Unification — all three theorems as generalized Stokes

4.4.34 · Maths › Multivariable Calculus

WHY karta hai ek theorem sabko rule?

WHAT hain woh chaar "costumes"?

Dim	$ω$ hai ek...	$d ω$ ban jaata hai	Classical naam
1	$0$ -form $f$	$f^{'} d x$	Fundamental Theorem of Calculus
2	$1$ -form $P d x + Q d y$	$(Q_{x} - P_{y}) d x \land d y$	Green's theorem
2 in 3D	$1$ -form (work form of $F$ )	curl form	Stokes' (curl) theorem
3	$2$ -form (flux form of $F$ )	divergence form	Divergence (Gauss) theorem

HOW karte hain hum har classical theorem recover? (scratch se derive karo)

Step 0 — Tools jo chahiye

Ek differential form woh cheez hai jise tum integrate karte ho. Wedge $\land$ antisymmetric hai: $d x \land d y = - d y \land d x$ , isliye $d x \land d x = 0$ . Exterior derivative kuch is tarah act karta hai: $d (f) = \sum_{i} \frac{\partial f}{\partial x _{i}} d x_{i}, d (α \land β) = d α \land β + (- 1)^{d e g α} α \land d β .$

Ye rules kyun? $d$ ko linear hona chahiye, functions par ordinary gradient se agree karna chahiye, aur $\land$ ki antisymmetry "orientation" (signed area/volume) encode karti hai.

Costume 1: FTC ( $k = 1$ )

Lo $M = [a, b]$ , $ω = f$ ek $0$ -form. Tab $\partial M = {b} - {a}$ (boundary points, orientation ke hisaab se signed), aur $d ω = f^{'} d x$ . $\int_{\partial M} f = f (b) - f (a), \int_{M} d ω = \int_{a}^{b} f^{'} (x) d x .$ Ye step kyun? Ek $0$ -form ko points par integrate karna bas use evaluate karta hai; orientation $a$ par minus sign deta hai. Stokes ⇒ FTC. ✅

Costume 2: Green's theorem ( $k = 2$ , flat)

Lo $ω = P d x + Q d y$ . Compute karo $d ω$ : $d ω = d P \land d x + d Q \land d y = (P_{x} d x + P_{y} d y) \land d x + (Q_{x} d x + Q_{y} d y) \land d y .$ $d x \land d x = 0$ , $d y \land d y = 0$ kill karo, aur $d y \land d x = - d x \land d y$ use karo: $d ω = (Q_{x} - P_{y}) d x \land d y .$ Ye step kyun? Antisymmetry chaar terms ko ek mein collapse kar deti hai — exactly curl-in-2D. Toh $\oint_{\partial M} (P d x + Q d y) = \iint_{M} (Q_{x} - P_{y}) d x d y . ✅$

Costume 3: Divergence theorem ( $k = 3$ )

Lo $F = (F_{1}, F_{2}, F_{3})$ ka flux 2-form: $ω = F_{1} d y \land d z + F_{2} d z \land d x + F_{3} d x \land d y .$ Tab $d ω = (\partial_{x} F_{1} + \partial_{y} F_{2} + \partial_{z} F_{3}) d x \land d y \land d z = (\nabla \cdot F) d V .$ Ye step kyun? Har term tabhi contribute karta hai jab $d$ missing variable ko hit kare (baaki wedge karke zero ho jaate hain), producing the divergence. Aur $\int_{\partial M} ω = \iint_{\partial M} F \cdot d S$ . Toh $\iint_{\partial M} F \cdot d S = ∭_{M} \nabla \cdot F d V . ✅$

Figure — Unification — all three theorems as generalized Stokes

Sabse gehri "why": $d^{2} = 0$ aur $\partial^{2} = 0$

Worked unification example

Common mistakes (steel-manned)

Flashcards

Generalized Stokes ek equation mein

\int_{\partial M} ω = \int_{M} d ω

Yahan

\partial M

ka matlab kya hai

Manifold

M

ki (oriented) boundary

Kaun sa classical theorem

k = 1

case hai

Fundamental Theorem of Calculus

ω = P d x + Q d y

ke liye,

d ω

kya hai

(Q_{x} - P_{y}) d x \land d y

→ Green's theorem

Divergence theorem kaun sa form deta hai

Flux 2-form

F_{1} d y \land d z + F_{2} d z \land d x + F_{3} d x \land d y

, jiska

d

hai

(\nabla \cdot F) d V

d x \land d x = 0

kyun hai

Wedge antisymmetric hai, toh

α \land α = - α \land α = 0

d^{2} = 0

ko words mein bolo

Exterior derivative do baar apply karne par zero milta hai (boundaries ki koi boundary nahi hoti)

d^{2} = 0

kaun si do vector identities se correspond karta hai

\nabla \times \nabla f = 0

aur

\nabla \cdot (\nabla \times F) = 0

Cell boundaries par sum karte waqt interior walls kyun cancel ho jaati hain

Unhe opposite orientation ke saath do baar traverse kiya jaata hai, toh contributions cancel ho jaate hain; sirf outer boundary bachti hai

Recall Feynman: ek 12-saal ke bachche ko samjhao

Socho tum jaanna chahte ho ki ek band fence ke andar kitni total "cheez" ban rahi hai. Ginaane ke do tarike hain: (1) sirf fence ke saath chalo aur add karo jo use cross kare, ya (2) andar khado aur add karo jo har chota patch banata hai. Dono ek hi number dete hain! Stokes' theorem bas yahi kehta hai: edge par jo hota hai = andar jo hota hai uska total. Agar andar ko tiny squares mein kaato, toh padosi squares ke beech ki shared fences cancel ho jaati hain (ek padosi ka "in" doosre ka "out" hai), aur sirf outer fence bachti hai. Fundamental Theorem of Calculus, Green's, aur Divergence theorem sab yahi exact fence-vs-inside trick hai, bas 1D, 2D, aur 3D mein.

Connections

Green's Theorem — $k = 2$ flat case
Stokes' Theorem (curl) — 3D mein surface
Divergence Theorem — $k = 3$ case
Fundamental Theorem of Calculus — $k = 1$ case
Differential Forms and the Wedge Product
Exterior Derivative aur d squared equals zero
Orientation and Induced Boundary Orientation