Line integrals — scalar and vector, work done: Exercises

Before we start, a one-line reminder of the two tools you will keep reaching for.

Figure — Line integrals — scalar and vector, work done

Level 1 — Recognition

Goal: pick the right tool and set up the integral. No cleverness yet.

Problem 1.1

A wire follows $r (t) = (3 t, 4 t)$ for $t \in [0, 1]$ with constant density $f = 1$ . Its total mass is just its length. Find that length.

Recall Solution 1.1

WHAT tool? Density spread along a wire → scalar line integral. With $f = 1$ this is Arc length.

Velocity: differentiate each component. $r^{'} (t) = (3, 4)$ .
Speed: $∣ r^{'} (t) ∣ = 3^{2} + 4^{2} = 25 = 5$ . Why this step? $d s = ∣ r^{'} ∣ d t$ converts a step in $t$ into a step in length.
$L = \int_{0}^{1} 1 \cdot 5 d t = 5 \cdot (1 - 0) = 5 .$ Sanity: the curve is a straight segment from $(0, 0)$ to $(3, 4)$ , whose ordinary distance is $3^{2} + 4^{2} = 5$ . ✓

Problem 1.2

For $F = (2, 3)$ (a constant force) and the straight path $r (t) = (t, t)$ , $t \in [0, 1]$ , set up and evaluate $\int_{C} F \cdot d r$ .

Recall Solution 1.2

WHAT tool? A force dragging along a path → vector line integral (work).

$r^{'} (t) = (1, 1)$ .
$F (r (t)) = (2, 3)$ — constant, so no substitution needed.
Dot product: $F \cdot r^{'} = 2 (1) + 3 (1) = 5$ . Why dot? Only the part of $F$ along the motion does work.
$W = \int_{0}^{1} 5 d t = 5 .$

Level 2 — Application

Goal: substitute a genuine parametrisation and integrate.

Problem 2.1

Density $f (x, y) = x + y$ along the quarter unit circle $r (t) = (cos t, sin t)$ , $t \in [0, π /2]$ . Find the mass.

Recall Solution 2.1

$r^{'} (t) = (- sin t, cos t)$ , so $∣ r^{'} (t) ∣ = sin^{2} t + cos^{2} t = 1$ . Unit circle → unit speed.
$f (r (t)) = x + y = cos t + sin t$ . Substitute the parametrisation into the density.
Mass $= \int_{0}^{π /2} (cos t + sin t) \cdot 1 d t = [sin t - cos t]_{0}^{π /2} = (1 - 0) - (0 - 1) = 2 .$

Problem 2.2

$F = (x, y)$ along $r (t) = (t, t^{2})$ , $t \in [0, 1]$ . Find the work $W$ .

Recall Solution 2.2

$r^{'} (t) = (1, 2 t)$ .
$F (r (t)) = (x, y) = (t, t^{2})$ .
$F \cdot r^{'} = t (1) + t^{2} (2 t) = t + 2 t^{3}$ .
$W = \int_{0}^{1} (t + 2 t^{3}) d t = [\frac{t ^{2}}{2} + \frac{t ^{4}}{2}]_{0}^{1} = \frac{1}{2} + \frac{1}{2} = 1 .$

Level 3 — Analysis

Goal: reason about what changes when the setup changes — orientation, speed, parametrisation.

Problem 3.1 (orientation)

For $F = (x, y)$ , compute the work along the reversed parabola: from $(1, 1)$ to $(0, 0)$ , i.e. $\tilde{r} (t) = (1 - t, (1 - t)^{2})$ , $t \in [0, 1]$ . Compare with Problem 2.2.

Recall Solution 3.1

$\tilde{r}^{'} (t) = (- 1, - 2 (1 - t))$ . Chain rule on each component.
$F (\tilde{r} (t)) = (1 - t, (1 - t)^{2})$ .
Dot: $(1 - t) (- 1) + (1 - t)^{2} \cdot (- 2 (1 - t)) = - (1 - t) - 2 (1 - t)^{3}$ .
Let $u = 1 - t$ , $d u = - d t$ ; as $t : 0 \to 1$ , $u : 1 \to 0$ . $W = \int_{0}^{1} [- (1 - t) - 2 (1 - t)^{3}] d t = \int_{1}^{0} [- u - 2 u^{3}] (- d u) = \int_{1}^{0} (u + 2 u^{3}) d u = - \int_{0}^{1} (u + 2 u^{3}) d u = - 1.$ So $W = - 1$ , exactly the negative of Problem 2.2. Why? Reversing orientation flips $d r$ , hence flips the whole dot-product integral.

Problem 3.2 (reparametrisation invariance of scalar integrals)

Redo Problem 2.1's mass but walk the quarter circle twice as fast: $r (t) = (cos 2 t, sin 2 t)$ , $t \in [0, π /4]$ . Show the mass is unchanged.

Recall Solution 3.2

$r^{'} (t) = (- 2 sin 2 t, 2 cos 2 t)$ , so $∣ r^{'} (t) ∣ = 4 sin^{2} 2 t + 4 cos^{2} 2 t = 2$ . Twice the speed.
$f (r (t)) = cos 2 t + sin 2 t$ .
Mass $= \int_{0}^{π /4} (cos 2 t + sin 2 t) \cdot 2 d t$ . Substitute $u = 2 t$ , $d u = 2 d t$ : $= \int_{0}^{π /2} (cos u + sin u) d u = [sin u - cos u]_{0}^{π /2} = 2.$ Same answer, $2$ . Why? The extra speed factor $2$ is exactly cancelled by covering the arc in half the $t$ -interval — $d s$ is the same physical length either way.

Level 4 — Synthesis

Goal: combine the tool with a bigger idea — conservative fields and the Fundamental Theorem.

Show that $F = (y, x)$ is conservative by finding a potential $ϕ$ with $F = \nabla ϕ$ , then use the Fundamental Theorem for Line Integrals to compute the work along any path from $(0, 0)$ to $(2, 3)$ .

Recall Solution 4.1

Find $ϕ$ . We need $\partial_{x} ϕ = y$ and $\partial_{y} ϕ = x$ .

Integrate the first in $x$ : $ϕ = x y + g (y)$ .
Differentiate in $y$ : $\partial_{y} ϕ = x + g^{'} (y)$ ; match to $x$ ⇒ $g^{'} (y) = 0$ ⇒ $g$ constant. Take $ϕ = x y$ .
Check via Gradient and conservative fields: $\nabla ϕ = (\partial_{x} ϕ, \partial_{y} ϕ) = (y, x) = F$ . ✓ Apply the theorem. For a conservative field, $\int_{C} F \cdot d r = ϕ (end) - ϕ (start)$ . $W = ϕ (2, 3) - ϕ (0, 0) = (2) (3) - 0 = 6 .$ Why is this legal for any path? Because the work depends only on endpoints — a direct consequence of $F = \nabla ϕ$ .

Problem 4.2

Verify Problem 4.1's answer directly along the straight segment $r (t) = (2 t, 3 t)$ , $t \in [0, 1]$ .

Recall Solution 4.2

$r^{'} (t) = (2, 3)$ .
$F (r (t)) = (y, x) = (3 t, 2 t)$ .
Dot: $3 t (2) + 2 t (3) = 12 t$ .
$W = \int_{0}^{1} 12 t d t = [6 t^{2}]_{0}^{1} = 6$ . ✓ Matches the endpoint computation.

Level 5 — Mastery

Goal: a closed-curve computation, two ways, revealing when the shortcut collapses.

Problem 5.1

Let $F = (- y, x)$ . Compute the work around the full unit circle counterclockwise, $r (t) = (cos t, sin t)$ , $t \in [0, 2 π]$ . Then explain, using Green's Theorem, why a conservative field would have given $0$ .

Recall Solution 5.1

Direct computation.

$r^{'} (t) = (- sin t, cos t)$ .
$F (r (t)) = (- sin t, cos t)$ .
Dot: $(- sin t) (- sin t) + (cos t) (cos t) = sin^{2} t + cos^{2} t = 1$ .
$W = \int_{0}^{2 π} 1 d t = 2 π .$

Green's Theorem check. For a positively oriented closed curve $C$ bounding region $D$ , $\oint_{C} P d x + Q d y = \iint_{D} (\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}) d A .$ Here $P = - y, Q = x$ , so $\partial_{x} Q - \partial_{y} P = 1 - (- 1) = 2$ . Thus $W = \iint_{D} 2 d A = 2 \cdot (area of unit disc) = 2 π .$ ✓ Same $2 π$ .

Why nonzero? A conservative field returns to the same potential value after a loop, giving $\oint = 0$ . Here $\partial_{x} Q - \partial_{y} P = 2 \neq = 0$ , so $F$ is not conservative — the loop genuinely does $2 π$ of net work. This is the swirling "rotation" field.

Problem 5.2 (mastery of the two-readings identity)

For $F = (- y, x)$ on the same circle, verify $\int_{C} F \cdot d r = \int_{C} (F \cdot T) d s$ by computing the right-hand side.

Recall Solution 5.2

Unit tangent: $T = \frac{r ^{'} ( t )}{∣ r ^{'} ( t ) ∣} = \frac{( - sin t , cos t )}{1} = (- sin t, cos t)$ .
$F \cdot T = (- sin t) (- sin t) + (cos t) (cos t) = 1$ . The field is entirely tangential here — it points exactly along the motion, magnitude $1$ .
$d s = ∣ r^{'} ∣ d t = 1 d t$ , so $\int_{C} (F \cdot T) d s = \int_{0}^{2 π} 1 d t = 2 π .$ ✓ Identical, as promised by $d r = T d s$ .

Recall

Recall Rapid self-test

Which integral did you use for "mass of a wire"? → scalar $\int_{C} f d s$ .
Reversing the curve changed which of your answers? → only the vector/work integral (P3.1, sign flip).
Speeding up the parametrisation changed the mass? → no (P3.2), $d s$ is invariant.
The shortcut $W = ϕ (end) - ϕ (start)$ needs what condition? → $F = \nabla ϕ$ conservative.
Why wasn't the loop of $(- y, x)$ zero? → curl $= 2 \neq = 0$ ; not conservative.

Connections

Arc length — Problem 1.1 is exactly $\int_{C} 1 d s$ .
Dot product — every vector integral above.
Parametric curves — every $r (t)$ .
Gradient and conservative fields — Problems 4.1–4.2.
Fundamental Theorem for Line Integrals — the endpoint shortcut.
Green's Theorem — Problem 5.1.