Questions: Conservative Vector Fields and Potential Functions

Question 1

A vector field F has zero curl everywhere in the plane except at the origin. Which statement best describes F?

• A. F is conservative everywhere in the plane
• B. F is conservative on any region that does not contain the origin
• C. F is conservative on any region that does not encircle the origin
• D. F cannot be conservative anywhere since its curl is not zero everywhere

Answer: (C)

Zero curl on a simply-connected region is sufficient for F to be conservative there. A region that does not *encircle* the origin is simply connected (no holes). A region that forms an annulus around the origin is not simply connected, and F may fail to be conservative on it even though curl F = 0. Option A is wrong because a circle around the origin gives a nonzero line integral. Option B is subtly wrong: not containing the origin is not the same as not encircling it.

Question 2

You compute ∫_C F · dr for a path C from A to B and get 5. A colleague takes a completely different path from A to B and also gets 5. What can you conclude?

• A. Nothing — two paths agreeing once is a coincidence
• B. F is likely conservative — the integrals would agree for any path from A to B
• C. F has zero curl at every point
• D. A potential function exists and f(B) − f(A) = 5

Answer: (B)

Two instances of path-independence are consistent with F being conservative but do not prove it. If F were conservative, *every* path from A to B would give the same value. You cannot conclude D (existence of a potential function) from two data points alone, and you cannot conclude C (zero curl) without either checking curl directly or verifying path-independence holds universally. The data is suggestive, not conclusive.

Question 3

For a vector field F with continuous partial derivatives on all of R³, F is conservative if and only if its curl is zero.

• T. True
• F. False

Answer: True

R³ is simply connected — it has no holes or tunnels — so the central equivalence theorem applies in full: zero curl is both necessary and sufficient for F to be conservative. This equivalence breaks down on domains with topological holes (such as R² minus the origin), but on all of R³ the four conditions (F = ∇f, path-independence, zero closed-loop integral, zero curl) are all equivalent.

Question 4

A vector field F satisfies curl F = 0 everywhere on R² except at the origin. Then F is conservative.

• T. True
• F. False

Answer: False

Zero curl away from the origin does not guarantee F is conservative on R² minus the origin. The classic counterexample is F = ⟨−y, x⟩/(x² + y²): its curl is zero everywhere except the origin, but ∮_C F · dr = 2π for any circle C enclosing the origin. R² minus the origin is not simply connected — closed curves that encircle the origin cannot be contracted to a point, which is exactly the topological obstruction that allows curl-free fields to fail path-independence.

Question 5

Why does the equivalence between 'zero curl' and 'conservative field' require the domain to be simply connected? Give an example illustrating what can go wrong.

Model answer: On a domain with holes, there exist closed curves that cannot be shrunk to a point. A field can have zero curl at every point yet still have nonzero circulation around a hole. The classic example: F = ⟨−y, x⟩/(x² + y²) has zero curl on R² \ {0}, but integrating F around a unit circle centered at the origin gives 2π ≠ 0. Simply-connected domains rule out such topological obstructions by ensuring every closed curve bounds a surface lying entirely in the domain, so Stokes' theorem can equate circulation to the (zero) surface integral of curl F.

Zero curl means the field is locally a gradient (no rotation at any point), but global path-independence also requires that every closed loop can be filled by a surface inside the domain. Holes prevent this for loops that encircle them. Simply connected is the precise topological condition guaranteeing that no such obstructing loops exist.