Fundamental theorem for line integrals: Worked examples

Before we start, one symbol we lean on constantly:

The scenario matrix

Every case class this topic can throw at you, and which example covers it:

Cell	Scenario	Covered by
A	Gradient given directly, endpoints in different quadrants (signs matter)	Example 1
B	Field given, must build the potential first (2D)	Example 2
C	Degenerate: closed loop / zero displacement ( $A = B$ )	Example 3
D	Field is NOT conservative — FTLI forbidden, path genuinely matters	Example 4
E	$P_{y} = Q_{x}$ but domain has a hole — the vortex trap	Example 5
F	3D field, reconstruct potential in three variables	Example 6
G	Real-world word problem (work / elevation)	Example 7
H	Exam twist: same endpoints reached two different ways to prove path independence	Example 8

We now hit every cell.

Cell A — gradient given, endpoints across quadrants

Worked example Example 1

Let $f (x, y) = x^{2} - y^{2}$ . Compute $\int_{C} \nabla f \cdot d r$ where $C$ runs from $A = (- 2, 1)$ (quadrant II) to $B = (3, - 4)$ (quadrant IV), along any path.

Forecast: guess whether the answer is positive or negative before reading on.

Step 1. Recognise the integrand is literally $\nabla f$ , so FTLI applies and the path is irrelevant. Why this step? FTLI's only requirement is that $F = \nabla f$ with $\nabla f$ continuous — here $f$ is a polynomial, smooth everywhere, so we may throw the path away. Step 2. Evaluate at $B = (3, - 4)$ : $f (3, - 4) = 3^{2} - (- 4)^{2} = 9 - 16 = - 7$ . Why this step? FTLI says the answer is $f (B) - f (A)$ ; we need each endpoint value. Note the signs: $x$ is squared so $x = 3$ and $x = - 2$ both contribute positively, but the $- y^{2}$ term is always subtracting. Step 3. Evaluate at $A = (- 2, 1)$ : $f (- 2, 1) = (- 2)^{2} - 1^{2} = 4 - 1 = 3$ . Why this step? Same reason — the quadrant of $A$ only affects things through the squares, which erase the sign of each coordinate. Step 4. Subtract: $\int_{C} \nabla f \cdot d r = - 7 - 3 = - 10$ . Why this step? "Gradient In, Endpoints Out" — $f (end) - f (start)$ .

Verify: parametrise the straight segment $r (t) = (- 2 + 5 t, 1 - 5 t)$ , $t \in [0, 1]$ , and grind the full line integral — it must also give $- 10$ (done in VERIFY). Sign sanity: we ended lower on the " $f$ -surface" than we started ( $- 7 < 3$ ), so a negative total is exactly right.

Figure — Fundamental theorem for line integrals

Cell B — build the potential first

Worked example Example 2

Evaluate $\int_{C} F \cdot d r$ where $F = (3 x^{2} y + y^{3}, x^{3} + 3 x y^{2})$ , from $A = (1, 0)$ to $B = (2, 1)$ .

Forecast: is this field even conservative? Guess yes/no.

Step 1 — Test conservativeness. Write $F = (P, Q)$ with $P = 3 x^{2} y + y^{3}$ , $Q = x^{3} + 3 x y^{2}$ . Compute $P_{y} = 3 x^{2} + 3 y^{2}$ and $Q_{x} = 3 x^{2} + 3 y^{2}$ . Why this step? A gradient field must satisfy $f_{x y} = f_{y x}$ (mixed partials commute), which forces $P_{y} = Q_{x}$ . They match, so a potential exists — see Conservative vector fields. Step 2 — Antidifferentiate $P$ in $x$ . $f = \int (3 x^{2} y + y^{3}) d x = x^{3} y + x y^{3} + h (y)$ . Why this step? We want $f$ with $f_{x} = P$ . Integrating $P$ over $x$ recovers $f$ up to a term $h (y)$ that vanishes under $\partial_{x}$ — so the "constant" may depend on $y$ . See Potential functions. Step 3 — Match the $y$ -component. Differentiate: $f_{y} = x^{3} + 3 x y^{2} + h^{'} (y)$ . Set equal to $Q = x^{3} + 3 x y^{2}$ . So $h^{'} (y) = 0 \Rightarrow h (y) = const$ ; take $h = 0$ . Thus $f = x^{3} y + x y^{3}$ . Why this step? $f$ must also satisfy $f_{y} = Q$ ; matching pins down the unknown $h$ . Step 4 — Apply FTLI. $f (2, 1) = 2^{3} \cdot 1 + 2 \cdot 1^{3} = 8 + 2 = 10$ ; $f (1, 0) = 1^{3} \cdot 0 + 1 \cdot 0 = 0$ . Answer $= 10 - 0 = 10$ .

Verify: $\nabla f = (3 x^{2} y + y^{3}, x^{3} + 3 x y^{2}) = F$ ✓ (checked in VERIFY), and $f (B) - f (A) = 10$ .

Cell C — degenerate closed loop

Cell D — field is NOT a gradient (FTLI forbidden)

Worked example Example 4

Consider $F = (- y, x)$ from $A = (1, 0)$ to $B = (- 1, 0)$ , along two paths: the upper semicircle $C_{1}$ and the lower semicircle $C_{2}$ of the unit circle. Show the integrals differ, so no potential can exist.

Forecast: if the answers disagree, what does that prove about $F$ ?

Step 1 — Conservativeness test. $P = - y$ , $Q = x$ . Then $P_{y} = - 1$ , $Q_{x} = + 1$ . Since $- 1 \neq = 1$ , $F$ is not conservative. Why this step? FTLI is only allowed for gradients; failing $P_{y} = Q_{x}$ means no $f$ exists and the path will matter — see the Curl test ( $curl = Q_{x} - P_{y} = 2 \neq = 0$ ). Step 2 — Upper path $C_{1}$ . Parametrise $r (t) = (cos t, sin t)$ , $t : 0 \to π$ . Then $r^{'} = (- sin t, cos t)$ and $F (r) = (- sin t, cos t)$ . Dot: $F \cdot r^{'} = sin^{2} t + cos^{2} t = 1$ . Why this step? With no potential we must fall back to the direct definition — substitute the parametrisation and integrate. $\int_{C_{1}} = \int_{0}^{π} 1 d t = π$ . Step 3 — Lower path $C_{2}$ . Go clockwise from $(1, 0)$ to $(- 1, 0)$ : $r (t) = (cos t, - sin t)$ , $t : 0 \to π$ . Then $r^{'} = (- sin t, - cos t)$ , $F (r) = (sin t, cos t)$ . Dot: $- sin^{2} t - cos^{2} t = - 1$ . So $\int_{C_{2}} = \int_{0}^{π} (- 1) d t = - π$ . Step 4 — Compare. $π \neq = - π$ . Two paths, same endpoints, different answers. Why this step? This is exactly the failure FTLI protects you from claiming — a non-gradient field is path-dependent.

Verify: $\int_{C_{1}} = π \approx 3.1416$ , $\int_{C_{2}} = - π$ ; difference $2 π$ . VERIFY confirms both. (This is the Green's theorem circulation of the rotation field.)

Cell E — $P_{y} = Q_{x}$ but the domain has a hole

Worked example Example 5

The vortex field $F = (\frac{- y}{x ^{2} + y ^{2}}, \frac{x}{x ^{2} + y ^{2}})$ . Compute $\oint_{C} F \cdot d r$ around the unit circle, then explain why FTLI does not force it to be $0$ .

Forecast: guess whether it's $0$ or $2 π$ — and why the "test" lies here.

Step 1 — Run the test. With $P = \frac{- y}{x ^{2} + y ^{2}}$ , $Q = \frac{x}{x ^{2} + y ^{2}}$ , both partials work out to $P_{y} = Q_{x} = \frac{y ^{2} - x ^{2}}{( x ^{2} + y ^{2} ) ^{2}}$ everywhere the field is defined. Why this step? Naively this says "conservative", tempting you to conclude $\oint = 0$ . The catch: $F$ is undefined at the origin, a hole the loop encircles. Step 2 — Integrate directly. On the unit circle $r (t) = (cos t, sin t)$ , $t : 0 \to 2 π$ : $x^{2} + y^{2} = 1$ , so $F = (- sin t, cos t)$ and $r^{'} = (- sin t, cos t)$ . Dot: $sin^{2} t + cos^{2} t = 1$ . Why this step? The test can't be trusted on a punctured domain, so compute from the definition. $\oint_{C} = \int_{0}^{2 π} 1 d t = 2 π$ . Step 3 — Diagnose. $2 π \neq = 0$ , so no single-valued potential exists on the punctured plane, even though $P_{y} = Q_{x}$ pointwise. Why this step? FTLI (and its corollary "closed loops vanish") needs the field defined on a simply connected region — no holes. Here the origin is a hole, so the theorem simply doesn't apply. See Green's theorem.

Verify: $\oint = 2 π \approx 6.2832$ , checked in VERIFY. Contrast with Cell C's honest zero.

Cell F — 3D potential reconstruction

Worked example Example 6

$F = (y z, x z, x y)$ in $R^{3}$ , from $A = (1, 1, 1)$ to $B = (2, 3, 4)$ . Find $\int_{C} F \cdot d r$ .

Forecast: guess the potential $f$ from the pattern of the components.

Step 1 — 3D conservativeness (curl $= 0$ ). A 3D field is a gradient iff its Curl vanishes. Check $\partial_{y} (y z) = z = \partial_{x} (x z)$ , $\partial_{z} (x z) = x = \partial_{y} (x y)$ , $\partial_{z} (y z) = y = \partial_{x} (x y)$ . All three pairs match — curl is $0$ . Why this step? In 3D the 2D " $P_{y} = Q_{x}$ " grows into three equalities; all must hold. Step 2 — Antidifferentiate the first component in $x$ . $f = \int y z d x = x y z + g (y, z)$ . Why this step? Same recipe as 2D, now the leftover $g$ may depend on the two remaining variables. Step 3 — Match $f_{y}$ and $f_{z}$ . $f_{y} = x z + g_{y} = x z \Rightarrow g_{y} = 0$ . $f_{z} = x y + g_{z} = x y \Rightarrow g_{z} = 0$ . So $g$ is constant; take $f = x y z$ . Why this step? $f$ must reproduce all three components of $F$ ; matching pins $g$ . Step 4 — Apply FTLI. $f (2, 3, 4) = 2 \cdot 3 \cdot 4 = 24$ , $f (1, 1, 1) = 1$ . Answer $= 24 - 1 = 23$ .

Verify: $\nabla (x y z) = (y z, x z, x y) = F$ ✓ and $24 - 1 = 23$ — checked in VERIFY.

Cell G — real-world word problem

Worked example Example 7

A hiker's elevation on a hillside is $f (x, y) = 100 - 2 x^{2} - 3 y^{2}$ metres (with $x, y$ in km). The wind assists them with force field $F = \nabla f = (- 4 x, - 6 y)$ newtons-per-unit (a "gravity/slope" style field). The hiker walks from the trailhead $A = (3, 2)$ to the summit-side lodge $B = (0, 1)$ along a winding trail. What is the work $W = \int_{C} F \cdot d r$ done by this field, and how does it relate to elevation?

Forecast: should the work be positive (field helps them climb) or negative?

Step 1 — The field is a gradient, so use FTLI. $F = \nabla f$ is given, path irrelevant. Why this step? Work by a conservative field equals the change in potential — the field bookkeeps only endpoints. Step 2 — Endpoint values. $f (3, 2) = 100 - 2 \cdot 9 - 3 \cdot 4 = 100 - 18 - 12 = 70$ m. $f (0, 1) = 100 - 0 - 3 = 97$ m. Why this step? $W = f (B) - f (A)$ ; we evaluate $f$ (the "elevation potential") at both ends. Step 3 — Work. $W = f (0, 1) - f (3, 2) = 97 - 70 = 27$ . Why this step? FTLI. The units: $f$ is in metres of "potential"; the work equals the rise in potential value, $+ 27$ (a gain), independent of the winding trail.

Verify: the hiker climbs from elevation $70$ m to $97$ m — a net rise of $27$ m — matching $W = + 27$ . Because it's a loop-free conservative walk, only start/end matter. Checked in VERIFY.

Cell H — exam twist: prove path independence by computing two ways

Worked example Example 8

$F = (2 x + y, x + 2 y)$ from $A = (0, 0)$ to $B = (1, 1)$ . Compute the integral two ways — via FTLI and via the straight-line parametrisation — and show they agree, demonstrating path independence.

Forecast: guess the single number both methods must produce.

Step 1 — Confirm conservative & find $f$ . $P = 2 x + y$ , $Q = x + 2 y$ : $P_{y} = 1 = Q_{x}$ ✓. Integrate $P$ in $x$ : $f = x^{2} + x y + h (y)$ ; then $f_{y} = x + h^{'} (y) = x + 2 y \Rightarrow h^{'} (y) = 2 y \Rightarrow h = y^{2}$ . So $f = x^{2} + x y + y^{2}$ . Why this step? Build the potential exactly as in Cell B. Step 2 — FTLI route. $f (1, 1) = 1 + 1 + 1 = 3$ , $f (0, 0) = 0$ . So $\int_{C} F \cdot d r = 3$ . Why this step? This is the "30-second" method the parent note advertises. Step 3 — Direct route (the exam's demand). Line $r (t) = (t, t)$ , $t : 0 \to 1$ , $r^{'} = (1, 1)$ . Then $F (r) = (2 t + t, t + 2 t) = (3 t, 3 t)$ , dot $r^{'}$ gives $6 t$ . Integrate: $\int_{0}^{1} 6 t d t = 3 t^{2}_{0}^{1} = 3$ . Why this step? Grinding the definition must reproduce the FTLI value — that agreement is path independence in action. Step 4 — Compare. Both give $3$ . ✓

Verify: FTLI $= 3$ , direct $= 3$ , identical — checked in VERIFY.

Recall One-line summary of the whole matrix

Gradient given ::: subtract endpoint values (Cell A) Field only ::: build $f$ , then subtract endpoints (Cell B, F, H) Closed loop, no holes ::: answer is $0$ (Cell C) $P_{y} \neq = Q_{x}$ ::: not conservative, path matters, integrate directly (Cell D) $P_{y} = Q_{x}$ but a hole inside ::: FTLI fails, loop may be nonzero (Cell E) Word problem ::: work $=$ change in potential $= f (B) - f (A)$ (Cell G)

Connections

Fundamental theorem for line integrals — the parent rule these examples exercise.
Line integrals of vector fields — the direct-parametrisation fallback (Cells D, E, H).
Conservative vector fields — the fields where Cells A–C, F–H live.
Potential functions — the $f$ we reconstruct in B, F, H.
Green's theorem — explains the hole in Cell E and circulation in Cell D.
Fundamental Theorem of Calculus — the 1D engine behind every subtraction.
Gradient — the $\nabla$ producing every conservative field here.
Curl — the 3D conservativeness test used in Cell F and diagnosing Cell D.