Green's theorem — proof sketch, both forms: Visual walkthrough

4.4.29 · D2 · Maths › Multivariable Calculus › Green's theorem — proof sketch, both forms

Step 0 — Dono sides pe kaun se objects hain?

KYA. Green's theorem ek equation hai jisme left side pe ek loop-cheez hai aur right side pe ek area-cheez hai: $edge ke around walk \oint_{C} P d x + Q d y = andar ka sum \iint_{D} (\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}) d A$

Ise use karne se pehle har piece ka naam rakh lete hain.

Definition Characters ki cast

$D$ ek region hai — flat plane ka ek bhara hua patch (socho: upar se dekha hua ek lake).
$C$ uski boundary hai — lake ke around ki fence, ek closed loop jo apne start pe wapas aati hai.
Ek vector field $F = (P, Q)$ har point $(x, y)$ ko ek chhota arrow assign karta hai. $P (x, y)$ batata hai arrow kitna right point karta hai; $Q (x, y)$ batata hai kitna upar.
$\oint_{C}$ ek line integral hai closed loop ke around — hum fence walk karte hain aur har step pe ek quantity add karte hain. $\oint$ pe chhota circle ka matlab hai "path ek closed loop hai."
$\iint_{D}$ ek double integral hai — hum $D$ ko tiny tiles $d A$ mein chop karte hain aur har tile pe ek quantity add karte hain.
$\frac{\partial Q}{\partial x}$ (padho "partial Q by x") ka matlab hai: $y$ ko freeze karo, $x$ ko thoda wiggle karo, aur dekhो $Q$ kitni tezi se change hoti hai. Yeh sirf $x$ -direction mein ek slope hai. Isi tarah $\frac{\partial P}{\partial y}$ $x$ ko freeze karta hai aur $y$ ko wiggle karta hai.

Figure — Green's theorem — proof sketch, both forms

Picture mein: blue arrows field $F$ hain, green patch $D$ hai, aur yellow loop $C$ hai jisme ek arrow dikhata hai hum kis taraf walk karte hain. Region ko apni left side pe rakho — yeh counterclockwise, "positive" direction hai.

Step 1 — Theorem ko do aasaan halves mein split karo

KYA. Hum $P$ aur $Q$ dono pe ek saath attack nahi karenge. Hum split karte hain: $\oint_{C} P d x = - \iint_{D} \frac{\partial P}{\partial y} d A (⋆)$ $\oint_{C} Q d y = + \iint_{D} \frac{\partial Q}{\partial x} d A (⋆ ⋆)$

KYUN. $(⋆)$ aur $(⋆ ⋆)$ ko add karne se full theorem rebuild ho jaata hai, kyunki $- P_{y} + Q_{x}$ exactly integrand $Q_{x} - P_{y}$ hai. Har half mein sirf ek field-component involved hai, isliye har ek disguise mein ek one-variable problem hai.

Left panel: sirf horizontal steps ( $d x$ ) matter karte hain, isliye hum top aur bottom edges shade karte hain. Right panel: sirf vertical steps ( $d y$ ) matter karte hain, isliye hum left aur right edges shade karte hain.

Step 2 — $(⋆)$ prove karo: ek "Type I" region setup karo

KYA. Ek Type I region woh hoti hai jo aisi shaped ho ki har $x$ ke liye $a$ aur $b$ ke beech, patch vertically ek bottom curve $y = g_{1} (x)$ se upar top curve $y = g_{2} (x)$ tak jaata hai: $D = {(x, y) : a \leq x \leq b, g_{1} (x) \leq y \leq g_{2} (x)} .$

KYUN. Is shape ke saath inside vertical strips ka stack hai. Har strip $y$ mein ek clean interval hai, toh hum pehle $y$ -integral kar sakte hain aur ise Fundamental Theorem of Calculus se finish kar sakte hain.

$a, b$ sabse leftmost aur rightmost $x$ -values hain (vertical dashed walls).
$g_{1} (x)$ (green, lower) aur $g_{2} (x)$ (yellow, upper) woh curves hain jinka boundary follow karta hai.
Ek red vertical strip ek single fixed $x$ dikhata hai: inner integral is strip ke upar run karta hai.

Step 3 — Fundamental Theorem of Calculus se inside collapse karo

KYA. $(⋆)$ ki right-hand side compute karo, pehle inner integral: $- \iint_{D} \frac{\partial P}{\partial y} d A = - \int_{a}^{b} (\int_{g_{1} (x)}^{g_{2} (x)} \frac{\partial P}{\partial y} d y) d x .$

Inner integrand $\frac{\partial P}{\partial y}$ $y$ mein ek derivative hai, aur hum $y$ mein integrate kar rahe hain — ek hi variable ki derivative aur integral ek doosre ko undo kar dete hain. Yahi Fundamental Theorem of Calculus hai: $\int_{g_{1} (x)}^{g_{2} (x)} rate \frac{\partial P}{\partial y} d y = top pe value P (x, g_{2} (x)) - bottom pe value P (x, g_{1} (x)) .$

KYUN yahi tool aur koi nahi. Humne andar ek perfect derivative engineer ki taaki FTC poora $y$ -integral evaporate kar sake, sirf boundary values $P$ top aur bottom curves pe chodta hai. Yahi Green's theorem ka poora secret hai: ek inside-derivative edge-values mein badal jaati hai.

Minus sign wapas laate hain: $-\iint_D \frac{\partial P}{\partial y}\,dA = \int_a^b\Big[\,\underbrace{P(x,g_1(x))}_{\text{bottom}} - \underbrace{P(x,g_2(x))}_{\text{top}}\,\Big]\,dx. \tag{RHS}$

Picture: ek red strip ke along, $P$ mein chhote changes ka tall stack sirf (top value) − (bottom value) tak add ho jaata hai — strip ke ends pe do dots. Beech wala telescope ho jaata hai.

Step 4 — $(⋆)$ ki left-hand side ke liye boundary walk karo

KYA. Ab left side compute karo, $\oint_{C} P d x$ . Fence ko char pieces mein split karo: bottom $C_{1}$ , right wall $C_{R}$ , top $C_{2}$ , left wall $C_{L}$ — counterclockwise walked (region left pe).

KYUN walls vanish hoti hain. Ek vertical wall pe, $x$ kabhi change nahi hota, toh $d x = 0$ . Kyunki integrand mein $d x$ ka factor hai, walls kuch contribute nahi karti: $\int_{C_{R}} P d x = 0, \int_{C_{L}} P d x = 0.$

Toh sirf bottom aur top survive karti hain: $\oint_{C} P d x = C_{1} : left \to right \int_{a}^{b} P (x, g_{1} (x)) d x + C_{2} : right \to left \int_{b}^{a} P (x, g_{2} (x)) d x .$

Top curve right se left walk ki jaati hai (yahi counterclockwise demand karta hai), toh uske limits $b$ se $a$ tak run karte hain. Limits flip karne se sign flip ho jaata hai: $\int_{b}^{a} P (x, g_{2} (x)) d x = - \int_{a}^{b} P (x, g_{2} (x)) d x .$

Isliye $\oint_C P\,dx = \int_a^b\Big[\,P(x,g_1(x)) - P(x,g_2(x))\,\Big]\,dx. \tag{LHS}$

Yellow arrows follow karo: bottom ke across rightward, right wall ke upar (koi $d x$ nahi), top ke across leftward, left wall ke neeche (koi $d x$ nahi). (LHS) bilkul (RHS) jaisa hai Step 3 se — toh $(⋆)$ prove ho gaya. ✅

Step 5 — Mirror argument se $(⋆ ⋆)$ prove karo

KYA. $(⋆ ⋆)$ ke liye hum $x$ aur $y$ ki roles flip karte hain. Ek Type II region horizontal strips ka stack hai: $D = {(x, y) : c \leq y \leq d, h_{1} (y) \leq x \leq h_{2} (y)} .$

KYUN. Ab $\oint Q d y$ sirf vertical motion dekhta hai, toh is baar horizontal top/bottom edges vanish hoti hain ( $d y = 0$ ) aur left/right curves $x = h_{1} (y), x = h_{2} (y)$ sab kuch carry karti hain. Pehle $x$ mein $\frac{\partial Q}{\partial x}$ integrate karo, ise FTC se hit karo, aur boundary pieces match karo — exactly Steps 3–4 $x$ aur $y$ swap ke saath. Result: $\oint_C Q\,dy = +\iint_D \frac{\partial Q}{\partial x}\,dA. \tag{$\star\star$}$

Poori wohi story hai diagonal line $y = x$ ke across reflect hui: horizontal strips, left curve $h_{1}$ , right curve $h_{2}$ . $(⋆) + (⋆ ⋆)$ add karne se kisi bhi region ke liye full theorem milta hai jo dono Type I aur Type II hai.

Step 6 — General regions: interior cuts cancel ho jaate hain

KYA. Bahut saare regions na to clean vertical stack hote hain na clean horizontal stack (socho ek bent "L" ya annulus). Hum aisi region ko sub-pieces mein chop karte hain jo Type I aur Type II hain, har ek pe theorem prove karte hain, aur add karte hain.

KYUN yeh phir bhi kaam karta hai. Har internal cut do neighbouring pieces share karte hain. Piece A us cut ko ek taraf walk karta hai; piece B usi cut ko opposite taraf walk karta hai (har ek apni region ko left pe rakhta hai). Us cut ke along do line integrals equal aur opposite hain, toh woh cancel ho jaate hain. Sirf outer fence survive karti hai.

Red internal cut left piece se upar aur right piece se neeche traverse hoti hai — unke contributions annihilate ho jaate hain. Double integrals sirf add ho jaate hain (areas overlap nahi karte), toh pieces ka sum = poore region ka double integral, surviving outer loop se bounded.

Step 7 — Flux form free mein lo

KYA. Humhare paas already circulation form hai. Flux / divergence form ke liye koi naya kaam nahi karna. Boundary ke along, unit tangent $T = (d x / d s, d y / d s)$ ke saath chalo (length 1 ka arrow jo us taraf point karta hai jis taraf tum walk karte ho). Ise clockwise $90°$ rotate karo outward unit normal paane ke liye: $n = (\frac{d y}{d s}, - \frac{d x}{d s}) .$

KYUN woh rotation outward point karta hai. Region ko left pe rakhte hue, "outward" "forward" se $90°$ clockwise hai. Phir $F \cdot n d s = (P, Q) \cdot (d y, - d x) = P d y - Q d x .$

Ab circulation form ko rotated field $(\tilde{P}, \tilde{Q}) = (- Q, P)$ pe apply karo: $\oint_{C} - Q d x + P d y = \iint_{D} (\tilde{Q}_{x} \frac{\partial P}{\partial x} - \tilde{P}_{y} \frac{\partial ( - Q )}{\partial y}) d A = \iint_{D} (P_{x} + Q_{y}) d A .$ Left side exactly $\oint_{C} F \cdot n d s$ hai. Toh $\oint_{C} F \cdot n d s = \iint_{D} (\nabla \cdot F P_{x} + Q_{y}) d A . ✅$

Blue tangent $T$ fence ke along point karta hai; ise clockwise $90°$ rotate karo red outward normal $n$ paane ke liye. Circulation fence ke along spin count karta hai; flux us se through flow count karta hai — same theorem, ek doosra rotated hai. Yeh Divergence Theorem ka 2D face hai aur dono Stokes' Theorem ki shadows hain; poora family curl/divergence bookkeeping ko generalise karta hai.

Ek-picture summary

Yeh single diagram saaton steps compress karta hai: ek inside-derivative ( $Q_{x} - P_{y}$ , curl) strip-by-strip telescope ho ke edge-values mein badal jaati hai (FTC), vertical/horizontal walls drop out ho jaate hain, interior cuts cancel ho jaate hain, aur sirf boundary walk bachti hai — edge poori inside ko yaad rakhti hai.

Recall Poore walkthrough ki Feynman retelling

Hum prove karna chahte the: fence walk karna = andar ki spin add karna. Pehle humne kaam ko ek sideways part ( $P d x$ ) aur ek up-down part ( $Q d y$ ) mein split kiya, kyunki ek sideways walk sirf horizontal steps feel karta hai aur ek up-down walk sirf vertical steps feel karta hai. Sideways part ke liye humne region ko thin vertical straws mein slice kiya; har straw ke andar chhote changes ka pile sirf (top minus bottom) tak collapse ho jaata hai — yeh Fundamental Theorem of Calculus hai, wohi magic jo derivative ko integral se undo karta hai. Fence ki vertical walls ne kuch nahi diya kyunki unke along tum sideways nahi chale. Saare straw-tops aur straw-bottoms add karna wahi hai jaise fence walk karna. Up-down part bilkul wohi trick hai ek taraf se turn karke. Weird shapes ke liye humne unhe nice pieces mein cut kiya; har cut opposite directions mein do baar walk hota hai aur cancel ho jaata hai, toh sirf outer fence bachti hai. Aakhir mein, har chhote walking-arrow ko quarter-turn deke outward point karne ke liye, "fence ke along spin" statement "fence se bahar flow" statement ban jaata hai — wahi sach, rotated. Edge hamesha jaanati hai andar kya ho raha hai.

Recall

Hum Type I region pe inner $y$ -integral closed form mein kyun kar sakte hain? ::: Integrand $\partial P / \partial y$ $y$ mein perfect derivative hai, toh FTC top-value minus bottom-value deta hai. Vertical walls $\oint P d x$ mein kuch contribute kyun nahi karti? ::: Vertical wall pe $x$ constant hai, toh $d x = 0$ . Top curve ko flipped integration limits kyun milte hain? ::: Counterclockwise top ko right-to-left walk karta hai, toh $x$ $b$ se $a$ tak run karta hai; $a \to b$ tak flip karne se minus sign aata hai. Pieces glue karte waqt interior cuts kyun cancel ho jaate hain? ::: Har cut apne do neighbouring pieces se opposite directions mein walk hota hai, toh do line integrals equal aur opposite hain. Tangent se outward normal kaise build hota hai? ::: Unit tangent ko clockwise $90°$ rotate karo: $n = (d y / d s, - d x / d s)$ .