Line integrals — scalar and vector, work done: Question bank

True or false — justify

A scalar line integral

\int_{C} f d s

never changes sign when you reverse the curve.

True —

d s = ∣ r^{'} (t) ∣ d t

is an unsigned length, so summing

f \cdot d s

gives the same total whichever way you walk the wire.

A vector line integral

\int_{C} F \cdot d r

never changes sign when you reverse the curve.

False — reversing flips

d r = r^{'} d t

, so every dot product flips, giving

- W

. Work has a sign because dragging against the field returns energy.

If two different parametrisations trace the same curve in the same direction, they give the same vector line integral.

True — the value depends only on the geometric path and its orientation, not on how fast the parameter runs; the speed factor cancels correctly in both flavours.

\int_{C} f d s

can be negative if

f

dips below zero somewhere on

C

True —

d s > 0

always, but

f

can be negative, so the integrand

f d s

can be negative. (For a physical density

f \geq 0

it won't, but abstractly it can.)

For any closed curve

C

\oint_{C} F \cdot d r = 0

False — only guaranteed when

F

is conservative (see Gradient and conservative fields). A rotational field like

F = (- y, x)

gives a nonzero loop integral.

F = \nabla ϕ

, then

\int_{C} F \cdot d r

depends only on the endpoints of

C

True — by the Fundamental Theorem for Line Integrals,

W = ϕ (end) - ϕ (start)

, so the shape of the path in between is irrelevant.

Setting

f = 1

\int_{C} f d s

gives the arc length of

C

True — you are summing

1 \cdot d s = d s

over the curve, which is exactly the definition of Arc length.

\int_{C} F \cdot d r

equals

\int_{C} (F \cdot T) d s

for the same orientation.

True — since

d r = T d s

with

T = r^{'} /∣ r^{'} ∣

, the vector integral is literally the scalar integral of the tangential component of

F

F

is everywhere perpendicular to the curve, the work is zero.

True — the tangential component

F \cdot T = 0

everywhere, so nothing accumulates; sideways force does no work.

Spot the error

"Mass of wire

= \int_{a}^{b} f (r (t)) d t

." What's wrong?

The

∣ r^{'} (t) ∣

factor is missing;

d t

is a step in parameter, not in length. Correct is

\int_{a}^{b} f (r (t)) ∣ r^{'} (t) ∣ d t

d r

and

d s

are the same tiny quantity, so I can swap them." Why wrong?

d s = ∣ r^{'} ∣ d t

is a positive scalar (length);

d r = r^{'} d t

is a vector (displacement). They are related by

d r = T d s

, not equal.

"I reversed the curve, so I flip the sign on my mass integral." Error?

Mass is a scalar line integral — it's orientation-independent, so no sign flip. Only the vector/work integral flips.

F \cdot d r

means multiply the length of

F

by the length of

d r

." Error?

The dot product keeps only the component of

F

along

d r

; it's

∣ F ∣∣ d r ∣ cos θ

, not a plain product. (See Dot product.)

"Since

∣ r^{'} (t) ∣ = 1

for the unit circle, I can always drop that factor." Error?

It's

1

only because that particular parametrisation has unit speed. A reparametrisation like

r (2 t)

has speed

2

; you must recompute

∣ r^{'} ∣

each time.

"To get work I integrate

∣ F ∣

over arc length." Error?

That would count the full force magnitude regardless of direction. Work uses only the tangential part:

\int_{C} (F \cdot T) d s

, so a force perpendicular to motion contributes nothing.

"Green's Theorem lets me convert any

\int_{C} F \cdot d r

into a double integral." Error?

Green's Theorem applies to a closed curve bounding a region (and needs

F

smooth on it); it doesn't apply to an open path.

Why questions

Why does the speed factor

∣ r^{'} (t) ∣

appear in the scalar integral but the raw velocity

r^{'} (t)

appear in the vector one?

The scalar integral needs only how much length each step covers (a magnitude), while the vector integral needs the direction of each step to dot against

F

, so it keeps the full velocity vector.

Why is

\int_{C} F \cdot d r

independent of how fast you parametrise, even though

r^{'}

gets bigger at higher speed?

A faster parametrisation shrinks the

t

-interval by exactly the same factor it grows

r^{'}

; the two effects cancel in

\int F \cdot r^{'} d t

, leaving the geometric value.

Why does a conservative field make loop integrals vanish?

Because

W = ϕ (end) - ϕ (start)

, and on a closed loop the start and end coincide, so the difference is zero.

Why do we take a dot product for work instead of just multiplying magnitudes?

Only the component of force pointing along the motion transfers energy; the dot product

F \cdot T

extracts exactly that forward-pushing part and discards the sideways part.

Why can we always collapse a line integral over a curve into an ordinary single integral in

t

A curve is 1-dimensional — one number

t

pins your position — so substituting the parametrisation turns everything into a standard integral over the interval

[a, b]

Why does reversing orientation flip the vector integral but a reparametrisation keeping direction does not?

Reversal literally negates every displacement

d r

, negating the sum; a same-direction reparametrisation only relabels the speed, which cancels out and leaves the answer unchanged.

Edge cases

What is

\int_{C} f d s

C

is a single point (degenerate, zero length)?

Zero — there is no arc length to integrate over, so the sum of

f \cdot d s

is empty.

What is the work if

F = 0

everywhere?

Zero — every

F \cdot d r = 0

, regardless of the path or its length.

For a closed curve, what does the scalar integral

\oint_{C} 1 d s

compute, and can it be zero?

It computes the total perimeter (arc length) of the loop; it's zero only if the loop is degenerate (a point), otherwise strictly positive.

F

is conservative, what is

\int_{C} F \cdot d r

over a path that starts and ends at the same point but wanders far away first?

Still zero — path-independence depends only on the endpoints coinciding, not on how far the path strays.

What happens to

\int_{C} f d s

if you traverse the same curve twice (once around, then again)?

It doubles — you accumulate the arc-length contribution a second time;

d s

is always positive so there's no cancellation.

What happens to

\oint_{C} F \cdot d r

(non-conservative

F

) if you loop the closed curve twice?

It doubles — you accumulate the same nonzero circulation on each lap since orientation is preserved both times.

If the curve has a corner (velocity undefined at one point), can the line integrals still be defined?

Yes — split

C

into smooth pieces and add the integrals; a single non-smooth point of zero length contributes nothing.

7. Connections

Line integrals — scalar and vector, work done — the parent this bank drills.
Dot product — the "component along the path" engine behind every work question.
Parametric curves — why one number $t$ collapses a curve integral.
Arc length — the $f = 1$ edge case.
Gradient and conservative fields and Fundamental Theorem for Line Integrals — path-independence traps.
Green's Theorem — the closed-curve conversion caveat.