Fundamental theorem for line integrals: Question bank

True or false — justify

Recall T/F: A line integral of

any vector field over a curve depends only on the endpoints. False ::: Only conservative fields ( $F = \nabla f$ ) are path independent; a general field like $F = (- y, x)$ gives different answers on different paths between the same two points.

Recall T/F: If

$\oint_{C} F \cdot d r = 0$ for one particular closed loop $C$ , then $F$ is conservative. False ::: Conservativeness requires every closed loop to give zero; a single vanishing loop can happen by accident (e.g. by symmetry) while other loops are nonzero.

Recall T/F: If

$F = \nabla f$ , then $\oint_{C} F \cdot d r = 0$ for every closed curve $C$ in the domain. True ::: A closed curve has start point $=$ end point, so $A = B$ and $f (B) - f (A) = 0$ ; this holds for all loops precisely because a potential exists everywhere on the domain.

Recall T/F: The condition

$P_{y} = Q_{x}$ guarantees $F = (P, Q)$ is conservative. False ::: It only guarantees conservativeness on a simply connected region; on a domain with a hole (like the plane minus the origin) $P_{y} = Q_{x}$ can hold yet a loop around the hole gives a nonzero integral.

Recall T/F: Two curves from

$A$ to $B$ that cross each other must give the same line integral for a conservative field. True ::: Path independence means all curves from $A$ to $B$ give $f (B) - f (A)$ ; whether they cross, wiggle, or share length is irrelevant.

Recall T/F: If

$f$ and $f + 7$ are both potentials for $F$ , they give different values of $\int_{C} F \cdot d r$ . False ::: Adding a constant cancels in the difference: $(f (B) + 7) - (f (A) + 7) = f (B) - f (A)$ , so the constant of integration never affects the integral.

Recall T/F: A conservative field can still have a nonzero line integral along an open path.

True ::: Path independence does not mean the integral is zero — it means it equals $f (B) - f (A)$ , which is nonzero whenever the endpoint potentials differ.

Spot the error

Recall "

$F = (- y, x)$ ; since it looks smooth and rotational, I'll use FTLI and just subtract endpoints." Error ::: This field is not conservative: $P_{y} = - 1 \neq = 1 = Q_{x}$ , so no potential exists and FTLI does not apply — the integral genuinely depends on the path.

Recall "I found

$f_{x} = P$ so $f = \int P d x$ ; done, that's the potential." Error ::: The constant of integration can depend on the other variable, so $f = \int P d x + h (y)$ ; you must still differentiate in $y$ and match $f_{y} = Q$ to pin down $h (y)$ .

Recall "

$P_{y} = Q_{x}$ for the vortex field $(\frac{- y}{x ^{2} + y ^{2}}, \frac{x}{x ^{2} + y ^{2}})$ , so its integral around the unit circle is $0$ ." Error ::: The domain (plane minus origin) is not simply connected; the loop encloses the hole and the integral is $2 π$ , not $0$ — the equal-partials test fails to guarantee conservativeness here.

Recall "The curve

$C$ is complicated, so I must parametrise $r (t)$ and grind the integral, even though $F = \nabla (x^{2} y)$ ." Error ::: Since $F$ is given as a gradient, FTLI lets you skip the parametrisation entirely and just compute $f (B) - f (A)$ — the messy path is irrelevant.

Recall "FTLI is

$\int_{C} \nabla f \cdot d r = f (A) - f (B)$ , end minus... wait, start minus end." Error ::: The correct orientation is $f (B) - f (A)$ = end minus start (matching FTC's $g (b) - g (a)$ ); reversing it flips the sign, which is exactly the effect of traversing $C$ backwards.

Recall "

$F$ is conservative, so $\int_{C} F \cdot d r = 0$ for this open path from $(0, 0)$ to $(1, 1)$ ." Error ::: Conservative does not mean zero on open paths; it equals $f (1, 1) - f (0, 0)$ , which is zero only if those two potential values happen to coincide.

Why questions

Recall Why does the proof of FTLI reduce to the ordinary

FTC? Because the integrand $\nabla f (r (t)) \cdot r^{'} (t)$ is exactly $\frac{d}{d t} f (r (t))$ by the chain rule ::: so it is a perfect derivative $g^{'} (t)$ , and FTC turns $\int_{a}^{b} g^{'} (t) d t$ into $g (b) - g (a) = f (B) - f (A)$ .

Recall Why must

$\nabla f$ be continuous on $C$ for FTLI to hold? Continuity ensures $g (t) = f (r (t))$ is differentiable with an integrable derivative ::: without it the FTC step can fail — a jump or singularity along $C$ breaks the "integral of a derivative" argument.

Recall Why does a closed-loop integral of a conservative field vanish, intuitively?

Because you return to your starting height ::: the potential is like elevation, and net elevation change on any loop back to the start is zero, so $f (B) - f (A) = 0$ .

Recall Why is the "hole in the domain" the real culprit in the vortex counterexample, not the formula?

Because a potential would have to be a single continuous "angle" function, but angle jumps by $2 π$ each time you circle the hole ::: no single-valued $f$ exists on the punctured plane, so FTLI's hypothesis silently fails.

Recall Why does the mixed-partials test

$P_{y} = Q_{x}$ come from potentials at all? If $F = \nabla f$ then $P = f_{x}$ and $Q = f_{y}$ , and continuous mixed partials commute ::: so $P_{y} = f_{x y} = f_{y x} = Q_{x}$ — the test is just Clairaut's equality, which is necessary for a potential to exist.

Recall Why can we ignore the specific path but never ignore the endpoints?

The path-dependent contributions cancel exactly (that is the theorem's content) ::: but $f (B) - f (A)$ still needs both endpoint values, so endpoints carry all the surviving information.

Recall Why is FTLI called the "multivariable twin" of the ordinary FTC?

| ::: Both say "integrate a derivative, get boundary values": FTC gives $\int_{a}^{b} g^{'} d x = g (b) - g (a)$ over an interval's endpoints, and FTLI gives $\int_{C} \nabla f \cdot d r = f (B) - f (A)$ over a curve's endpoints.

Edge cases

Recall Edge case: What is

$\int_{C} \nabla f \cdot d r$ when $C$ is a single point (start = end, zero length)? Zero ::: with $A = B$ we get $f (B) - f (A) = 0$ , consistent with the degenerate "loop" of no length; there is nothing to integrate.

Recall Edge case:

$F = 0$ (the zero field). Is it conservative, and what is any line integral of it? Yes, conservative with constant potential ::: $0 = \nabla (c)$ for any constant $c$ , and every line integral is $c - c = 0$ regardless of the path or endpoints.

Recall Edge case: The curve

$C$ leaves the domain of $f$ partway (e.g. passes through a point where $\nabla f$ is undefined). Can we still apply FTLI? No ::: FTLI needs $\nabla f$ continuous on all of $C$ ; if $C$ crosses a singularity the theorem does not apply and the integral may not even converge.

Recall Edge case: What if we traverse the same conservative field along

$C$ and then along $C$ reversed? The two integrals are negatives and sum to zero ::: reversing swaps $A$ and $B$ , giving $f (A) - f (B) = - (f (B) - f (A))$ ; together they form a closed loop with total $0$ .

Recall Edge case: A field is conservative on the right half-plane

$x > 0$ but you integrate along a path staying in $x > 0$ . Does the global "hole" elsewhere matter? No ::: what matters is that $C$ and a potential live in a simply connected sub-region; since the right half-plane is simply connected and contains $C$ , FTLI applies there even if the field misbehaves far away.

Recall Edge case: Is a constant nonzero field like

$F = (3, 4)$ conservative, and what is its potential? Yes ::: $F = \nabla (3 x + 4 y)$ , so $f = 3 x + 4 y$ is a potential and any line integral equals $f (B) - f (A)$ ; constant fields are always conservative on $R^{2}$ .

Connections

Fundamental theorem for line integrals — the parent theorem every trap here tests.
Conservative vector fields — the exact class where path independence lives.
Potential functions — the $f$ whose endpoint values are the answer.
Green's theorem — explains why the "hole" breaks the $P_{y} = Q_{x}$ test.
Fundamental Theorem of Calculus — the 1D ancestor the proof reduces to.
Gradient — the operator $\nabla$ producing conservative fields.
Curl — the 3D generalisation of the $P_{y} = Q_{x}$ test.