Fundamental theorem for line integrals: Worked examples

4.4.28 · D3 · Maths › Multivariable Calculus › Fundamental theorem for line integrals

Shuru karne se pehle, ek symbol jo hum baar baar use karte hain:

The scenario matrix

Is topic mein jo bhi case class aa sakti hai, aur kaun sa example use cover karta hai:

Cell	Scenario	Covered by
A	Gradient directly diya gaya, endpoints alag-alag quadrants mein (signs matter karte hain)	Example 1
B	Field diya gaya, pehle potential banana padega (2D)	Example 2
C	Degenerate: closed loop / zero displacement ( $A = B$ )	Example 3
D	Field NOT conservative hai — FTLI forbidden, path genuinely matter karta hai	Example 4
E	$P_{y} = Q_{x}$ hai lekin domain mein hole hai — vortex trap	Example 5
F	3D field, teen variables mein potential reconstruct karo	Example 6
G	Real-world word problem (work / elevation)	Example 7
H	Exam twist: same endpoints do alag tareekon se reach karo taaki path independence prove ho sake	Example 8

Ab hum har cell hit karte hain.

Cell A — gradient diya gaya hai, endpoints alag quadrants mein

Worked example Example 1

Maano $f (x, y) = x^{2} - y^{2}$ . Compute karo $\int_{C} \nabla f \cdot d r$ jahan $C$ , $A = (- 2, 1)$ (quadrant II) se $B = (3, - 4)$ (quadrant IV) tak, kisi bhi path ke saath jaata hai.

Forecast: aage padhne se pehle andaza lagao ki answer positive hoga ya negative.

Step 1. Pehchano ki integrand literally $\nabla f$ hai, isliye FTLI apply hoti hai aur path irrelevant hai. Ye step kyun? FTLI ki sirf yahi requirement hai ki $F = \nabla f$ ho aur $\nabla f$ continuous ho — yahan $f$ ek polynomial hai, smooth everywhere, isliye hum path ko ignore kar sakte hain. Step 2. $B = (3, - 4)$ par evaluate karo: $f (3, - 4) = 3^{2} - (- 4)^{2} = 9 - 16 = - 7$ . Ye step kyun? FTLI kehta hai answer $f (B) - f (A)$ hai; hume har endpoint ki value chahiye. Signs dhyan se: $x$ squared hai isliye $x = 3$ aur $x = - 2$ dono positively contribute karte hain, lekin $- y^{2}$ term hamesha subtract kar raha hai. Step 3. $A = (- 2, 1)$ par evaluate karo: $f (- 2, 1) = (- 2)^{2} - 1^{2} = 4 - 1 = 3$ . Ye step kyun? Same wajah — $A$ ka quadrant sirf squares ke through effect karta hai, jo har coordinate ka sign mita dete hain. Step 4. Subtract karo: $\int_{C} \nabla f \cdot d r = - 7 - 3 = - 10$ . Ye step kyun? "Gradient In, Endpoints Out" — $f (end) - f (start)$ .

Verify: straight segment ko parametrise karo $r (t) = (- 2 + 5 t, 1 - 5 t)$ , $t \in [0, 1]$ , aur poora line integral grind karo — isko bhi $- 10$ dena chahiye (VERIFY mein kiya gaya hai). Sign sanity: hum start se zyada neeche end hue " $f$ -surface" par ( $- 7 < 3$ ), isliye negative total bilkul sahi hai.

Figure — Fundamental theorem for line integrals

Cell B — pehle potential banao

Worked example Example 2

Evaluate karo $\int_{C} F \cdot d r$ jahan $F = (3 x^{2} y + y^{3}, x^{3} + 3 x y^{2})$ , $A = (1, 0)$ se $B = (2, 1)$ tak.

Forecast: kya ye field conservative bhi hai? Haan/nahi guess karo.

Step 1 — Conservativeness test. $F = (P, Q)$ likho jahan $P = 3 x^{2} y + y^{3}$ , $Q = x^{3} + 3 x y^{2}$ . Compute karo $P_{y} = 3 x^{2} + 3 y^{2}$ aur $Q_{x} = 3 x^{2} + 3 y^{2}$ . Ye step kyun? Ek gradient field ko $f_{x y} = f_{y x}$ satisfy karna hoga (mixed partials commute karte hain), jo force karta hai $P_{y} = Q_{x}$ . Ye match karte hain, isliye potential exist karta hai — dekho Conservative vector fields. Step 2 — $P$ ko $x$ mein antidifferentiate karo. $f = \int (3 x^{2} y + y^{3}) d x = x^{3} y + x y^{3} + h (y)$ . Ye step kyun? Hume $f$ chahiye jiska $f_{x} = P$ ho. $P$ ko $x$ par integrate karne se $f$ milta hai ek $h (y)$ term tak, jo $\partial_{x}$ ke under vanish ho jaata hai — isliye "constant" $y$ par depend kar sakta hai. Dekho Potential functions. Step 3 — $y$ -component match karo. Differentiate karo: $f_{y} = x^{3} + 3 x y^{2} + h^{'} (y)$ . $Q = x^{3} + 3 x y^{2}$ ke equal set karo. To $h^{'} (y) = 0 \Rightarrow h (y) = const$ ; $h = 0$ lo. Isliye $f = x^{3} y + x y^{3}$ . Ye step kyun? $f$ ko $f_{y} = Q$ bhi satisfy karna hoga; match karne se unknown $h$ pin ho jaata hai. Step 4 — FTLI apply karo. $f (2, 1) = 2^{3} \cdot 1 + 2 \cdot 1^{3} = 8 + 2 = 10$ ; $f (1, 0) = 1^{3} \cdot 0 + 1 \cdot 0 = 0$ . Answer $= 10 - 0 = 10$ .

Verify: $\nabla f = (3 x^{2} y + y^{3}, x^{3} + 3 x y^{2}) = F$ ✓ (VERIFY mein check kiya gaya), aur $f (B) - f (A) = 10$ .

Cell C — degenerate closed loop

Worked example Example 3

$F = \nabla (e^{x} cos y)$ . Compute karo $\oint_{C} F \cdot d r$ jahan $C$ ellipse $\frac{x ^{2}}{9} + \frac{y ^{2}}{4} = 1$ hai jo ek baar traverse hoti hai, $(3, 0)$ se shuru aur khatam hoti hai.

Forecast: answer ek number mein guess karo.

Step 1. Potential identify karo: $f (x, y) = e^{x} cos y$ , aur $F = \nabla f$ diya gaya hai. Ye step kyun? FTLI ko $F = \nabla f$ chahiye; yahan ye seedha diya gaya hai. Step 2. Note karo ki loop closed hai: start point aur end point same hain, $A = B = (3, 0)$ . Ye step kyun? Closed curve ke liye dono endpoints coincide karte hain, isliye FTLI deta hai $f (B) - f (A) = f (A) - f (A)$ . Step 3. Isliye $\oint_{C} F \cdot d r = f (3, 0) - f (3, 0) = 0$ . Ye step kyun? Koi bhi cheez apne aap se minus hone par $0$ hoti hai — ye degenerate/zero-displacement case hai.

Verify: $f (3, 0) = e^{3} cos 0 = e^{3} \approx 20.09$ , aur $20.09 - 20.09 = 0$ . Domain poora $R^{2}$ hai (koi holes nahi), isliye Cell E ke unlike ye zero genuine hai. VERIFY mein check kiya gaya.

Cell D — field gradient nahi hai (FTLI forbidden)

Worked example Example 4

Maano $F = (- y, x)$ , $A = (1, 0)$ se $B = (- 1, 0)$ tak, do paths ke saath: upper semicircle $C_{1}$ aur unit circle ka lower semicircle $C_{2}$ . Dikhao ki integrals differ karte hain, isliye koi potential exist nahi kar sakta.

Forecast: agar answers disagree karte hain, to ye $F$ ke baare mein kya prove karta hai?

Step 1 — Conservativeness test. $P = - y$ , $Q = x$ . To $P_{y} = - 1$ , $Q_{x} = + 1$ . Kyunki $- 1 \neq = 1$ , $F$ conservative nahi hai. Ye step kyun? FTLI sirf gradients ke liye allowed hai; $P_{y} = Q_{x}$ fail karne ka matlab hai koi $f$ exist nahi karta aur path matter karega — dekho Curl test ( $curl = Q_{x} - P_{y} = 2 \neq = 0$ ). Step 2 — Upper path $C_{1}$ . Parametrise karo $r (t) = (cos t, sin t)$ , $t : 0 \to π$ . To $r^{'} = (- sin t, cos t)$ aur $F (r) = (- sin t, cos t)$ . Dot: $F \cdot r^{'} = sin^{2} t + cos^{2} t = 1$ . Ye step kyun? Koi potential nahi hone ki wajah se hume direct definition par wapas aana padega — parametrisation substitute karo aur integrate karo. $\int_{C_{1}} = \int_{0}^{π} 1 d t = π$ . Step 3 — Lower path $C_{2}$ . $(1, 0)$ se $(- 1, 0)$ tak clockwise jao: $r (t) = (cos t, - sin t)$ , $t : 0 \to π$ . To $r^{'} = (- sin t, - cos t)$ , $F (r) = (sin t, cos t)$ . Dot: $- sin^{2} t - cos^{2} t = - 1$ . Isliye $\int_{C_{2}} = \int_{0}^{π} (- 1) d t = - π$ . Step 4 — Compare karo. $π \neq = - π$ . Do paths, same endpoints, alag answers. Ye step kyun? Ye exactly wahi failure hai jis claim se FTLI tumhe protect karta hai — ek non-gradient field path-dependent hota hai.

Verify: $\int_{C_{1}} = π \approx 3.1416$ , $\int_{C_{2}} = - π$ ; difference $2 π$ . VERIFY dono confirm karta hai. (Ye rotation field ka Green's theorem circulation hai.)

Cell E — $P_{y} = Q_{x}$ hai lekin domain mein hole hai

Worked example Example 5

Vortex field $F = (\frac{- y}{x ^{2} + y ^{2}}, \frac{x}{x ^{2} + y ^{2}})$ . Unit circle ke around $\oint_{C} F \cdot d r$ compute karo, phir explain karo ki FTLI ye kyun force nahi karta ki ye $0$ ho.

Forecast: guess karo ki ye $0$ hai ya $2 π$ — aur "test" yahan kyun jhooth bolta hai.

Step 1 — Test chalao. $P = \frac{- y}{x ^{2} + y ^{2}}$ , $Q = \frac{x}{x ^{2} + y ^{2}}$ ke saath, dono partials jahan field defined hai wahan $P_{y} = Q_{x} = \frac{y ^{2} - x ^{2}}{( x ^{2} + y ^{2} ) ^{2}}$ nikalta hai. Ye step kyun? Naive tarike se ye kehta hai "conservative", jo tumhe conclude karne ka laalach deta hai ki $\oint = 0$ . Catch ye hai: $F$ origin par undefined hai, ek hole jo loop encircle karta hai. Step 2 — Directly integrate karo. Unit circle par $r (t) = (cos t, sin t)$ , $t : 0 \to 2 π$ : $x^{2} + y^{2} = 1$ , isliye $F = (- sin t, cos t)$ aur $r^{'} = (- sin t, cos t)$ . Dot: $sin^{2} t + cos^{2} t = 1$ . Ye step kyun? Test ko punctured domain par trust nahi kiya ja sakta, isliye definition se compute karo. $\oint_{C} = \int_{0}^{2 π} 1 d t = 2 π$ . Step 3 — Diagnose karo. $2 π \neq = 0$ , isliye punctured plane par koi single-valued potential exist nahi karta, chahe $P_{y} = Q_{x}$ pointwise ho. Ye step kyun? FTLI (aur iski corollary "closed loops vanish") ko field simply connected region par defined chahiye — koi holes nahi. Yahan origin ek hole hai, isliye theorem simply apply nahi hota. Dekho Green's theorem.

Verify: $\oint = 2 π \approx 6.2832$ , VERIFY mein check kiya gaya. Cell C ke honest zero se compare karo.

Cell F — 3D potential reconstruction

Worked example Example 6

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

Forecast: components ke pattern se potential $f$ guess karo.

Step 1 — 3D conservativeness (curl $= 0$ ). 3D field gradient hoga tabhi jab iski Curl vanish ho. Check karo $\partial_{y} (y z) = z = \partial_{x} (x z)$ , $\partial_{z} (x z) = x = \partial_{y} (x y)$ , $\partial_{z} (y z) = y = \partial_{x} (x y)$ . Teeno pairs match karte hain — curl $0$ hai. Ye step kyun? 3D mein 2D " $P_{y} = Q_{x}$ " teen equalities mein grow ho jaata hai; sab hold karni chahiye. Step 2 — Pehle component ko $x$ mein antidifferentiate karo. $f = \int y z d x = x y z + g (y, z)$ . Ye step kyun? Same recipe jaisi 2D, ab leftover $g$ do remaining variables par depend kar sakta hai. Step 3 — $f_{y}$ aur $f_{z}$ match karo. $f_{y} = x z + g_{y} = x z \Rightarrow g_{y} = 0$ . $f_{z} = x y + g_{z} = x y \Rightarrow g_{z} = 0$ . Isliye $g$ constant hai; $f = x y z$ lo. Ye step kyun? $f$ ko $F$ ke teeno components reproduce karne chahiye; match karne se $g$ pin ho jaata hai. Step 4 — FTLI apply karo. $f (2, 3, 4) = 2 \cdot 3 \cdot 4 = 24$ , $f (1, 1, 1) = 1$ . Answer $= 24 - 1 = 23$ .

Verify: $\nabla (x y z) = (y z, x z, x y) = F$ ✓ aur $24 - 1 = 23$ — VERIFY mein check kiya gaya.

Cell G — real-world word problem

Worked example Example 7

Ek hiker ki hillside par elevation hai $f (x, y) = 100 - 2 x^{2} - 3 y^{2}$ metres (jahan $x, y$ km mein hain). Wind unhe force field $F = \nabla f = (- 4 x, - 6 y)$ newtons-per-unit se help karta hai (ek "gravity/slope" style field). Hiker trailhead $A = (3, 2)$ se summit-side lodge $B = (0, 1)$ tak ek winding trail ke saath chalta hai. Is field dwara kiya gaya work $W = \int_{C} F \cdot d r$ kya hai, aur ye elevation se kaise relate karta hai?

Forecast: work positive hona chahiye (field climb mein help karta hai) ya negative?

Step 1 — Field gradient hai, isliye FTLI use karo. $F = \nabla f$ diya gaya hai, path irrelevant hai. Ye step kyun? Conservative field dwara kiya gaya work potential mein change ke barabar hota hai — field sirf endpoints bookkeep karta hai. 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. Ye step kyun? $W = f (B) - f (A)$ ; hum dono ends par $f$ (jo "elevation potential" hai) evaluate karte hain. Step 3 — Work. $W = f (0, 1) - f (3, 2) = 97 - 70 = 27$ . Ye step kyun? FTLI. Units: $f$ "potential" ke metres mein hai; work potential value mein rise ke barabar hai, $+ 27$ (ek gain), winding trail se independent.

Verify: hiker elevation $70$ m se $97$ m tak climb karta hai — net rise $27$ m — $W = + 27$ se match karta hai. Kyunki ye loop-free conservative walk hai, sirf start/end matter karte hain. VERIFY mein check kiya gaya.

Cell H — exam twist: do tareekon se compute karke path independence prove karo

Worked example Example 8

$F = (2 x + y, x + 2 y)$ , $A = (0, 0)$ se $B = (1, 1)$ tak. Integral do tareekon se compute karo — FTLI se aur straight-line parametrisation se — aur dikhao ki dono agree karte hain, jo path independence demonstrate karta hai.

Forecast: dono methods ko jo single number produce karna chahiye, use guess karo.

Step 1 — Conservative confirm karo & $f$ nikalo. $P = 2 x + y$ , $Q = x + 2 y$ : $P_{y} = 1 = Q_{x}$ ✓. $P$ ko $x$ mein integrate karo: $f = x^{2} + x y + h (y)$ ; phir $f_{y} = x + h^{'} (y) = x + 2 y \Rightarrow h^{'} (y) = 2 y \Rightarrow h = y^{2}$ . Isliye $f = x^{2} + x y + y^{2}$ . Ye step kyun? Potential exactly Cell B ki tarah banao. Step 2 — FTLI route. $f (1, 1) = 1 + 1 + 1 = 3$ , $f (0, 0) = 0$ . Isliye $\int_{C} F \cdot d r = 3$ . Ye step kyun? Ye wahi "30-second" method hai jo parent note advertise karta hai. Step 3 — Direct route (exam ki demand). Line $r (t) = (t, t)$ , $t : 0 \to 1$ , $r^{'} = (1, 1)$ . To $F (r) = (2 t + t, t + 2 t) = (3 t, 3 t)$ , $r^{'}$ se dot karne par $6 t$ milta hai. Integrate karo: $\int_{0}^{1} 6 t d t = 3 t^{2}_{0}^{1} = 3$ . Ye step kyun? Definition ko grind karne par FTLI value reproduce honi chahiye — wahi agreement hi path independence in action hai. Step 4 — Compare karo. Dono $3$ dete hain. ✓

Verify: FTLI $= 3$ , direct $= 3$ , identical — VERIFY mein check kiya gaya.

Recall Poore matrix ka one-line summary

Gradient diya gaya hai ::: endpoint values subtract karo (Cell A) Sirf field diya gaya hai ::: $f$ banao, phir endpoints subtract karo (Cell B, F, H) Closed loop, koi holes nahi ::: answer $0$ hai (Cell C) $P_{y} \neq = Q_{x}$ ::: conservative nahi, path matters, directly integrate karo (Cell D) $P_{y} = Q_{x}$ hai lekin andar hole hai ::: FTLI fails, loop nonzero ho sakta hai (Cell E) Word problem ::: work $=$ potential mein change $= f (B) - f (A)$ (Cell G)

Connections

Fundamental theorem for line integrals — parent rule jise ye examples exercise karte hain.
Line integrals of vector fields — direct-parametrisation fallback (Cells D, E, H).
Conservative vector fields — fields jahan Cells A–C, F–H rehte hain.
Potential functions — woh $f$ jo hum B, F, H mein reconstruct karte hain.
Green's theorem — Cell E mein hole explain karta hai aur Cell D mein circulation.
Fundamental Theorem of Calculus — 1D engine jo har subtraction ke peeche hai.
Gradient — woh $\nabla$ jo yahan har conservative field produce karta hai.
Curl — Cell F mein use hone wala 3D conservativeness test aur Cell D diagnose karne ke liye.