region it encloses contains that point. Direct integration along the curve can still be valid in that situation, but an integral theorem requiring a smooth field throughout the region cannot be used.

Key idea: Compute boundary integrals directly whenever the field is defined along the boundary. Use the divergence theorem or Green’s theorem only when the field is smooth on the whole region, including its boundary.

A Mixed Source-and-Vortex Field

Use the planar vector field

$$ \mathbf{F}(x,y) = \alpha\frac{(x,y)}{x^2+y^2} + \beta\frac{(-y,x)}{x^2+y^2}, \qquad (x,y)\ne(0,0), $$

where \(\alpha\) and \(\beta\) are constants. Write \(r=\sqrt{x^2+y^2}\). The first term points radially outward and has source strength \(\alpha\). The second points in the direction of rotation around the origin and has vortex strength \(\beta\). Both terms are undefined at the origin.

For an oriented curve \(C\), the circulation is the line integral \(\int_C \mathbf{F}\cdot d\mathbf{r}\). For a plane region with positively oriented boundary, the outward flux is \(\int_{\partial D}\mathbf{F}\cdot\mathbf{n}\,ds\), where \(\mathbf{n}\) is the outward unit normal. Positive orientation means counterclockwise along an outer boundary; on the inner boundaries of a region with holes, it means clockwise.

If \(C\) is traversed counterclockwise and is parametrized by \(x=x(t)\), \(y=y(t)\), then \(d\mathbf{r}=(dx,dy)\), and the outward normal element is \(\mathbf{n}\,ds=(dy,-dx)\). Thus the circulation and flux contributions can be written in differential form as

$$ \mathbf{F}\cdot d\mathbf{r} = \alpha\frac{x\,dx+y\,dy}{r^2} + \beta\frac{x\,dy-y\,dx}{r^2}, \qquad \mathbf{F}\cdot\mathbf{n}\,ds = \alpha\frac{x\,dy-y\,dx}{r^2} - \beta\frac{x\,dx+y\,dy}{r^2}. $$

Along a curve avoiding the origin, the expressions \((x\,dx+y\,dy)/r^2\) and \((x\,dy-y\,dx)/r^2\) are respectively \(d(\ln r)\) and the differential change in polar angle \(d\theta\). These identities separate the radial and rotational effects.

Why the Integral Theorems Give Zero Away from the Origin

For a planar field \(\mathbf{F}=(P,Q)\), the divergence is \(P_x+Q_y\) and the scalar curl is \(Q_x-P_y\). Direct differentiation of the mixed field gives

$$ \operatorname{div}\mathbf{F}=0, \qquad \operatorname{curl}\mathbf{F}=0 \qquad\text{for }r>0. $$

For example, the source part has components \(P=\alpha x/r^2\) and \(Q=\alpha y/r^2\). Its divergence is

$$ \alpha\left(\frac{1}{r^2}-\frac{2x^2}{r^4}\right) + \alpha\left(\frac{1}{r^2}-\frac{2y^2}{r^4}\right) = \alpha\left(\frac{2r^2-2x^2-2y^2}{r^4}\right) =0. $$

The vortex part has components \(-\beta y/r^2\) and \(\beta x/r^2\); differentiating gives zero divergence as well. The source part has zero curl, and differentiation of the vortex part gives zero curl. These calculations apply only where \(r>0\).

Theorem conditions: The planar divergence theorem identifies outward flux with the integral of divergence over a region when the field is continuously differentiable on the region and its boundary. Green’s theorem for circulation identifies the positively oriented boundary circulation with the integral of scalar curl over the region, under the corresponding smoothness conditions. A singularity anywhere in the region prevents these hypotheses from being met.

Therefore, if a bounded region and its entire boundary avoid the origin, both theorem-based integrals are zero. The phrase “the boundary avoids the origin” alone is not enough: the region enclosed by that boundary must avoid it too.

Worked Example: Flux and Circulation on an Annulus

Worked Example: Flux and Circulation on an Annulus

Take \(\alpha=2\), \(\beta=1\), and the annulus \(D=\{(x,y):1\le r\le3\}\). Compute the flux and circulation on each boundary component, then compare their totals with the integral theorems.

On a counterclockwise circle of radius \(R\), use \(x=R\cos t\), \(y=R\sin t\), for \(0\le t\le2\pi\). Then \(dx=-R\sin t\,dt\), \(dy=R\cos t\,dt\), and

$$ x\,dy-y\,dx=R^2\,dt, \qquad x\,dx+y\,dy=0. $$

The counterclockwise circulation is consequently \(\int_0^{2\pi}\beta\,dt=2\pi\beta\), and the outward flux from the disk is \(\int_0^{2\pi}\alpha\,dt=2\pi\alpha\). With the given strengths, these are \(2\pi\) and \(4\pi\), respectively.

The outer boundary of the annulus is counterclockwise, so its outward flux is \(4\pi\) and its circulation is \(2\pi\). The inner boundary must be clockwise for positive orientation of the annulus. Reversing the direction reverses each line integral, so the inner boundary contributes flux \(-4\pi\) and circulation \(-2\pi\).

$$ \text{Total flux}=4\pi+(-4\pi)=0, \qquad \text{total circulation}=2\pi+(-2\pi)=0. $$

The whole annulus, including both boundary circles, avoids the origin. The divergence theorem gives total flux \(\iint_D 0\,dA=0\), and Green’s theorem gives total circulation \(\iint_D 0\,dA=0\). Both results agree with the direct boundary calculations.

A Direct Calculation Is Not the Same as Theorem Eligibility

The preceding example uses an annulus because the region between its boundary circles excludes the singularity. A different situation arises when a boundary itself avoids the origin but encloses it. The boundary integrals remain meaningful, but the divergence theorem and Green’s theorem cannot be applied to the enclosed disk for this field.

Worked Example: A Circle Whose Disk Contains the Singularity

Again take \(\alpha=2\), \(\beta=1\). Directly calculate the flux and circulation on the counterclockwise circle \(C\) of radius \(2\) centered at the origin.

For this circle, \(r=2\) is constant, so \(d(\ln r)=0\). The polar angle increases from \(0\) to \(2\pi\). The boundary formulas therefore give

$$ \int_C\mathbf{F}\cdot\mathbf{n}\,ds = \int_C\bigl(\alpha\,d\theta-\beta\,d(\ln r)\bigr) = 2(2\pi)-1(0) = 4\pi, $$

and

$$ \int_C\mathbf{F}\cdot d\mathbf{r} = \int_C\bigl(\alpha\,d(\ln r)+\beta\,d\theta\bigr) = 2(0)+1(2\pi) = 2\pi. $$

These are valid direct boundary integrals because every point on \(C\) has \(r=2\), not zero. However, the disk bounded by \(C\) contains the origin, where \(\mathbf{F}\) is undefined. Neither theorem can be used on that disk to derive these values. In particular, it would be incorrect to integrate the zero divergence or curl over the disk and conclude that either boundary integral is zero.

Worked Example: A Disk That Does Not Enclose the Origin

Worked Example: A Disk That Does Not Enclose the Origin

Use \(\alpha=2\), \(\beta=1\), and let \(D\) be the disk of radius \(1\) centered at \((3,0)\). Its boundary is parametrized counterclockwise by \(x=3+\cos t\), \(y=\sin t\), for \(0\le t\le2\pi\). Since \(x\ge2\) throughout this disk, it and its boundary avoid the origin.

To check the direct boundary integrals, first note that

$$ r^2=(3+\cos t)^2+\sin^2t=10+6\cos t. $$

Along the parametrization, \(x\,dx+y\,dy=-3\sin t\,dt\), whose quotient by \(r^2\) integrates to zero over \(0\le t\le2\pi\): the denominator is unchanged under \(t\mapsto2\pi-t\), while \(\sin t\) changes sign. Also,

$$ x\,dy-y\,dx=(1+3\cos t)\,dt. $$

The angular contribution integrates to zero. Indeed, since \(x>0\), the angle has the single-valued expression \(\theta(t)=\arctan(y/x)\), and its derivative is \(\theta'(t)=(1+3\cos t)/(10+6\cos t)\). The angle starts and ends at zero, so \(\int_0^{2\pi}(1+3\cos t)/(10+6\cos t)\,dt=0\). Thus both \(\oint_C d(\ln r)\) and \(\oint_C d\theta\) are zero. The boundary formulas give flux \(2(0)-1(0)=0\) and circulation \(2(0)+1(0)=0\).

The theorem-based checks are also valid here: the field is smooth on the disk and boundary, and divergence and curl are both zero throughout. Hence the divergence theorem and Green’s theorem each give zero, matching the direct calculations.

A Reliable Method for Mixed Flow Problems

When source and vortex components appear together, use a short sequence of checks before calculating. The order matters: a correct integral evaluated using an invalid theorem is not a valid solution.

1
Locate singularities.
Find where the field is undefined. Here the only singular point is the origin.
2
Check the whole region.
Do not check only the boundary. Determine whether the enclosed region or annular region contains a singularity.
3
Choose the quantity and orientation.
Use outward normals for flux and positive boundary orientation for circulation. On an annulus, the inner boundary is clockwise.
4
Calculate and verify.
Integrate directly when appropriate, or apply an integral theorem only when its smoothness conditions hold. Compare results when both approaches are available.

Common Mistake: Checking Only the Boundary

A tempting but incorrect argument says: “The circle \(r=2\) does not pass through the origin, so Green’s theorem applies to the disk it encloses.” The calculation would then be written as

$$ \int_C\mathbf{F}\cdot d\mathbf{r} \overset{\text{incorrect}}{=} \iint_{\text{disk}} \operatorname{curl}\mathbf{F}\,dA = \iint_{\text{disk}}0\,dA =0. $$

The error is the theorem application: the field is undefined at the origin inside the disk, so it is not continuously differentiable throughout the region. The direct calculation in the previous example gives \(2\pi\), not zero. Correct the argument by either integrating the boundary directly or choosing a region, such as an annulus excluding the origin, on which the field is smooth throughout.

Why Separating the Components Helps

Flux and circulation respond to the two parts of this field in complementary ways. On a centered circle, the source produces outward flux \(2\pi\alpha\), while the vortex produces circulation \(2\pi\beta\). The other two contributions vanish on that circle because its radius is constant. On an annulus, the inner boundary has the opposite orientation from the outer boundary, so those contributions cancel in the total.

This separation is useful beyond these particular formulas. It helps diagnose whether a nonzero boundary integral is caused by a source, a rotational component, a hole in the region, or an invalid theorem application. A zero divergence or curl away from a singularity does not mean every boundary integral around that singularity is zero.

Takeaway: For a mixed source-and-vortex field, first locate the singularity, then check whether it lies in the region. Direct boundary integration may be valid even when an integral theorem is not. If the entire region avoids the singularity, the divergence and curl calculations can be used safely, with orientations handled consistently.

Check Your Understanding

Use the definitions and checks in this tutorial to answer the following questions.

  1. For a centered circle traversed counterclockwise, which component contributes to flux, and which contributes to circulation?
  2. Why may the flux on a circle centered at the origin be calculated directly even though the divergence theorem cannot be applied to its disk?
  3. What is the positive direction on the inner boundary of an annulus, and why does it matter for the total flux and circulation?
  4. A disk lies entirely to the right of the origin. What must be checked before applying Green’s theorem to its boundary?
  5. For the annulus in the first worked example, why do the outer and inner boundary contributions cancel even though each is nonzero?

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