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Work and Vector Line Integrals · Tutorial 761 of 999

Circulation Around Circles in Rotational Fields

Use a circular parametrization to calculate circulation from a field’s tangential component, and see precisely how reversing the direction changes the result.

Intermediate 10 min read

What You'll Learn

  • Parametrize a circle centered at the origin in counterclockwise and clockwise directions.
  • Identify radial and tangential components of a planar field along a circle.
  • Derive a circulation formula for rotationally symmetric fields.
  • Calculate circulation for common rotational fields, including one singular at the origin.
  • Track the sign of circulation when the traversal direction is reversed.
  • Avoid confusing a unit tangent with the velocity of the parametrized curve.

Why Circles Make Circulation Easier to See

A vector field assigns a vector to each point in a region of the plane. A line integral around a closed curve measures how much the field points along the direction of travel, accumulated over the whole trip. That quantity is called circulation.

For a circle centered at the origin, the direction of travel is particularly simple: the tangent always points around the circle. A field whose vectors also have a tangential component is therefore well suited to a direct calculation. The key is to distinguish the field’s tangential strength from the speed of the circular parametrization, and to keep track of whether the circle is traversed counterclockwise or clockwise.

Definition: If a planar curve \(C\) is parametrized by \(\mathbf{r}(t)=(x(t),y(t))\), for \(a\leq t\leq b\), and a vector field \(\mathbf{F}=(F_1,F_2)\) is defined along the curve, its work integral is $$ \int_C \mathbf{F}\cdot d\mathbf{r} = \int_a^b \mathbf{F}(\mathbf{r}(t))\cdot \mathbf{r}'(t)\,dt. $$ When \(C\) is closed, this integral is called the circulation of \(\mathbf{F}\) around \(C\). The symbol \(\oint_C\) emphasizes that the curve is closed.

Parametrizing a Centered Circle

Let the circle have radius \(R>0\). A counterclockwise parametrization, starting at \((R,0)\), is

$$ \mathbf{r}(t)=(R\cos t,R\sin t), \qquad 0\leq t\leq 2\pi. $$

Differentiating gives the velocity vector along the parametrization:

$$ \mathbf{r}'(t)=(-R\sin t,R\cos t). $$

The factor \(R\) matters: \(\mathbf{r}'(t)\) is not just a unit tangent vector. It gives both the direction of travel and the distance traveled per unit change in \(t\). For clockwise traversal, one can instead let \(t\) run from \(0\) to \(-2\pi\), or use a parametrization whose angular coordinate decreases.

Radial and Tangential Components

At a point with distance \(\rho>0\) from the origin, define the outward radial unit vector and the counterclockwise tangential unit vector by

$$ \mathbf{e}_{\rho}=(\cos\theta,\sin\theta), \qquad \mathbf{e}_{\theta}=(-\sin\theta,\cos\theta), $$

where \(x=\rho\cos\theta\) and \(y=\rho\sin\theta\). These vectors are perpendicular, with \(\mathbf{e}_{\rho}\cdot\mathbf{e}_{\theta}=0\). A rotationally symmetric field with radial component \(P(\rho)\) and tangential component \(Q(\rho)\) has the form

$$ \mathbf{F}(x,y)=P(\rho)\mathbf{e}_{\rho}+Q(\rho)\mathbf{e}_{\theta}. $$

Here rotational symmetry means the two component strengths depend on distance from the origin, not on the angle. On the circle of radius \(R\), the outward direction is radial and the counterclockwise direction of travel is tangential. Thus only \(Q(R)\) contributes to the circulation; the radial component is perpendicular to the direction of travel.

Theorem: If \(\mathbf{F}=P(\rho)\mathbf{e}_{\rho}+Q(\rho)\mathbf{e}_{\theta}\) is defined on the circle of radius \(R>0\), then its circulation around that circle is \(2\pi RQ(R)\) for counterclockwise traversal and \(-2\pi RQ(R)\) for clockwise traversal.

Why the Formula Works

For the counterclockwise parametrization, the position on the circle has \(\rho=R\) and angle \(\theta=t\). Therefore the field along the curve is \(P(R)\mathbf{e}_{\rho}+Q(R)\mathbf{e}_{\theta}\). The velocity is \(R\mathbf{e}_{\theta}\). Substituting these into the definition of the line integral gives

$$ \begin{aligned} \oint_C \mathbf{F}\cdot d\mathbf{r} &=\int_0^{2\pi} \bigl(P(R)\mathbf{e}_{\rho}+Q(R)\mathbf{e}_{\theta}\bigr) \cdot R\mathbf{e}_{\theta}\,dt\\ &=\int_0^{2\pi} RQ(R)\,dt\\ &=2\pi RQ(R). \end{aligned} $$

The second equality uses \(\mathbf{e}_{\rho}\cdot\mathbf{e}_{\theta}=0\) and \(\mathbf{e}_{\theta}\cdot\mathbf{e}_{\theta}=1\). Since \(R\) and \(Q(R)\) are constant around this circle, the final integral is the constant \(RQ(R)\) accumulated over an angular interval of length \(2\pi\).

For clockwise traversal, the tangent points in the direction \(-\mathbf{e}_{\theta}\). Equivalently, the angular parameter runs through a total change of \(-2\pi\). The tangential dot product therefore changes sign, yielding \(-2\pi RQ(R)\). This also agrees with the general orientation rule: reversing a curve’s direction reverses the sign of its line integral.

Worked Examples

Worked Example: The Field \(\mathbf{F}(x,y)=(-y,x)\)

Find the counterclockwise circulation around the circle of radius \(R=2\).

At a point with polar coordinates \((\rho,\theta)\), we have \((-y,x)=(-\rho\sin\theta,\rho\cos\theta)=\rho\mathbf{e}_{\theta}\). Thus the field is purely tangential, with \(Q(\rho)=\rho\). On the circle, \(Q(2)=2\), so the formula gives

$$ \oint_C \mathbf{F}\cdot d\mathbf{r} =2\pi(2)(2)=8\pi. $$

Check directly. Use \(\mathbf{r}(t)=(2\cos t,2\sin t)\), so \(\mathbf{r}'(t)=(-2\sin t,2\cos t)\). Along the curve, \(\mathbf{F}(\mathbf{r}(t))=(-2\sin t,2\cos t)\). Hence

$$ \begin{aligned} \mathbf{F}(\mathbf{r}(t))\cdot\mathbf{r}'(t) &=(-2\sin t)(-2\sin t)+(2\cos t)(2\cos t)\\ &=4\sin^2 t+4\cos^2 t\\ &=4. \end{aligned} $$

Integrating the constant gives \(\int_0^{2\pi}4\,dt=8\pi\), matching the result.

Worked Example: A Field with Both Radial and Tangential Components

Let \(\mathbf{F}(x,y)=(3x-2y,3y+2x)\). Find its circulation around the circle of radius \(2\), first counterclockwise and then clockwise.

In radial and tangential form, this field is \(3\rho\mathbf{e}_{\rho}+2\rho\mathbf{e}_{\theta}\). Therefore \(P(\rho)=3\rho\) and \(Q(\rho)=2\rho\). At \(R=2\), the tangential strength is \(Q(2)=4\). The radial strength is \(P(2)=6\), but it contributes zero because it is perpendicular to the direction of travel.

$$ \begin{aligned} \text{Counterclockwise circulation} &=2\pi RQ(R)=2\pi(2)(4)=16\pi,\\ \text{Clockwise circulation} &=-2\pi RQ(R)=-2\pi(2)(4)=-16\pi. \end{aligned} $$

Check the counterclockwise result by parametrization. On \(\mathbf{r}(t)=(2\cos t,2\sin t)\), the field is \((6\cos t-4\sin t,6\sin t+4\cos t)\), and the velocity is \((-2\sin t,2\cos t)\). Their dot product is

$$ \begin{aligned} &(6\cos t-4\sin t)(-2\sin t) +(6\sin t+4\cos t)(2\cos t)\\ &=-12\cos t\sin t+8\sin^2 t +12\sin t\cos t+8\cos^2 t\\ &=8. \end{aligned} $$

Thus the integral is \(\int_0^{2\pi}8\,dt=16\pi\). Reversing the direction changes this to \(-16\pi\), as required.

Worked Example: A Tangential Field Singular at the Origin

Consider \(\mathbf{F}(x,y)=\left(-\frac{y}{x^2+y^2},\frac{x}{x^2+y^2}\right)\). Find its circulation around a circle of radius \(R=3\), centered at the origin.

This field is not defined at the origin, but the circle of radius \(3\) does not pass through the origin, so the field is defined everywhere along the curve and the line integral is valid. Since \(x^2+y^2=\rho^2\), its radial component is zero and its tangential component is \(Q(\rho)=1/\rho\). Therefore \(Q(3)=1/3\), and

$$ \oint_C\mathbf{F}\cdot d\mathbf{r} =2\pi(3)\left(\frac{1}{3}\right)=2\pi. $$

Check by direct substitution. Along the parametrized circle, \(x=3\cos t\), \(y=3\sin t\), and \(x^2+y^2=9\). Thus \(\mathbf{F}(\mathbf{r}(t))=(-\sin t/3,\cos t/3)\), while \(\mathbf{r}'(t)=(-3\sin t,3\cos t)\). Their dot product is

$$ \left(-\frac{\sin t}{3}\right)(-3\sin t) +\left(\frac{\cos t}{3}\right)(3\cos t) =\sin^2 t+\cos^2 t=1. $$

Consequently, the integral is \(\int_0^{2\pi}1\,dt=2\pi\), which agrees with the formula.

A Useful Special Form

Some rotational fields are written using the vector \((-y,x)\) rather than the unit tangent. If

$$ \mathbf{F}(x,y)=g(\rho)(-y,x), \qquad \rho=\sqrt{x^2+y^2}, $$

then \((-y,x)=\rho\mathbf{e}_{\theta}\). The tangential component is therefore \(Q(\rho)=\rho g(\rho)\), not just \(g(\rho)\). For counterclockwise travel around radius \(R\), this gives

$$ \oint_C\mathbf{F}\cdot d\mathbf{r} =2\pi R^2g(R). $$

This form makes the two radius factors visible: one comes from the field vector \((-y,x)\), and one comes from the circular velocity \(\mathbf{r}'(t)\).

Common Pitfall: Dropping a Radius Factor

A frequent error is to use the unit tangent \(\mathbf{e}_{\theta}\) in place of the parametrization velocity \(R\mathbf{e}_{\theta}\). For \(\mathbf{F}(x,y)=(-y,x)\) on the circle of radius \(2\), the field has tangential strength \(2\). If one incorrectly computes the dot product using only the unit tangent, one gets \(2\) per unit angle and the incorrect circulation \(2(2\pi)=4\pi\).

The line integral uses \(\mathbf{r}'(t)\), not just a unit tangent. At radius \(2\), the velocity is \(2\mathbf{e}_{\theta}\), so the correct dot product is \(2\cdot2=4\) at every parameter value. Therefore the correct integral is \(4(2\pi)=8\pi\). The missing factor was the radius in the velocity vector.

Takeaway: For a rotationally symmetric field on a centered circle, find its tangential component \(Q(R)\), multiply by the circle’s radius, and accumulate over the full angle: \(2\pi RQ(R)\) counterclockwise. Reverse the sign for clockwise traversal. The field must be defined along the entire circle.

Why Direction and Tangential Strength Matter

Circulation measures the field’s contribution along the direction in which the curve is traversed. A radial force does not contribute on a centered circle because it points perpendicular to the motion. A tangential force contributes according to its signed strength: positive in the counterclockwise direction and negative in the clockwise direction. Thus two fields can have substantial magnitudes along the same circle but different circulations if their tangential components differ.

The simple formula depends on the symmetry condition that the radial and tangential strengths are constant around each circle centered at the origin. For a field whose tangential strength varies with angle, the pointwise tangential contribution must instead be accumulated around the circle; one cannot replace it with a single value \(Q(R)\). The parametrized definition remains the reliable starting point in that case.

Check Your Understanding

Use the definition of the line integral and the role of the tangential component to answer these questions.

  1. For \(\mathbf{F}=P(\rho)\mathbf{e}_{\rho}+Q(\rho)\mathbf{e}_{\theta}\), why does the radial component contribute zero to circulation around a centered circle?
  2. What is the counterclockwise circulation when the circle has radius \(4\) and \(Q(4)=3\)?
  3. For \(\mathbf{F}(x,y)=g(\rho)(-y,x)\), what is the tangential component in terms of \(g\) and \(\rho\)?
  4. What changes in the circulation when the same circle is traversed clockwise instead of counterclockwise?
  5. Why is a field singular at the origin still eligible for a line integral around a circle of positive radius, provided it is defined at every point of that circle?

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