A curl-free field that is not conservative: the whole story
The vector field F(x, y) = (−y/(x² + y²), x/(x² + y²)) is defined on the plane with the origin removed. A candidate computes ∂/∂x [x/(x² + y²)] = (y² − x²)/(x² + y²)² and ∂/∂y [−y/(x² + y²)] = (y² − x²)/(x² + y²)², observes that the two are equal, and concludes that ∮_C F·dr = 0 for every closed curve C. The tutor proposes three test curves: C₁ the unit circle traversed counterclockwise, C₂ the circle of radius 3 centred at (5,0) traversed counterclockwise, and C₃ a simple closed curve, winding twice counterclockwise around the origin. Direct parametrisation of C₁ gives ∫₀^{2π} dt = 2π, contradicting the conclusion, while F is perfectly smooth on and inside C₂. The candidate suspects an error in the partial derivatives and wants to know where the real gap lies.
1. Compute ∮ F·dr over C₁ and over C₂ from parametrisations, and explain why Green's theorem is legitimate for one curve and not the other.
Read the worked answer
Both integrals are of the form ∮ (x dy − y dx)/(x² + y²).
C₁: x = cos t, y = sin t, so x dy − y dx = (cos²t + sin²t) dt = dt and x² + y² = 1. Hence the integral is ∫₀^{2π} dt = 2π.
C₂: x = 5 + 3cos t, y = 3sin t, so x² + y² = 34 + 30cos t and x dy − y dx = (15cos t + 9) dt. The integral is ∫₀^{2π} (9 + 15cos t)/(34 + 30cos t) dt = ∫₀^{2π} [1/2 − 8/(34 + 30cos t)] dt. Using ∫₀^{2π} dt/(a + b cos t) = 2π/√(a² − b²) = 2π/√(34² − 30²) = π/8, the value is π − 8(π/8) = 0.
Green's theorem applies to C₂ because F is C¹ on the closed disk bounded by C₂ (its closest point to the origin is (2,0), so the singular point is excluded). For C₁ the origin lies inside the region, so F is not defined, let alone C¹, on the whole interior, and the hypothesis fails.
2. What is the integral over C₃, and what is the general rule? On which domains is F conservative, and what potential can be written down?
Read the worked answer
General rule: ∮_C F·dr = 2π·wind(C, 0), where wind is the winding number about the origin. So C₁ gives 2π, and C₃, winding twice, gives 4π. The value is always an integer multiple of 2π; it is not a defect of computation but a measure of how the curve links the missing point.
F is conservative on any simply connected domain that avoids the origin. On such a domain one may take the polar angle as a potential: F = ∇θ with θ = arctan(y/x) on x > 0 (and θ = arctan(y/x) + π on x < 0), or equivalently the angle restricted to a slit plane such as R² minus the ray {(x, 0) : x ≤ 0}. Then every closed line integral in that domain vanishes.
θ cannot be defined continuously on the whole punctured plane - going once around adds 2π - which is exactly why the global integral is 2π rather than 0. The field is locally a gradient everywhere but has no global single-valued potential.
3. The candidate's computation of the two partial derivatives is correct. State the precise theorem being misapplied and why its hypothesis is the only thing that fails.
Read the worked answer
The theorem is: if F = (P, Q) is C¹ on a simply connected domain D and ∂Q/∂x = ∂P/∂y at every point of D, then F is conservative on D, and ∮_C F·dr = 0 for every closed piecewise-smooth curve in D.
Here D = R² \ {(0,0)}, which is not simply connected: the unit circle cannot be shrunk to a point without passing through the removed origin. The curl condition is satisfied at every point of D, so the algebraic hypothesis holds; only the topological hypothesis fails, and it fails definitively, as the value 2π shows.
The curl condition is local: it controls integrals over small loops that enclose no hole, which is why the same computation correctly predicts that the integral over C₂ vanishes. A useful sanity rule is that the value of a closed integral depends only on the winding number and the singularities enclosed, so one should never conclude 'curl zero hence integral zero' without first checking the shape of the domain.