Fundamental theorem for line integrals

4.4.28 · Maths › Multivariable Calculus

YEH HAI KYA?

Key words cloze karne ke liye:

$F = \nabla f$ matlab $F$ conservative hai.
Integral sirf endpoints pe depend karta hai, path pe nahi.
Ise path independence kehte hain.

YEH SACH KYU HAI? (Derivation from scratch)

Hum ise sirf chain rule aur ordinary FTC se derive karte hain. Kuch yaad nahi karna.

Step 1 — Line integral ko 1D integral ki tarah likho. Ek parametrised curve pe line integral ki definition se, $\int_{C} \nabla f \cdot d r = \int_{a}^{b} \nabla f (r (t)) \cdot r^{'} (t) d t .$ Yeh step kyun? Line integral matlab "field ko tiny displacement $d r = r^{'} (t) d t$ se dot karo", curve ke saath sum karo. Yeh sirf parametrisation substitute karta hai.

Step 2 — Multivariable chain rule pehchano. Maano $g (t) = f (r (t))$ , ek variable ka plain function. Chain rule kehta hai $\frac{d g}{d t} = \nabla f (r (t)) \cdot r^{'} (t) .$ Yeh step kyun? Jab $t$ change hota hai, $f$ har coordinate ke through change hota hai: $\frac{d}{d t} f (x (t), y (t), z (t)) = f_{x} x^{'} + f_{y} y^{'} + f_{z} z^{'}$ , jo exactly dot product $\nabla f \cdot r^{'}$ hai. Yeh proof ka dil hai — integrand secretly ek perfect derivative hai.

Step 3 — Ordinary FTC apply karo. $\int_{a}^{b} \frac{d g}{d t} d t = g (b) - g (a) .$ Yeh step kyun? Jab hum dikh chuke hain ki integrand $g^{'} (t)$ hai, single-variable FTC kaam khatam kar deta hai.

Step 4 — Substitute back karo. $g (b) - g (a) = f (r (b)) - f (r (a)) = f (B) - f (A) . ■$

Figure — Fundamental theorem for line integrals

KAISE USE KARTE HAIN (worked examples)

Worked example Example 2 — Pehle potential dhundho

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

Step 1 — Conservativeness test karo. 2D mein, $F = (P, Q)$ conservative hai (simply connected region pe) iff $\partial P / \partial y = \partial Q / \partial x$ . Yahan $P_{y} = 2 x$ , $Q_{x} = 2 x$ . ✓ Equal hain, toh potential exist karta hai. Kyun? Agar potential $f$ exist karta hai, toh $P = f_{x}, Q = f_{y}$ , aur mixed partials commute karte hain: $f_{x y} = f_{y x}$ . Step 2 — $f$ banao. $f_{x} = 2 x y$ se, $x$ mein integrate karo: $f = x^{2} y + h (y)$ . Kyun? $x$ -component ko "antidifferentiate" karo; integration ka constant $y$ pe depend kar sakta hai. Step 3 — $y$ -component match karo. $f_{y} = x^{2} + h^{'} (y)$ ko $Q = x^{2} + 2 y$ ke barabar hona chahiye, toh $h^{'} (y) = 2 y \Rightarrow h (y) = y^{2}$ . Is tarah $f = x^{2} y + y^{2}$ . Step 4 — FTLI apply karo. $f (1, 2) = 1 \cdot 2 + 4 = 6$ , $f (0, 0) = 0$ . Answer $= 6 - 0 = 6$ .

Common mistakes

Recall Feynman: ek 12-saal ke bacche ko samjhao

Socho tum ek pahaad pe hiking kar rahe ho. Tumhari "height" function $f$ hai. Chalna elevation add aur subtract karta hai jab tum upar neeche jaate ho. Vector field $\nabla f$ har jagah "steepest slope arrow" ki tarah hai. Fundamental Theorem kehta hai: chahe tum neeche ke cabin se upar ke lodge tak kitna bhi ghoomdar raasta lo, total elevation gained sirf (lodge ki height) − (cabin ki height) hai. Ghoomdar path matter nahi karta — sirf wahan matter karta hai jahan tum shuru aur khatam hote ho. Aur agar tum apne cabin pe wapas loop karo, tumhara net elevation change exactly zero hai.

Active-recall flashcards

Fundamental Theorem for Line Integrals kya kehta hai?

Agar

F = \nabla f

aur

\nabla f

curve

C

A

B

tak continuous hai, toh

\int_{C} \nabla f \cdot d r = f (B) - f (A)

FTLI ke proof mein kaun se do ingredients use hote hain?

Multivariable chain rule (

\frac{d}{d t} f (r (t)) = \nabla f \cdot r^{'}

) aur ordinary single-variable FTC.

Woh field jiska line integral sirf endpoints pe depend karta hai use kya kehte hain?

Conservative (yeh kisi potential

f

ke liye

\nabla f

ke barabar hota hai); is property ko path independence kehte hain.

Kisi closed loop ke around conservative field ka line integral kya hota hai?

Zero, kyunki start aur end points ek hi hain toh

f (B) - f (A) = 0

2D mein

F = (P, Q)

conservative hai ya nahi check karne ka test?

Simply connected region pe,

\partial P / \partial y = \partial Q / \partial x

Akela

P_{y} = Q_{x}

hamesha kyun sufficient nahi hota?

Domain simply connected hona chahiye; hole wale region pe (e.g. origin ke around vortex field) closed-loop integral nonzero ho sakta hai.

F = (P, Q)

se potential

f

kaise reconstruct karte hain?

P

x

mein integrate karo

f = \int P d x + h (y)

pane ke liye, phir

y

mein differentiate karo aur

f_{y} = Q

match karo

h (y)

dhundhne ke liye.

Connections

Line integrals of vector fields — general case jiska FTLI specialisation hai.
Conservative vector fields — fields jahan FTLI apply hota hai.
Potential functions — $\nabla f$ mein woh $f$ .
Green's theorem — simply-connected subtlety ko circulation ke through explain karta hai.
Fundamental Theorem of Calculus — 1D ancestor.
Gradient — operator $\nabla$ jo sab ke centre mein hai.
Curl — vanishing curl 3D conservativeness test hai.