Closed-Curve Integrals and Circulation

Closed Paths and Circulation

A particle can return to its starting point without the force acting on it having done zero work. The endpoint displacement is zero, but along the path the force may point partly in the direction of motion at some places and partly against it at others. A vector line integral keeps track of this accumulated tangential effect.

When the path is closed, the resulting work is called circulation. The word emphasizes that the field is being accumulated around a loop, with a specified direction of travel. The direction matters: traversing the same loop in reverse changes the sign of the integral.

Definition: Let \(\mathbf{F}(x,y)=(P(x,y),Q(x,y))\) be a vector field, and let \(C\) be a piecewise smooth oriented curve. If \(C\) is parametrized by \(\mathbf{r}(t)=(x(t),y(t))\), \(a\le t\le b\), its vector line integral is \(\int_C \mathbf{F}\cdot d\mathbf{r}=\int_a^b \mathbf{F}(\mathbf{r}(t))\cdot\mathbf{r}'(t)\,dt=\int_a^b [P(x(t),y(t))x'(t)+Q(x(t),y(t))y'(t)]\,dt\). If \(C\) is closed, so that \(\mathbf{r}(a)=\mathbf{r}(b)\), this integral is the circulation of \(\mathbf{F}\) around \(C\) in the chosen orientation.

For the integral to be defined by this formula, the parametrization must be piecewise continuously differentiable, and the field must be defined and continuous along the curve. At the finitely many joining points of a piecewise smooth parametrization, the derivative may be defined separately on each segment. The integrals over the segments are then added.

A closed curve has the same starting and ending point. A simple closed curve has no self-intersections, apart from the matching start and end. For a simple planar loop, counterclockwise travel is called the positive orientation, and clockwise travel is the negative orientation. The loop’s shape alone does not specify its orientation; the direction of traversal must also be given or made clear.

Why the Parametrized Formula Measures Circulation

Over a small portion of a path, the displacement is approximately \(\Delta\mathbf{r}\), and the field is approximately constant at some value \(\mathbf{F}\) on that portion. The work over that small displacement is then approximately \(\mathbf{F}\cdot\Delta\mathbf{r}\). Adding these contributions along the curve and taking the limit gives the line integral.

To see the formula in coordinates, write \(\mathbf{r}(t)=(x(t),y(t))\). Its velocity is \(\mathbf{r}'(t)=(x'(t),y'(t))\), so the field’s component in the direction of motion contributes \(\mathbf{F}(\mathbf{r}(t))\cdot\mathbf{r}'(t)\). Substituting \(\mathbf{F}=(P,Q)\) gives the two component contributions \(Px'\) and \(Qy'\), each evaluated along the path. Thus

\mathbf{F}(\mathbf{r}(t))\cdot\mathbf{r}'(t) = P(x(t),y(t))x'(t)+Q(x(t),y(t))y'(t).

For a polygonal loop, apply this formula separately to every straight segment, in its direction of travel, and add the results. This is why a closed-curve calculation requires more than comparing the initial and final positions: the field can take different values on different parts of the route.

Orientation and the Sign of Circulation

Reversing orientation does not change which geometric points lie on the curve, but it reverses the direction of every displacement. The sign change follows directly from the parametrized formula. If \(\mathbf{r}(t)\), \(a\le t\le b\), parametrizes a curve forward, a parametrization in reverse is \(\widetilde{\mathbf{r}}(s)=\mathbf{r}(a+b-s)\). Then \(\widetilde{\mathbf{r}}'(s)=-\mathbf{r}'(a+b-s)\). Substituting \(u=a+b-s\) in the reversed integral gives

\begin{aligned} \int_a^b \mathbf{F}(\widetilde{\mathbf{r}}(s))\cdot\widetilde{\mathbf{r}}'(s)\,ds &= -\int_a^b \mathbf{F}(\mathbf{r}(a+b-s))\cdot\mathbf{r}'(a+b-s)\,ds\\ &= -\int_a^b \mathbf{F}(\mathbf{r}(u))\cdot\mathbf{r}'(u)\,du. \end{aligned}

Therefore, reversing a curve changes the sign of its line integral. For a closed loop, the circulation is positive in one direction precisely when it is negative in the opposite direction, unless its value is zero.

Orientation rule: If \(C^{-1}\) denotes the same curve traversed in the opposite direction, then \(\int_{C^{-1}}\mathbf{F}\cdot d\mathbf{r}=-\int_C\mathbf{F}\cdot d\mathbf{r}\). Always follow the stated direction of travel when setting the limits on each segment.

Computing Circulation Around Polygonal Loops

The practical method is to list the vertices in traversal order, parametrize each edge, and evaluate the line integral on that edge. For an edge \(\mathbf{r}(t)=(x(t),y(t))\), the required integrand is \(P(x(t),y(t))x'(t)+Q(x(t),y(t))y'(t)\). The total circulation is the sum of the edge integrals. The next examples show this process for loops traversed counterclockwise.

Worked Example: Circulation Around a Rectangle

Compute the circulation of \(\mathbf{F}(x,y)=(y,2x)\) around the rectangle with vertices \((0,0),(2,0),(2,1),(0,1)\), traversed in that order.

The order of the vertices gives a counterclockwise traversal. Write \(\mathbf{F}=(P,Q)\), where \(P=y\) and \(Q=2x\), so the integrand on a parametrized edge is \(y\,dx+2x\,dy\).

Bottom edge: Parametrize from \((0,0)\) to \((2,0)\) by \(x=t,\ y=0,\ 0\le t\le2\). Then \(dx=dt\) and \(dy=0\). Its contribution is

\int_0^2 (0\cdot 1+2t\cdot 0)\,dt=0.

Right edge: Parametrize from \((2,0)\) to \((2,1)\) by \(x=2,\ y=t,\ 0\le t\le1\). Then \(dx=0\) and \(dy=dt\). Its contribution is

\int_0^1 (t\cdot 0+2(2)\cdot 1)\,dt=\int_0^1 4\,dt=4.

Top edge: Parametrize from \((2,1)\) to \((0,1)\) by \(x=2-t,\ y=1,\ 0\le t\le2\). Then \(dx=-dt\) and \(dy=0\). Its contribution is

\int_0^2 (1\cdot(-1)+2(2-t)\cdot0)\,dt=-2.

Left edge: Parametrize from \((0,1)\) to \((0,0)\) by \(x=0,\ y=1-t,\ 0\le t\le1\). Then \(dx=0\) and \(dy=-dt\). Since \(2x=0\), its contribution is

\int_0^1 ((1-t)\cdot0+0\cdot(-1))\,dt=0.

Adding the four contributions gives the circulation:

$$0+4-2+0=2.$$

As a check, the edge-by-edge values sum to \(2\), and reversing the traversal would give \(-2\) by the orientation rule.

Worked Example: Circulation Around a Triangle

Compute the circulation of \(\mathbf{F}(x,y)=(y,x^2)\) around the triangle with vertices \((0,0),(2,0),(0,1)\), traversed in that order.

This vertex order is counterclockwise. Here \(P=y\) and \(Q=x^2\), so the integral along a parametrized segment is \(\int(y\,dx+x^2\,dy)\).

Bottom edge: Take \(x=t,\ y=0,\ 0\le t\le2\). Then \(dx=dt\), \(dy=0\), and the contribution is

\int_0^2 (0\cdot1+t^2\cdot0)\,dt=0.

Sloping edge: From \((2,0)\) to \((0,1)\), take \(x=2-2t,\ y=t,\ 0\le t\le1\). Then \(dx=-2\,dt\) and \(dy=dt\). Substitution gives

\begin{aligned} \int_0^1 \bigl(yx'+x^2y'\bigr)\,dt &=\int_0^1\left(t(-2)+(2-2t)^2(1)\right)\,dt\\ &=\int_0^1(4-10t+4t^2)\,dt\\ &=\left[4t-5t^2+\frac{4}{3}t^3\right]_0^1 =4-5+\frac43=\frac13. \end{aligned}

Vertical edge: From \((0,1)\) to \((0,0)\), take \(x=0,\ y=1-t,\ 0\le t\le1\). Then \(dx=0\), \(dy=-dt\), and \(x^2=0\), so the contribution is

\int_0^1 \bigl((1-t)\cdot0+0\cdot(-1)\bigr)\,dt=0.

The circulation is therefore \(0+\frac13+0=\frac13\). The result is consistent with the orientation rule: if the same three edges were traversed in reverse, the circulation would be \(-\frac13\).

Worked Example: Circulation Around a Trapezoid

Compute the circulation of \(\mathbf{F}(x,y)=(xy,x+y)\) around the trapezoid with vertices \((0,0),(3,0),(2,1),(0,1)\), in that order.

The loop is counterclockwise. In this example \(P=xy\) and \(Q=x+y\), so each segment contributes \(\int(xy\,dx+(x+y)\,dy)\).

Bottom edge: Use \(x=t,\ y=0,\ 0\le t\le3\). Both \(y\) and \(dy\) are zero, so the contribution is \(0\).

Sloping edge: From \((3,0)\) to \((2,1)\), take \(x=3-t,\ y=t,\ 0\le t\le1\). Then \(dx=-dt\), \(dy=dt\), \(xy=3t-t^2\), and \(x+y=3\). The contribution is

\begin{aligned} \int_0^1\bigl((3t-t^2)(-1)+3(1)\bigr)\,dt &=\int_0^1(3-3t+t^2)\,dt\\ &=\left[3t-\frac32t^2+\frac13t^3\right]_0^1 =3-\frac32+\frac13=\frac{11}{6}. \end{aligned}

Top edge: From \((2,1)\) to \((0,1)\), take \(x=2-t,\ y=1,\ 0\le t\le2\). Then \(dx=-dt\), \(dy=0\), and \(xy=2-t\). The contribution is

\int_0^2 (2-t)(-1)\,dt=\left[-2t+\frac12t^2\right]_0^2=-2.

Left edge: From \((0,1)\) to \((0,0)\), take \(x=0,\ y=1-t,\ 0\le t\le1\). Then \(dx=0,\ dy=-dt\), and \(x+y=1-t\). The contribution is

\int_0^1 (1-t)(-1)\,dt=\left[-t+\frac12t^2\right]_0^1=-\frac12.

Adding the four contributions gives \(0+\frac{11}{6}-2-\frac12=-\frac23\). This is the circulation in the specified counterclockwise direction; the clockwise traversal would give \(\frac23\). The negative value does not mean that the orientation was wrong: it records the signed accumulation of the field along the chosen direction.

A Common Mistake: Using Only the Endpoints

A closed path has zero net displacement, since its final position equals its initial position. It is tempting to conclude that the work around every closed path must therefore be zero. That conclusion is valid for a constant force, but not in general for a field whose value changes from point to point.

For instance, on the rectangle in the first example, an incorrect calculation would be

\mathbf{F}\cdot\bigl(\mathbf{r}(b)-\mathbf{r}(a)\bigr) = \mathbf{F}\cdot\mathbf{0} = 0.

This uses a single \(\mathbf{F}\) as if the field were constant along the path. But in that example \(\mathbf{F}(x,y)=(y,2x)\) varies with position. The correct segment calculation gave \(0+4-2+0=2\). Endpoint displacement alone does not capture the field’s changing contribution around the loop.

A second frequent error is to use the correct edges but parameterize one of them in the wrong direction. For example, the top edge of the rectangle is traversed from \(x=2\) to \(x=0\), so \(dx\) is negative. Replacing that contribution \(-2\) by \(+2\) changes the integral and breaks the specified orientation. Recording the vertices in order before calculating helps prevent this mistake.

Why Circulation Is Useful

Circulation gives a signed measure of how a field acts along an oriented loop. In applications, the field may represent a force and the circulation is work around a route. In a fluid interpretation, the integral records the field’s tangential contribution as one moves around a closed path. In each case, both the field and the orientation are part of the quantity being measured.

Direct calculation is especially useful when the loop consists of a few simple segments and the field is explicitly given. It also provides a reliable way to check signs: verify the direction of each edge, calculate its contribution, and only then add the contributions. A zero result is possible, but closedness alone does not guarantee it.

Check Your Understanding

Use the definitions and procedures from this tutorial to answer the following questions.

What condition on the endpoints of a parametrized curve makes it closed?
For a counterclockwise polygonal loop, how do you combine the integrals over its edges to obtain the circulation?
Why does reversing the orientation change the sign of a vector line integral?
What is the circulation around the rectangle in the first worked example if it is traversed clockwise?
Why does zero endpoint displacement not imply zero circulation for a variable vector field?