Visual walkthrough — Unification — all three theorems as generalized Stokes

4.4.34 · D2 · Maths › Multivariable Calculus › Unification — all three theorems as generalized Stokes

Hum sirf do bilkul naye pieces of notation use karte hain, aur inhe hum wahan define karte hain jahan woh pehli baar aate hain:

• $\int$ — "ek region pe add up karo" ka sign. Socho bahut lamba S for "Sum".
• $\partial M$ — padho "$M$ ki boundary": region $M$ ka baahri edge. Symbol $\partial$ ek curly d hai; matlab hai "ka edge".

Baki sab kuch hum pictures se build karte hain.

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Step 1 — "Edge ke around add up karna" ka matlab kya hai

KYA: hum left-hand side ${\int }_{\partial M}\omega$ ko sabse simple tarike se define kar rahe hain — edge ke around ek walk, ek push ka sum.

KYUN: pehle hum claim kar sakein "edge equals inside", hume precisely pata hona chahiye ki "edge ke along add up karna" kya hai. Yeh area nahi hai; yeh ek signed walk hai.

PICTURE: neeche, blue loop humara path hai. Chhote coral arrows wind $\mathbf{F}$ hain. Green tick woh direction dikhata hai jisme hum chalte hain (counterclockwise, hamari chosen orientation). Jahan wind aur walk agree karte hain hum positive amount add karte hain; jahan oppose karte hain, negative amount.

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Step 2 — Ek tiny square, uske around walk kiya

KYA: loop ko ek single tiny square mein chhota karo jiske corners $\Delta x$ wide aur $\Delta y$ tall hain. Uske chaar sides counterclockwise chalko: bottom ke along right, right wali side se upar, top ke along left, left wali side se neeche.

KYUN: agar hum ek tiny square ka circulation samajh saken, toh hum kisi bhi region ko tiny squares se tile kar ke unhe add kar sakte hain. Yahi poori strategy hai.

PICTURE: chaar coloured arrows walk ke chaar legs hain. Sign dhyan se padho: bottom leg +x walk karta hai, top leg −x (leftward) walk karta hai, isliye top leg ka contribution minus ke saath aata hai.

Maan lo wind $\mathbf{F}=(P,Q)$ hai, matlab $P$ uska rightward part hai aur $Q$ uska upward part. Chaar legs add karo:

$P$ terms aur $Q$ terms group karo:

$$[Q(x+\Delta x,y)-Q(x,y)]\Delta y-[P(x,y+\Delta y)-P(x,y)]\Delta x.$$

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Step 3 — Kyun ek derivative ka aana zaruri hai

KYA: har bracket hai "yahan value minus ek step dur value". Step size se divide aur multiply karo:

$$\frac{Q(x+\Delta x,y)-Q(x,y)}{\Delta x}\Delta x\Delta y-\frac{P(x,y+\Delta y)-P(x,y)}{\Delta y}\Delta y\Delta x.$$

KYUN derivative, aur koi aur tool nahi? Ek derivative exactly woh machine hai jo jawab deta hai "yeh quantity kitni tezi se change hoti hai jab main ek given direction mein tiny step leta hoon?" Hamare brackets literally wohi pooch rahe hain. Toh derivative koi clever trick nahi hai — yeh woh ek tool hai jiska definition match karta hai "yahan value minus ek step dur value, per unit step". Hum ${\partial }_{x}Q$ likhte hain "$Q$ ka rate of change as $x$ increases" ke liye (curly $\partial$ phir se: change with respect to).

Jaise square chhota hota hai ($\Delta x,\Delta y\to 0$) har ratio ek derivative ban jaata hai:

PICTURE: tiny square mein ek paddlewheel. Right-hand side edge ko left se zyada/kam up-wind feel hoti hai (${\partial }_{x}Q$); top edge ko bottom se zyada/kam right-wind feel hoti hai ($-{\partial }_{y}P$). Unka difference wheel ko spin karta hai.

Toh ek square ke liye: edge-ke-around-walk $=$ (local spin) × (andar ka area). Yeh already puri theorem miniature mein hai.

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Step 4 — Region tile karo, aur walls cancel hoti dekho

KYA: poori region $M$ ko in tiny squares ki grid se cover karo. Har square ka circulation add karo. Yeh equals hai (local spin × area) ka sum sab squares pe — right-hand side.

KYUN: hume outer boundary chahiye, tiny squares ka mess nahi. Jaadu yeh hai ki andar ki walls khud ko delete kar leti hain.

PICTURE: do neighbouring squares ek wall share karte hain. Left square us shared wall ko upward walk karta hai; right square wahi wall downward walk karta hai. Same wind, opposite steps — woh zero mein add hote hain. Yeh har interior wall par hota hai. Sirf woh edges jinke koi neighbour nahi — sach muchi outer boundary $\partial M$ — bach jaati hain.

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Step 5 — Pieces ko naam do: form $\omega$ aur uska derivative $d\omega$

KYA: walk-integrand aur spin ko do objects mein package karo taaki equation chhoti ho jaaye.

• Jo cheez hum edge ke along integrate karte hain woh hai $\omega =Pdx+Qdy$, ise 1-form kehte hain (ek "cheez jo tum curve ke along integrate karte ho"). Yahan $dx$ matlab hai "bacha hua rightward step" aur $dy$ "upward step" — wahi tiny steps Step 2 se.
• Exterior derivative $d$ apply karna (agle mein define) $\omega$ ko $d\omega =({\partial }_{x}Q-{\partial }_{y}P)dx\wedge dy$ mein badle — exactly Step 3 ka spin-times-area.

KYUN wedge $\wedge$? Area signed hota hai: $x$-then-$y$ sweep kiya hua square $y$-then-$x$ se opposite orientation rakhta hai. Wedge ise ek rule ke saath build karta hai, $dx\wedge dy=-dy\wedge dx$. Dono equal set karne se $dx\wedge dx=0$ milta hai: zero width wale "square" ka zero area hoga.

PICTURE: wedge ek oriented parallelogram ke roop mein — do edge-arrows swap karo aur shaded area apna sign flip karta hai (colour flip dikhake show kiya).

Ab $d\omega$ expand karo aur dekho ki Step 2 ke chaar terms wapas appear aur collapse hote hain:

$$d\omega =\underset{d\text{ of the }Pdx\text{ piece}}{\underbrace{(P_{x}dx+P_{y}dy)\wedge dx}}+\underset{d\text{ of the }Qdy\text{ piece}}{\underbrace{(Q_{x}dx+Q_{y}dy)\wedge dy}}.$$

$dx\wedge dx=0$ aur $dy\wedge dy=0$ kill karo; $dy\wedge dx=-dx\wedge dy$ use karo:

$$d\omega =(Q_x-P_y)dx\wedge dy.$$

Tiny-square walk ke chaar terms (Step 2) yahi chaar terms hain. Theorem ke dono sides ek hi computation hain — bahar se vs andar se dekhi gayi.

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Step 6 — Har dimension mein same picture hai

KYA: exactly wahi "sum around edge = sum of derivative inside" kisi bhi dimension mein chalta hai. Sirf props change hote hain.

KYUN: Step 4 ka cancellation argument ne kabhi number 2 use nahi kiya. Shared walls hamesha cancel hoti hain — points, walls, ya faces ho.

PICTURE: teen stacked panels — ek segment (uske do endpoints boundary hain), ek patch (uska loop boundary hai), ek solid (uski skin boundary hai) — sab pe caption "edge = derivative inside".

$k$	$M$ hai ek...	edge $\partial M$	$d$ ban jaata hai	classical naam
1	segment $[a,b]$	do endpoints $\{b\}-\{a\}$	slope $f^{\prime }$	FTC
2	flat patch	ek loop	curl $Q_x-P_y$	Green
2-in-3D	surface	uska rim	curl of $\mathbf{F}$	Stokes
3	solid	uski skin	divergence $\nabla \cdot \mathbf{F}$	Gauss

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Step 7 — Degenerate & edge cases (taaki tumhe kabhi surprise na mile)

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Ek-picture summary

Upar sab kuch ek single image mein compress hota hai: ek region tiny squares se tiled, interior walls greyed-out (cancelled), surviving outer edge glowing, aur caption ${\oint }_{\partial M}\omega ={\iint }_{M}d\omega$. Left side glowing rim hai; right side andar ke sab chhote spins ka sum hai.

Recall Feynman retelling — poora walkthrough plain words mein

Tum jaanna chahte ho ki wind ek fenced field ke andar kitna total "swirl" bana rahi hai. Ginine ke do honest tarike hain. Tarika ek (inside): field ko tiny squares mein kaato, har mein ek paddlewheel daalo, add karo ki har wheel kitna spin karta hai. Tarika do (edge): bas outer fence ke around ek baar chalo aur add karo ki wind tumhe kitna push karti hai. Dono same number dete hain — aur yahan kyun hai: jab tum har tiny square ke around chalte ho, do neighbouring squares ke beech ki fence ek baar har direction mein chali jaati hai, isliye woh inner fences sab cancel ho jaati hain. Sirf outer fence bachti hai un-cancelled. Toh "andar sab spin add karo" $=$ "outer fence ek baar chalo". Yahi poori theorem hai: bahar ka edge $=$ andar ka derivative. Ab is idea ko stretch karo. 1-D mein "field" ek line segment hai, uski "fence" sirf uske do endpoints hain, aur spin slope ban jaata hai — yeh Fundamental Theorem of Calculus hai. 3-D mein "field" ek solid lump hai, uski "fence" uski skin hai, aur spin divergence ban jaata hai (per unit volume kitna stuff create hota hai) — yeh Divergence theorem hai. Same sentence, alag-alag size ke rooms. Ise ek baar aur hamesha ke liye likho: ${\int }_{\partial M}\omega ={\int }_{M}d\omega$.

Connections

• Green's Theorem — $k=2$ case yahan Steps 2–5 mein build kiya
• Stokes' Theorem (curl) — same tiling ek curved surface par
• Divergence Theorem — Step 6, $k=3$ case
• Fundamental Theorem of Calculus — Step 6/7, $k=1$ case
• Differential Forms and the Wedge Product — Step 5
• Exterior Derivative aur d squared equals zero
• Orientation and Induced Boundary Orientation — Steps 1, 4, 7