Fundamental theorem for line integrals: Exercises

Before we start, one picture to keep in your head the whole way down — the only idea being tested is "the integral is the height difference between two points on a landscape $f$ ".

Figure — Fundamental theorem for line integrals

The coloured surface is $f$ (a "height"), the arrows are $\nabla f$ (the "steepest-uphill" field), and the two black dots are the start $A$ and end $B$ . Every problem below is secretly asking "what is (height at $B$ ) − (height at $A$ )?"

Level 1 — Recognition

Goal: recognise when FTLI applies and read off endpoints. No integration.

Exercise 1.1

$f (x, y) = x^{2} y$ . Compute $\int_{C} \nabla f \cdot d r$ where $C$ runs from $A = (1, 0)$ to $B = (3, 2)$ along any path.

Recall Solution 1.1

What we notice: the integrand is already written as $\nabla f$ . By FTLI the answer is $f (B) - f (A)$ — the path is irrelevant. Why: FTLI says $\int_{C} \nabla f \cdot d r = f (B) - f (A)$ .

$f (3, 2) = 3^{2} \cdot 2 = 18$ .
$f (1, 0) = 1^{2} \cdot 0 = 0$ . $\int_{C} \nabla f \cdot d r = 18 - 0 = 18 .$

Exercise 1.2

$f (x, y, z) = x y z$ . Find $\int_{C} \nabla f \cdot d r$ from $A = (1, 1, 1)$ to $B = (2, 1, 3)$ .

Recall Solution 1.2

What we do: endpoints only. $f (2, 1, 3) = 2 \cdot 1 \cdot 3 = 6$ , and $f (1, 1, 1) = 1$ . $\int_{C} \nabla f \cdot d r = 6 - 1 = 5 .$

Exercise 1.3

$F = \nabla (cos (x y))$ . Find $\oint_{C} F \cdot d r$ around the ellipse $\frac{x ^{2}}{4} + y^{2} = 1$ (start = end).

Recall Solution 1.3

What we notice: the little circle $\oint$ on the integral means the curve is closed — you finish where you started, so $A = B$ . Why it collapses: $f (B) - f (A) = f (A) - f (A) = 0$ regardless of what $f$ is. $\oint_{C} F \cdot d r = 0 .$ The specific ellipse and the specific $f$ never mattered.

Level 2 — Application

Goal: build the potential $f$ from a raw field $F = (P, Q)$ , then apply FTLI.

Exercise 2.1

$F = (2 x y, x^{2})$ , curve $C$ from $(0, 0)$ to $(2, 3)$ . Evaluate $\int_{C} F \cdot d r$ .

Recall Solution 2.1

Step 1 — Test conservativeness. Write $F = (P, Q)$ with $P = 2 x y$ , $Q = x^{2}$ . On the whole plane (simply connected) we need $\frac{\partial P}{\partial y} = \frac{\partial Q}{\partial x}$ . $P_{y} = 2 x, Q_{x} = 2 x . ✓$ Why: if a potential $f$ exists then $P = f_{x}, Q = f_{y}$ , and mixed partials commute, forcing $P_{y} = Q_{x}$ . Step 2 — Integrate $P$ in $x$ . $f = \int 2 x y d x = x^{2} y + h (y)$ . The "constant" of integration can depend on $y$ because we froze $y$ while integrating. Step 3 — Match $Q$ . $f_{y} = x^{2} + h^{'} (y)$ must equal $Q = x^{2}$ , so $h^{'} (y) = 0 \Rightarrow h (y) =$ const. Take $h = 0$ : $f = x^{2} y$ . Step 4 — FTLI. $f (2, 3) = 4 \cdot 3 = 12$ , $f (0, 0) = 0$ . $\int_{C} F \cdot d r = 12 - 0 = 12 .$

Exercise 2.2

$F = (2 x y, x^{2} + 2 y)$ , $C$ from $(0, 0)$ to $(1, 2)$ .

Recall Solution 2.2

Step 1 — Test. $P = 2 x y, Q = x^{2} + 2 y$ ; $P_{y} = 2 x$ , $Q_{x} = 2 x$ . Equal ✓. Step 2 — Integrate $P$ : $f = x^{2} y + h (y)$ . Step 3 — Match $Q$ : $f_{y} = x^{2} + h^{'} (y) = x^{2} + 2 y \Rightarrow h^{'} (y) = 2 y \Rightarrow h (y) = y^{2}$ . So $f = x^{2} y + y^{2}$ . Step 4 — FTLI: $f (1, 2) = 2 + 4 = 6$ , $f (0, 0) = 0$ . $\int_{C} F \cdot d r = 6 - 0 = 6 .$

Exercise 2.3

$F = (y cos x, sin x + 2 y)$ , $C$ from $(0, 0)$ to $(\frac{π}{2}, 1)$ .

Recall Solution 2.3

Step 1 — Test. $P = y cos x, Q = sin x + 2 y$ . $P_{y} = cos x$ , $Q_{x} = cos x$ . Equal ✓. Step 2 — Integrate $P$ in $x$ : $\int y cos x d x = y sin x + h (y)$ . Step 3 — Match $Q$ : $f_{y} = sin x + h^{'} (y) = sin x + 2 y \Rightarrow h^{'} (y) = 2 y \Rightarrow h (y) = y^{2}$ . So $f = y sin x + y^{2}$ . Step 4 — FTLI: $f (\frac{π}{2}, 1) = 1 \cdot sin \frac{π}{2} + 1 = 1 + 1 = 2$ ; $f (0, 0) = 0$ . $\int_{C} F \cdot d r = 2 - 0 = 2 .$

Level 3 — Analysis

Goal: decide whether FTLI even applies, and if not, why.

Exercise 3.1

Is $F = (- y, x)$ conservative? If yes, find $f$ ; if no, compute $\oint_{C} F \cdot d r$ around the unit circle traversed once counterclockwise.

Recall Solution 3.1

Step 1 — Test. $P = - y, Q = x$ ; $P_{y} = - 1$ , $Q_{x} = 1$ . Not equal ⇒ no potential exists ⇒ FTLI does not apply. The integral genuinely depends on the path. Step 2 — Compute directly since we can't shortcut. Parametrise $x = cos t, y = sin t, t \in [0, 2 π]$ , so $d x = - sin t d t, d y = cos t d t$ . $\oint_{C} F \cdot d r = \int_{0}^{2 π} (P d x + Q d y) = \int_{0}^{2 π} ((- sin t) (- sin t) + (cos t) (cos t)) d t .$ The bracket is $sin^{2} t + cos^{2} t = 1$ , so $\oint = \int_{0}^{2 π} 1 d t = 2 π \neq = 0.$ A nonzero loop integral is the fingerprint of a non-conservative field.

Exercise 3.2

Two paths from $(0, 0)$ to $(1, 1)$ : path $C_{1}$ is the straight line, path $C_{2}$ is the parabola $y = x^{2}$ . For $F = (2 x y, x^{2})$ (from 2.1), show both give the same value by direct integration, confirming path independence.

Recall Solution 3.2

We already know from FTLI (2.1's potential $f = x^{2} y$ ) the answer should be $f (1, 1) - f (0, 0) = 1 - 0 = 1$ . Let's verify the hard way.

Path $C_{1}$ (line): $x = t, y = t, t \in [0, 1]$ ; $d x = d t, d y = d t$ . $\int_{0}^{1} (2 t \cdot t d t + t^{2} d t) = \int_{0}^{1} 3 t^{2} d t = [t^{3}]_{0}^{1} = 1.$ Path $C_{2}$ (parabola): $x = t, y = t^{2}, t \in [0, 1]$ ; $d x = d t, d y = 2 t d t$ . $\int_{0}^{1} (2 t \cdot t^{2} d t + t^{2} \cdot 2 t d t) = \int_{0}^{1} (2 t^{3} + 2 t^{3}) d t = \int_{0}^{1} 4 t^{3} d t = [t^{4}]_{0}^{1} = 1.$ Both equal $1$ — path independence, confirmed. Because $F = \nabla (x^{2} y)$ , FTLI guaranteed this before any integration.

Exercise 3.3

The vortex field $F = (\frac{- y}{x ^{2} + y ^{2}}, \frac{x}{x ^{2} + y ^{2}})$ satisfies $P_{y} = Q_{x}$ . Yet $\oint$ around the unit circle is not $0$ . Compute it and explain the paradox.

Recall Solution 3.2 has nothing to do with this — read carefully.

Recall Solution 3.3

Check $P_{y} = Q_{x}$ (so it looks conservative): with $P = \frac{- y}{x ^{2} + y ^{2}}$ , $P_{y} = \frac{- ( x ^{2} + y ^{2} ) - ( - y ) ( 2 y )}{( x ^{2} + y ^{2} ) ^{2}} = \frac{y ^{2} - x ^{2}}{( x ^{2} + y ^{2} ) ^{2}},$ and with $Q = \frac{x}{x ^{2} + y ^{2}}$ , $Q_{x} = \frac{( x ^{2} + y ^{2} ) - x ( 2 x )}{( x ^{2} + y ^{2} ) ^{2}} = \frac{y ^{2} - x ^{2}}{( x ^{2} + y ^{2} ) ^{2}} .$ They match everywhere the field is defined ✓. Now compute the loop. $x = cos t, y = sin t$ so $x^{2} + y^{2} = 1$ . Then $P = - sin t, Q = cos t$ , $d x = - sin t d t, d y = cos t d t$ : $\oint = \int_{0}^{2 π} (sin^{2} t + cos^{2} t) d t = \int_{0}^{2 π} 1 d t = 2 π .$ The paradox resolved: $P_{y} = Q_{x}$ guarantees a potential only on a simply connected region — one with no holes. This field is undefined at the origin, so the disk it encloses has a puncture. FTLI's hypothesis fails, and the loop need not vanish. (Away from the origin, on any region avoiding the hole, a local potential $f = arctan (y / x)$ does exist — but it can't be defined consistently all the way around.)

Level 4 — Synthesis

Goal: combine potential-finding, endpoint algebra, and 3D.

Exercise 4.1

$F = (y z, x z, x y)$ in 3D. Find a potential $f$ and evaluate $\int_{C} F \cdot d r$ from $(1, 1, 1)$ to $(2, 3, 4)$ .

Recall Solution 4.1

Step 1 — Integrate $P = y z$ in $x$ : $f = x y z + h (y, z)$ (constant may depend on the other two variables). Step 2 — Match $Q$ : $f_{y} = x z + h_{y} = Q = x z \Rightarrow h_{y} = 0$ , so $h = h (z)$ only. Step 3 — Match $R$ : $f_{z} = x y + h^{'} (z) = R = x y \Rightarrow h^{'} (z) = 0$ , so $h =$ const. Take $f = x y z$ . Step 4 — FTLI: $f (2, 3, 4) = 24$ , $f (1, 1, 1) = 1$ . $\int_{C} F \cdot d r = 24 - 1 = 23 .$

Exercise 4.2

For which constant $a$ is $F = (a y^{2}, 2 x y + 3 y)$ conservative? For that $a$ , find $f$ and $\int_{C} F \cdot d r$ from $(1, 0)$ to $(0, 2)$ .

Recall Solution 4.2

Step 1 — Force the test. $P = a y^{2}, Q = 2 x y + 3 y$ . Need $P_{y} = Q_{x}$ : $P_{y} = 2 a y$ , $Q_{x} = 2 y$ . Equal for all $y$ ⇒ $2 a = 2 \Rightarrow a = 1$ . Step 2 — Build $f$ with $a = 1$ . $P = y^{2}$ ; integrate in $x$ : $f = x y^{2} + h (y)$ . Step 3 — Match $Q$ : $f_{y} = 2 x y + h^{'} (y) = 2 x y + 3 y \Rightarrow h^{'} (y) = 3 y \Rightarrow h (y) = \frac{3}{2} y^{2}$ . So $f = x y^{2} + \frac{3}{2} y^{2}$ . Step 4 — FTLI: $f (0, 2) = 0 \cdot 4 + \frac{3}{2} \cdot 4 = 6$ ; $f (1, 0) = 0 + 0 = 0$ . $\int_{C} F \cdot d r = 6 - 0 = 6 .$

Exercise 4.3

A field is conservative with potential $f (x, y) = e^{x} sin y$ . A particle travels from $A = (0, π /2)$ out to $B = (1, π)$ and then back to $A$ along a different curve. What is the total work $\oint F \cdot d r$ ?

Recall Solution 4.3

Key insight: the round trip starts and ends at $A$ . So it is a closed loop, and for a conservative field $\oint \nabla f \cdot d r = f (A) - f (A) = 0$ — no matter that the two legs are different curves. $\oint F \cdot d r = 0 .$ (Sanity check on the outbound leg alone: $f (1, π) = e^{1} sin π = 0$ , $f (0, \frac{π}{2}) = e^{0} sin \frac{π}{2} = 1$ , so out = $0 - 1 = - 1$ ; the return leg must give $+ 1$ , and $- 1 + 1 = 0$ .)

Level 5 — Mastery

Goal: prove structure, not just crunch numbers.

Exercise 5.1

Prove that if $F$ is conservative on a connected region, then $\int_{C} F \cdot d r$ is the same for every curve $C$ joining fixed points $A$ and $B$ (path independence). Then prove the converse-flavoured corollary: path independence ⇒ every closed-loop integral is zero.

Recall Solution 5.1

Forward direction (conservative ⇒ path independence). Since $F = \nabla f$ , FTLI applies to any curve $C$ from $A$ to $B$ : $\int_{C} F \cdot d r = \int_{C} \nabla f \cdot d r = f (B) - f (A) .$ The right-hand side contains no reference to $C$ — only the endpoints. So two curves $C_{1}, C_{2}$ both give $f (B) - f (A)$ ; hence $\int_{C_{1}} F \cdot d r = \int_{C_{2}} F \cdot d r . ■$ Corollary (path independence ⇒ loops vanish). Take any closed loop $C$ with base point $A = B$ . Split it into two arcs $C_{1}$ (from $A$ to some point $M$ ) and $C_{2}$ (from $M$ back to $A$ ). Reversing $C_{2}$ gives a curve $- C_{2}$ from $A$ to $M$ . Path independence says $\int_{C_{1}} F \cdot d r = \int_{- C_{2}} F \cdot d r = - \int_{C_{2}} F \cdot d r .$ Therefore $\int_{C_{1}} + \int_{C_{2}} = 0$ , i.e. $\oint_{C} F \cdot d r = 0$ . $■$

Exercise 5.2

A hiker's altitude is $f (x, y) = 100 - x^{2} - 2 y^{2}$ (metres). The wind pushes with force field $F = \nabla f$ . Compute the work done by the wind as the hiker walks from the summit-adjacent point $(1, 1)$ to the trailhead $(4, 3)$ along a switchback trail. Interpret the sign.

Recall Solution 5.2

Step 1 — FTLI. Work $= \int_{C} \nabla f \cdot d r = f (4, 3) - f (1, 1)$ , path ignored.

$f (4, 3) = 100 - 16 - 2 \cdot 9 = 100 - 16 - 18 = 66$ .
$f (1, 1) = 100 - 1 - 2 = 97$ . $Work = 66 - 97 = - 31 J (per unit) .$ Interpretation: the hiker moved to lower altitude ( $97 \to 66$ ), i.e. downhill. The field $\nabla f$ points uphill (toward increasing $f$ ), so moving against it costs the field negative work. The switchbacks are irrelevant — only the $31$ -metre net descent matters.

Exercise 5.3

Show that for $F = (P, Q, R)$ in 3D, being conservative forces $\nabla \times F = 0$ (zero curl). Then verify on $F = (y z, x z, x y)$ from 4.1.

Recall Solution 5.3

Why curl must vanish. If $F = \nabla f$ then $P = f_{x}, Q = f_{y}, R = f_{z}$ . The curl's first component is $(\nabla \times F)_{x} = R_{y} - Q_{z} = f_{z y} - f_{y z} = 0,$ because mixed second partials commute (Clairaut). The same cancellation gives the other two components: $Q_{x} - P_{y} = f_{y x} - f_{x y} = 0, P_{z} - R_{x} = f_{x z} - f_{z x} = 0.$ So $\nabla \times F = 0$ . See Curl — vanishing curl is the 3D conservativeness test (on simply connected domains). Verify on $F = (y z, x z, x y)$ :

$R_{y} - Q_{z} = x - x = 0$ .
$P_{z} - R_{x} = y - y = 0$ .
$Q_{x} - P_{y} = z - z = 0$ . $\nabla \times F = (0, 0, 0) . ✓$ Consistent with 4.1, where a potential $f = x y z$ was found.

Connections

Fundamental theorem for line integrals — the parent theorem these drills exercise.
Conservative vector fields — the "does FTLI apply?" question of L2–L3.
Potential functions — the $f$ you reconstruct in L2 and L4.
Line integrals of vector fields — the direct method used in L3 when FTLI fails.
Green's theorem — explains the $2 π$ vortex anomaly of 3.3.
Curl — the 3D test proved in 5.3.
Gradient — the operator at the heart of every problem here.