Line integrals — scalar and vector, work done: Worked examples

The scenario matrix

Every line-integral problem you will ever meet lands in one of these cells. Each worked example below is tagged with the cell(s) it covers, so you can see the whole board fills up.

#	Cell (the scenario class)	What makes it tricky	Example
A	Scalar $\int_{C} f d s$ , non-constant speed	must keep $∣ r^{'} ∣$ , it isn't 1	Ex 1
B	Scalar integral, orientation reversed	answer must not change	Ex 2
C	Vector integral, non-conservative field	genuinely path-dependent	Ex 3
D	Vector integral, orientation reversed	answer flips sign	Ex 3 (part b)
E	Conservative field, use endpoints only	shortcut via a potential	Ex 4
F	Closed loop, non-conservative	nonzero circulation	Ex 5
G	Closed loop, conservative	must be zero	Ex 6
H	Degenerate curve (a single point)	zero length ⇒ zero integral	Ex 7
I	Piecewise path (corner)	split into pieces, add	Ex 8
J	Real-world word problem + units	translate physics to symbols	Ex 9

Read across: A–B exhaust the scalar flavour (speed matters, orientation doesn't). C–G exhaust the vector flavour by field type (conservative vs not) and curve type (open vs closed). H–J catch the degenerate, piecewise, and applied twists that exams love.

Two tools recur, so let's re-earn them in one line each:

Dot product $a \cdot b = a_{1} b_{1} + a_{2} b_{2}$ — measures "how much of $a$ points along $b$ ". We need it because work only counts the force along the motion.
Parametric curves — describing $C$ by one number $t$ , so every integral becomes an ordinary $\int_{a}^{b} \dots d t$ .

Ex 1 — Scalar integral, speed is NOT constant · Cell A

Worked example Mass of a wire along

$y = x^{2}$ Wire follows $r (t) = (t, t^{2})$ for $t \in [0, 1]$ , with density $f (x, y) = x$ . Find the total mass $\int_{C} f d s$ .

Forecast: the wire steepens as it climbs, so it gets "longer per unit $t$ " near the top. Do you expect the speed factor $∣ r^{'} ∣$ to be $1$ , or bigger than $1$ and growing? Guess before reading.

Step 1. Velocity: $r^{'} (t) = (1, 2 t)$ . Why this step? $d s$ comes from the length of a tiny step, and a tiny step is $r^{'} (t) d t$ ; we differentiate each component to get it.

Step 2. Speed: $∣ r^{'} (t)∣ = 1^{2} + (2 t)^{2} = 1 + 4 t^{2}$ . Why this step? $d s = ∣ r^{'} ∣ d t$ . Notice it is not $1$ — it grows with $t$ , exactly the steepening we forecast. (See figure: the step arrows get longer.)

Step 3. Density on the curve: $f (r (t)) = x = t$ . Why this step? Substitute the parametrisation into $f$ ; here $x = t$ .

Step 4. Assemble and integrate: $\int_{C} f d s = \int_{0}^{1} t 1 + 4 t^{2} d t .$ Let $u = 1 + 4 t^{2}$ , so $d u = 8 t d t$ , i.e. $t d t = \frac{1}{8} d u$ . Limits: $t = 0 \to u = 1$ , $t = 1 \to u = 5$ . $= \frac{1}{8} \int_{1}^{5} u d u = \frac{1}{8} \cdot \frac{2}{3} [u^{3/2}]_{1}^{5} = \frac{1}{12} (5^{3/2} - 1) .$ Why this step? Substitution turns the messy $t 1 + 4 t^{2}$ into a clean power of $u$ .

Answer: $\frac{5 5 - 1}{12} \approx 0.8844.$

Verify: $5^{3/2} = 55 \approx 11.18$ , so $(11.18 - 1) /12 \approx 0.848$ … let's be exact: $(11.1803 - 1) /12 = 0.8484$ . Positive, sensible (density and length both positive), and dimensionally a mass. ✓

Worked example Same wire, walked backwards

Take Ex 1's wire but traverse it from $(1, 1)$ to $(0, 0)$ : $r (t) = (1 - t, (1 - t)^{2})$ , $t \in [0, 1]$ . Show the mass is unchanged.

Forecast: density is unsigned, arc-length is unsigned. Do you expect the same $0.8484$ or its negative?

Step 1. $r^{'} (t) = (- 1, - 2 (1 - t))$ , so $∣ r^{'} ∣ = 1 + 4 (1 - t)^{2}$ . Why this step? Same machine; the minus signs are inside a square, so they vanish — length can't be negative.

Step 2. $f = x = 1 - t$ . Integral: $\int_{0}^{1} (1 - t) 1 + 4 (1 - t)^{2} d t .$ Sub $s = 1 - t$ (so $d s = - d t$ , and swapping limits kills the minus) gives $\int_{0}^{1} s 1 + 4 s^{2} d s$ — identical to Ex 1. Why this step? The change of variable exposes that the two integrals are literally the same object.

Answer: $\frac{5 5 - 1}{12} \approx 0.8484$ , same as Ex 1.

Verify: matches Ex 1 exactly ⇒ confirms the rule: scalar line integrals are orientation-independent. ✓

Ex 3 — Vector integral, non-conservative, both directions · Cells C & D

Worked example Work of a swirling field, forward then back

Field $F = (- y, x)$ along the unit circle $r (t) = (cos t, sin t)$ . (a) Forward quarter, $t \in [0, π /2]$ . (b) The same quarter reversed.

Forecast: this field circulates counter-clockwise (arrows point along the direction of motion). Forward should give positive work; will (b) be its exact negative?

(a) Step 1. $r^{'} (t) = (- sin t, cos t)$ . Why? Displacement direction — differentiate each component.

Step 2. $F (r (t)) = (- sin t, cos t)$ (since $x = cos t, y = sin t$ ). Why? Substitute the point into the field.

Step 3. Dot product: $F \cdot r^{'} = (- sin t) (- sin t) + (cos t) (cos t) = sin^{2} t + cos^{2} t = 1$ . Why? Work counts only the tangential part; here the field is perfectly aligned with the motion, so the dot product is a constant $1$ .

Step 4. $W = \int_{0}^{π /2} 1 d t = \frac{π}{2} \approx 1.5708.$

(b) Reverse: by the orientation rule, $W \to - \frac{π}{2} \approx - 1.5708.$ Why? $d r$ flips sign, the dot product flips, the integral flips — the vector asymmetry.

Verify: forward $+ π /2$ , backward $- π /2$ ; they sum to $0$ as they must (out and back cancels). Contrast Ex 2, where reversing did nothing. ✓

Ex 4 — Conservative field: use endpoints only · Cell E

Worked example Skip the integral with a potential

$F = (2 x y, x^{2})$ along any path from $(0, 0)$ to $(2, 3)$ . Find the work.

Forecast: if we can find a scalar $ϕ$ with $\nabla ϕ = F$ , the Fundamental Theorem for Line Integrals says $W = ϕ (end) - ϕ (start)$ regardless of the path. Guess: does the answer depend on the route?

Step 1. Seek $ϕ$ with $ϕ_{x} = 2 x y$ and $ϕ_{y} = x^{2}$ . Why? Gradient and conservative fields: work of a gradient field is a potential difference. Integrate $ϕ_{x}$ in $x$ : $ϕ = x^{2} y + g (y)$ . Then $ϕ_{y} = x^{2} + g^{'} (y)$ must equal $x^{2}$ , so $g^{'} (y) = 0$ , $g =$ const. Take $ϕ = x^{2} y$ .

Step 2. Endpoint check on conservativeness: $\partial_{y} (2 x y) = 2 x = \partial_{x} (x^{2})$ — matches, so $F$ really is conservative (needed for the shortcut to be legal). Why? If the "cross-derivatives" disagreed, no $ϕ$ would exist and we'd have to integrate the long way.

Step 3. $W = ϕ (2, 3) - ϕ (0, 0) = (2^{2} \cdot 3) - 0 = 12.$

Verify (integrate along a straight line to be sure): $r (t) = (2 t, 3 t)$ , $t \in [0, 1]$ : $r^{'} = (2, 3)$ , $F = (2 (2 t) (3 t), (2 t)^{2}) = (12 t^{2}, 4 t^{2})$ , dot $= 24 t^{2} + 12 t^{2} = 36 t^{2}$ , $\int_{0}^{1} 36 t^{2} d t = 12$ . ✓ Same as the endpoint shortcut.

Ex 5 — Closed loop, non-conservative: nonzero circulation · Cell F

Worked example Full lap of the swirl

$F = (- y, x)$ around the whole unit circle, $t \in [0, 2 π]$ , counter-clockwise.

Forecast: Ex 3 gave $1$ per unit $t$ of tangential push. A full lap is $2 π$ of $t$ . Also — a closed loop of a conservative field would give $0$ ; is this field conservative?

Step 1. From Ex 3, $F \cdot r^{'} = 1$ on the unit circle. Why? Same field, same curve; the dot product is still the constant $1$ .

Step 2. $\oint_{C} F \cdot d r = \int_{0}^{2 π} 1 d t = 2 π \approx 6.2832.$

Step 3. Cross-check conservativeness: $\partial_{y} (- y) = - 1$ , $\partial_{x} (x) = 1$ . They disagree ( $- 1 \neq = 1$ ), so $F$ is not conservative — which is exactly why a closed loop gives a nonzero answer. Why? A conservative field would force $0$ on every loop (Ex 6). The nonzero $2 π$ is a signature of genuine swirl.

Verify with Green's Theorem: $\oint P d x + Q d y = \iint_{D} (Q_{x} - P_{y}) d A = \iint_{D} (1 - (- 1)) d A = 2 \cdot (area of disc) = 2 π$ . ✓ Matches.

Ex 6 — Closed loop, conservative: must vanish · Cell G

Ex 7 — Degenerate curve: a single point · Cell H

Ex 8 — Piecewise path with a corner · Cell I

Worked example Work over an L-shaped route

$F = (y, x)$ from $(0, 0)$ to $(1, 0)$ (segment $C_{1}$ ) then up to $(1, 1)$ (segment $C_{2}$ ). Find total work.

Forecast: $F = (y, x) = \nabla (x y)$ is conservative (Ex from parent), so the answer should equal $ϕ (1, 1) - ϕ (0, 0) = 1$ . Let's confirm by grinding each piece.

Step 1 — $C_{1}$ : $r (t) = (t, 0)$ , $t \in [0, 1]$ , $r^{'} = (1, 0)$ , $F = (y, x) = (0, t)$ . Dot $= 0 \cdot 1 + t \cdot 0 = 0$ . So $\int_{C_{1}} = 0$ . Why? Along the bottom edge the " $y$ -part" of the force is $0$ and the motion is purely horizontal.

Step 2 — $C_{2}$ : $r (t) = (1, t)$ , $t \in [0, 1]$ , $r^{'} = (0, 1)$ , $F = (t, 1)$ . Dot $= t \cdot 0 + 1 \cdot 1 = 1$ . So $\int_{C_{2}} = \int_{0}^{1} 1 d t = 1$ . Why? Now motion is vertical and the " $x$ -part" of the force is $1$ everywhere on this edge.

Step 3. Total $= 0 + 1 = 1$ . Why? Line integrals add over concatenated pieces: $\int_{C_{1} + C_{2}} = \int_{C_{1}} + \int_{C_{2}}$ .

Verify: potential shortcut $ϕ (1, 1) - ϕ (0, 0) = 1 - 0 = 1$ . ✓ Corner handled correctly.

Ex 9 — Real-world word problem with units · Cell J

Worked example Dragging a crate through a headwind

A crate moves along $r (t) = (t, t)$ metres for $t \in [0, 4]$ . A wind force (newtons) is $F = (- 1, 0)$ N — it blows in the $- x$ direction the whole time. How much work does the wind do on the crate?

Forecast: the crate moves in the $+ x, + y$ direction; the wind opposes the $+ x$ part. Expect negative work (the wind fights the motion).

Step 1. $r^{'} (t) = (1, 1)$ m per unit $t$ . Why? Displacement direction; here motion is a $4 5^{\circ}$ line.

Step 2. Dot product: $F \cdot r^{'} = (- 1) (1) + (0) (1) = - 1$ . Why? Only the component of wind along the motion does work; the wind's $x$ -component opposes the crate's $x$ -motion, giving $- 1$ .

Step 3. $W = \int_{0}^{4} (- 1) d t = - 4$ joules (N·m). Why? Constant integrand times the parameter range.

Answer: $W = - 4 J$ — the wind removes $4$ joules.

Verify (units + sign): N × m $=$ J. ✓ Sign is negative, matching "wind opposes motion", as forecast. ✓

Recall

Recall Which cells flip sign under reversal, which don't?

Scalar (Cells A–B, H) ::: orientation-independent — no sign flip. Vector (Cells C–J) ::: flips sign when the curve is reversed.

Recall Fast test: is a closed-loop integral automatically zero?

Only if the field is conservative ::: closed loop of $\nabla ϕ$ gives $0$ (Ex 6); a swirl gives nonzero circulation (Ex 5).

Connections

Parametric curves — every example started here.
Dot product — the engine of every vector example.
Arc length — the $f = 1$ scalar case (Ex 7's check).
Gradient and conservative fields · Fundamental Theorem for Line Integrals — Ex 4, 6, 8 shortcuts.
Green's Theorem — Ex 5 cross-check.

Worked examples — Line integrals — scalar and vector, work done