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Conservative Fields and Path Independence · Tutorial 778 of 999

Closed-Curve Integrals as a Conservativeness Test

Use closed-loop line integrals to test whether a planar vector field can be conservative, with careful attention to parametrization, orientation, and what a zero result does—and does not—show.

Intermediate 9 min read

What You'll Learn

  • Define a closed-curve line integral for a planar vector field.
  • Explain why every closed-curve integral of a conservative field is zero.
  • Calculate circulation around an oriented rectangle by adding its side integrals.
  • Calculate circulation around a circle using a parametrization.
  • Use a nonzero loop integral to rule out a potential on the domain.
  • Explain why a zero integral on one selected loop does not prove conservativeness.

Why Integrate Around a Closed Curve?

A vector field may do different amounts of work along different routes between the same two points. A closed curve provides a direct way to test this: travel around a loop and add up the work along every part of the route. If the total is nonzero, the field cannot have a potential on a domain containing that loop.

This gives a useful one-sided test. A single nonzero closed-curve integral disproves conservativeness. By contrast, obtaining zero on one selected loop proves very little: other loops may have nonzero integrals.

Closed-Curve Line Integrals

Let \(D\) be a domain in the plane, and let \(\mathbf{F}(x,y)=(P(x,y),Q(x,y))\) be a vector field defined on \(D\). A piecewise-smooth parametrized curve \(\mathbf{r}(t)=(x(t),y(t))\), for \(a\leq t\leq b\), is closed if \(\mathbf{r}(a)=\mathbf{r}(b)\). Its line integral is

$$ \int_C \mathbf{F}\cdot d\mathbf{r} = \int_a^b \left(P(x(t),y(t))x'(t)+Q(x(t),y(t))y'(t)\right)\,dt. $$

The curve’s orientation is the direction in which the parameter traces it. Reversing the direction changes the sign of the integral. For a loop made of several smooth pieces, calculate the integral on each piece in its direction of travel, then add the results.

Definition: A vector field \(\mathbf{F}\) is conservative on \(D\) if there is a differentiable scalar function \(\phi\) on \(D\) such that \(\mathbf{F}=\nabla\phi\) throughout \(D\). The function \(\phi\) is called a potential for \(\mathbf{F}\).

Why Conservative Fields Have Zero Closed-Curve Integrals

Suppose \(\mathbf{F}=\nabla\phi\) on \(D\), and let \(C\) be a piecewise-smooth closed curve contained in \(D\). On each smooth piece, the chain rule gives the rate of change of \(\phi\) along the curve:

$$ \frac{d}{dt}\phi(\mathbf{r}(t)) = \nabla\phi(\mathbf{r}(t))\cdot\mathbf{r}'(t) = \mathbf{F}(\mathbf{r}(t))\cdot\mathbf{r}'(t). $$

Integrating over each piece gives the change in \(\phi\) between that piece’s endpoints. Adding the pieces makes all intermediate endpoint values cancel: each piece ends where the next begins. The only remaining difference is the value at the end of the whole curve minus the value at its start. Because the curve is closed, those points coincide, so the difference is zero.

$$ \int_C \mathbf{F}\cdot d\mathbf{r} = \phi(\mathbf{r}(b))-\phi(\mathbf{r}(a)) = 0. $$
Theorem: If \(\mathbf{F}\) is conservative on \(D\), then its line integral around every piecewise-smooth closed curve contained in \(D\) is zero. Therefore, if even one such curve has a nonzero integral, \(\mathbf{F}\) is not conservative on \(D\).

The conclusion is a contradiction argument: if a potential existed on \(D\), the closed-curve integral would have to be zero. A computed nonzero value rules out that potential. No calculation of a potential is needed.

Worked Example: A Rectangle in a Planar Field

Worked Example: A Rectangle in a Planar Field

Test whether \(\mathbf{F}(x,y)=(-y,0)\) is conservative on \(\mathbb{R}^2\) by integrating counterclockwise around the rectangle with vertices \((1,0)\), \((3,0)\), \((3,2)\), and \((1,2)\).

Split the boundary into four directed sides. On the bottom side, \(y=0\) and \(dy=0\). Since \(\mathbf{F}\cdot d\mathbf{r}=-y\,dx\), this side contributes

$$ \int_{x=1}^{3} -0\,dx=0. $$

On the right side, \(x=3\) and \(dx=0\). The field has zero second component, so the contribution is

$$ \int_{y=0}^{2} 0\,dy=0. $$

On the top side, the direction is from \(x=3\) to \(x=1\), and \(y=2\). Its contribution is

$$ \int_{x=3}^{1} -2\,dx = -2(1-3) = 4. $$

On the left side, again \(dx=0\), so the contribution is \(0\). Adding the four side integrals gives

$$ \int_C \mathbf{F}\cdot d\mathbf{r} = 0+0+4+0 = 4. $$

The result is nonzero, so \(\mathbf{F}\) is not conservative on \(\mathbb{R}^2\). As a check on the orientation, reversing the direction of the entire rectangle would change the integral from \(4\) to \(-4\), not to \(4\) or \(0\).

Worked Example: A Circle in a Rotational Field

Worked Example: A Circle in a Rotational Field

Calculate the counterclockwise line integral of \(\mathbf{F}(x,y)=(-y,x)\) around the circle of radius \(2\) centered at the origin. The circle lies in the field’s domain, \(\mathbb{R}^2\). Parametrize it by

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

The velocity vector and field along the curve are

$$ \mathbf{r}'(t)=(-2\sin t,2\cos t), \qquad \mathbf{F}(\mathbf{r}(t))=(-2\sin t,2\cos t). $$

Taking their dot product gives the integrand:

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

Therefore,

$$ \int_C\mathbf{F}\cdot d\mathbf{r} = \int_0^{2\pi}4\,dt = 4(2\pi)-4(0) = 8\pi. $$

The integral is nonzero, so the field is not conservative on \(\mathbb{R}^2\). The parametrization traces the circle counterclockwise; tracing it clockwise would give \(-8\pi\). Substituting the dot product back into the line-integral formula verifies that the computed value is exactly the integral around the chosen curve.

A Nonzero Loop Also Shows Path Dependence

The closed-loop test is closely related to comparing two paths with the same endpoints. If \(C_1\) goes from \(A\) to \(B\) and \(C_2\) also goes from \(A\) to \(B\), follow \(C_1\) and then traverse \(C_2\) backwards. The result is a closed route. Reversing a curve changes the sign of its line integral, so the integral around this route is

$$ \int_{C_1}\mathbf{F}\cdot d\mathbf{r} - \int_{C_2}\mathbf{F}\cdot d\mathbf{r}. $$

Thus, a nonzero closed-loop integral means that the two path integrals differ. In particular, the integral is not path independent. The converse reasoning is also useful for constructing a loop: if two paths between the same endpoints give different integrals, joining one path to the reverse of the other produces a loop with nonzero integral.

Worked Example: Why One Zero Loop Is Not Enough

Worked Example: Why One Zero Loop Is Not Enough

Consider \(\mathbf{F}(x,y)=(xy,0)\). First integrate around the unit circle centered at the origin, counterclockwise. Use \(\mathbf{r}(t)=(\cos t,\sin t)\), for \(0\leq t\leq 2\pi\). Then \(dx=-\sin t\,dt\), and along the circle \(xy=\cos t\sin t\). Hence

$$ \int_C\mathbf{F}\cdot d\mathbf{r} = \int_0^{2\pi}(\cos t\sin t)(-\sin t)\,dt = -\int_0^{2\pi}\cos t\sin^2t\,dt = -\left[\frac{\sin^3t}{3}\right]_0^{2\pi} = 0. $$

This one circle gives zero. Now test the counterclockwise rectangle with vertices \((1,0)\), \((2,0)\), \((2,1)\), and \((1,1)\). The bottom side contributes \(0\), since \(y=0\). The right and left sides contribute \(0\), since \(dx=0\) and the field’s second component is zero. On the top side, \(y=1\) and \(x\) runs from \(2\) to \(1\), so

$$ \int_{x=2}^{1}xy\,dx = \int_2^1x\,dx = \left[\frac{x^2}{2}\right]_2^1 = \frac{1}{2}-2 = -\frac{3}{2}. $$

Adding the four sides gives \(0+0-\frac{3}{2}+0=-\frac{3}{2}\), which is nonzero. The field is therefore not conservative on \(\mathbb{R}^2\), despite the zero integral around the unit circle. The circle calculation checks as zero, but it says nothing decisive about the integral around every other loop.

How to Use the Test Carefully

For a proposed test, first ensure that the entire curve lies in the domain where conservativeness is being considered. Then choose an orientation, parametrize the curve or split it into simpler pieces, and calculate the integral using that direction. A nonzero answer is decisive: the field cannot be conservative on that domain. A zero answer is only the result for the particular curve tested.

1
Choose a closed curve in the domain.
Make sure every point of the loop is included in the field’s domain.
2
Set its orientation and parametrization.
For a piecewise curve, record the direction of each segment.
3
Compute and add the line integrals.
Use the parametrized formula or evaluate each segment separately.
4
Interpret the result.
A nonzero total disproves conservativeness on the domain; a zero total does not establish it.

A common mistake is to treat one zero loop integral as proof that the field is conservative. The calculation for \((xy,0)\) shows why that inference fails: the unit-circle integral is zero, but a rectangle integral is \(-\frac{3}{2}\). The reliable conclusion from this test is asymmetric: one nonzero loop is enough to disprove conservativeness, while proving conservativeness requires more than checking one loop.

Check Your Understanding

Use the definitions and calculations above to answer these questions.

  1. Why must a conservative field have zero line integral around every piecewise-smooth closed curve in its domain?
  2. What does a nonzero integral around one closed curve establish about the field on a domain containing that curve?
  3. When a loop is traversed in the reverse direction, how does its line integral change?
  4. Why is it invalid to conclude that \((xy,0)\) is conservative from its zero integral around the unit circle?
  5. If two paths with the same endpoints have unequal line integrals, how can they be combined to form a closed curve?

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