Conservative vector fields — potential functions: Question bank

Before we start, three words we lean on constantly:

A conservative field is one that equals $\nabla f$ (the gradient of some scalar hill $f$ ) — see Gradient and directional derivative.
A loop integral $\oint_{C} F \cdot d r$ is the total "push along" a closed path — see Line integrals of vector fields.
The curl measures local swirl; curl $= 0$ is the cross-partial test $P_{y} = Q_{x}$ — see Curl of a vector field.

Spot the error

"I integrated

P

x

and got

f = x^{2} y + 3 x

, so that's the potential."

Error: dropped the ==

+ g (y)

==. Integrating with respect to

x

leaves an unknown function of $y$ , which the equation

f_{y} = Q

must still pin down.

P_{y} = Q_{x}

held, so the field is conservative — no need to check the domain."

Error: skipped the simply-connected requirement. On a domain with a hole the test can pass while a loop around the hole gives a nonzero integral.

"Every vector field is the gradient of something, so I'll just find

f

Error: only curl-free fields on nice domains have potentials. For

(y, - x)

P_{y} = 1 \neq = - 1 = Q_{x}

, so no

f

exists.

"To find

f

I integrated

P

x

, then integrated

Q

y

, then added them."

Error: that double-counts shared terms. The correct method integrates

P

once, then differentiates the result in

y

and matches

Q

to recover only the missing piece

g (y)

"The work along

C

from

A

B

was hard, so I picked a straight-line path to make it easier — but only because I got lucky."

Not an error, but the reasoning is wrong: it isn't luck. Once the field is confirmed conservative, every path gives the same answer, so choosing the easy path is always legal.

g^{'} (y) = - 4 y

g (y) = - 4 y

Error: forgot to integrate.

g (y) = \int - 4 y d y = - 2 y^{2} + C

; you must antidifferentiate, not copy.

"Curl is zero in 2D means the single number

P_{y} - Q_{x}

is zero, so

F

is automatically conservative anywhere."

Error: "anywhere" hides the domain assumption again; the implication needs no holes enclosed by the loops you care about.

Why questions

Why does the cross-partial test only involve mixed partials and not

P_{x}

Q_{y}

Because conservativeness compares

f_{x y}

with

f_{y x}

; the "straight" partials

f_{xx}, f_{y y}

carry no cross-consistency information and are unconstrained.

Why is Clairaut's theorem (equality of mixed partials) the engine behind the test?

It guarantees

f_{x y} = f_{y x}

for smooth

f

; reading that backwards, if

F = \nabla f

then

P_{y} = Q_{x}

must hold, so a failure of

P_{y} = Q_{x}

rules out any potential.

Why does a hole in the domain break the "curl-zero ⟹ conservative" step?

A hole lets a loop encircle a point where the field isn't defined; the recovery of

f

needs to fill in that missing point, and no single-valued

f

can, so path-independence fails — this is the punctured-plane counterexample.

Why is a conservative force said to "conserve energy"?

Work done equals only the potential difference between endpoints, so a round trip returns you to the same potential — no energy is created or milked by looping, matching Conservation of energy in physics.

Why can Green's theorem be used to confirm a loop integral is zero for a curl-free field on a filled region?

Green's theorem converts the loop integral into a double integral of

Q_{x} - P_{y}

over the enclosed area; if that integrand is zero throughout a hole-free region, the loop integral is zero.

Why does the Fundamental Theorem of Line Integrals need a potential, not just any scalar function?

The derivation uses the chain rule

\frac{d}{d t} f (r (t)) = \nabla f \cdot r^{'}

; this only turns the integrand into a perfect derivative when

F

actually equals

\nabla f

Why does knowing

f

let you skip the parametrization of

C

entirely?

Because the integral collapses to

f (B) - f (A)

, which depends only on the two endpoints — the curve's shape and speed never enter.

Edge cases

Is the zero field

F = 0

conservative?

Yes — it is

\nabla f

for any constant

f

; trivially curl-free and every integral is zero.

What is the loop integral of the punctured field

(- y, x) / (x^{2} + y^{2})

around a loop that does not enclose the origin?

Zero — with the hole outside, the region is effectively simply connected there, so curl-free gives a vanishing loop, and only loops circling the origin give

2 π

Can a field be conservative on one region but not on a larger one containing a hole?

Yes — restricted to a hole-free patch it may be

\nabla f

, but extending across a puncture can make the potential multivalued and break conservativeness globally.

P_{y} = Q_{x}

fails at even a single point, is the field still non-conservative?

Yes — a genuine gradient field must satisfy

P_{y} = Q_{x}

everywhere it is defined and smooth, so one violating point rules out any potential on any region containing it.

Does adding a constant

C

to the potential change any physically measurable quantity?

No — work and loop integrals depend on differences

f (B) - f (A)

, and the

C

cancels; this is why potentials are defined only up to a constant.

What happens to the recovery method if the field is not conservative — where does it break?

The step

f_{y} = Q

becomes impossible: after integrating

P

you get an

f_{y}

that still contains

x

(or another variable it shouldn't), so

g^{'} (y)

can't be an honest function of

y

alone, signalling no potential exists.

Is a field defined only along a single curve (not a 2D region) meaningfully "conservative"?

The concept needs an open domain to take partial derivatives and compare paths; on a lone curve there are no alternative paths, so the cross-partial test isn't even defined.

Recall One-line litmus for the whole page

Curl zero plus no holes ⟹ conservative ⟹ path-independent ⟹ loop integrals vanish ⟹ a potential $f$ exists, unique up to $+ C$ .

Question bank — Conservative vector fields — potential functions

True or false — justify

Spot the error

Why questions

Edge cases