Green's theorem — proof sketch, both forms: Worked examples

4.4.29 · D3 · Maths › Multivariable Calculus › Green's theorem — proof sketch, both forms

Do objects hum poori jagah use karte hain. Dono parent note mein define the; yahan ek jagah phir se hain taaki koi symbol unexplained na rahe:

Scenario matrix

Har woh problem jo Green's theorem pose kar sakta hai, in cells mein se ek mein aati hai. Neeche diye examples mein har ek par [cell] tag laga hai taaki aap dekh sako ki coverage complete hai.

Cell	Kya cheez isse alag banati hai	Example(s)
A. Curl, positive	$Q_{x} - P_{y} > 0$ har jagah → nonzero circulation	Ex 1
B. Curl, zero by symmetry	integrand odd over a symmetric region → answer $0$	Ex 2
C. Wrong orientation	curve clockwise di gayi → sign flip	Ex 3
D. Non-standard region	annulus / hole wala region	Ex 4
E. Singularity (hypothesis breaks)	field andar $D$ mein blow up karta hai → theorem ka galat use	Ex 5
F. Flux form	divergence use karna, curl nahi	Ex 6
G. Degenerate / limiting	shrinking loop, curl as a limit	Ex 7
H. Word problem	real-world flux (fluid crossing a boundary)	Ex 8
I. Exam twist	non-circular boundary, area trick	Ex 9

Example 1 — Curl, positive `[cell A]`

Worked example Ek asli swirl ki circulation

$\oint_{C} (- y) d x + x d y$ compute karo jahan $C$ radius $R$ ke disk ki counterclockwise boundary hai.

Forecast: yeh field $F = (- y, x)$ circles ke saath point karta hai (ek pure swirl). Counterclockwise chalte waqt aap hamesha aage push hote ho, isliye answer positive hona chahiye aur area ke saath badhna chahiye. Aage padhne se pehle guess karo.

Step 1. $P = - y$ , $Q = x$ identify karo. Yeh step kyun? Green's theorem ke liye $P$ aur $Q$ ko differentiate karne se pehle name karna zaroori hai.

Step 2. Curl integrand compute karo: $Q_{x} - P_{y} = \frac{\partial x}{\partial x} - \frac{\partial ( - y )}{\partial y} = 1 - (- 1) = 2.$ Yeh step kyun? Circulation form loop integral ko $\iint_{D} (Q_{x} - P_{y}) d A$ mein convert karta hai. Hume woh integrand chahiye.

Step 3. Disk par constant integrate karo: $\iint_{D} 2 d A = 2 \cdot (disk ka area) = 2 π R^{2} .$ Yeh step kyun? Constant $2$ ko $D$ par integrate karna sirf $2 \times$ area hai, aur disk ka area $π R^{2}$ hai.

Result: $\oint_{C} (- y) d x + x d y = 2 π R^{2}$ .

Verify: $x = R cos t, y = R sin t$ parametrise karo. Tab $- y d x + x d y = R^{2} sin^{2} t + R^{2} cos^{2} t = R^{2}$ , aur $\int_{0}^{2 π} R^{2} d t = 2 π R^{2}$ . ✅ Match karta hai, aur area ke saath exactly forecast ki tarah badha.

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

Red arrows dekho: woh counterclockwise curl kar rahe hain, hamare chalne ki direction ke aligned — isliye circulation positive hai.

Example 2 — Curl zero by symmetry `[cell B]`

Worked example Woh swirl jo cancel ho jaata hai

Unit circle (counterclockwise) ke around $\oint_{C} y^{2} d x + 3 x y d y$ compute karo.

Forecast: yeh field ek clean swirl nahi hai. Kya answer nonzero hoga? Symmetry ke liye dhyan rakho.

Step 1. $P = y^{2}$ , $Q = 3 x y$ . Yeh step kyun? Pehle components name karo.

Step 2. Curl integrand: $Q_{x} - P_{y} = 3 y - 2 y = y .$ Yeh step kyun? Yahi woh quantity hai jo hum disk par integrate karte hain.

Step 3. Unit disk par $y$ integrate karo: $\iint_{D} y d A = 0.$ Yeh step kyun / kyun zero? Disk $y = 0$ line ke baare mein symmetric hai. Integrand $y$ odd hai: $+ y$ height par har tile ke liye ek mirror tile $- y$ par hai jo opposite contribute karti hai. Woh pairs mein cancel ho jaate hain. Koi brute-force ki zaroorat nahi.

Result: $0$ .

Verify: polar mein switch karo, $y = r sin θ$ , $d A = r d r d θ$ : $\int_{0}^{2 π} \int_{0}^{1} (r sin θ) r d r d θ = (\int_{0}^{1} r^{2} d r) (\int_{0}^{2 π} sin θ d θ) = \frac{1}{3} \cdot 0 = 0. ✅$

Example 3 — Wrong orientation `[cell C]`

Worked example Clockwise loop

$\oint_{C} (- y) d x + x d y$ compute karo jahan $C$ unit circle hai jo clockwise traverse ki gayi hai.

Forecast: yeh Ex 1 wala hi field hai ( $R = 1$ ke saath), lekin ab hum galat taraf chal rahe hain. Green's theorem counterclockwise loops ke liye stated hai, isliye hum expect karte hain ki answer Ex 1 ka negative hoga.

Step 1. Green's theorem ko positive (counterclockwise) orientation chahiye. Hamari curve clockwise hai, isliye maano $C^{+}$ counterclockwise version hai aur note karo ki $C = - C^{+}$ . Yeh step kyun? Line integral ki direction reverse karna uska sign flip karta hai; hume theorem jo orientation assume karta hai uspe convert karna hoga.

Step 2. $C^{+}$ par theorem apply karo (Ex 1 se $R = 1$ ke saath): $\oint_{C^{+}} (- y) d x + x d y = 2 π (1)^{2} = 2 π .$

Step 3. Clockwise curve ke liye sign flip karo: $\oint_{C} (- y) d x + x d y = - \oint_{C^{+}} = - 2 π .$ Yeh step kyun? $C = - C^{+}$ , aur $\int_{- C} = - \int_{C}$ .

Result: $- 2 π$ .

Verify: clockwise ka matlab hai $x = cos t, y = - sin t$ (angle decreasing). Tab $- y d x + x d y = - sin^{2} t - cos^{2} t = - 1$ , aur $\int_{0}^{2 π} (- 1) d t = - 2 π$ . ✅

Red arrow dikhata hai ki chalne ki direction ab field ke swirl ke against hai — region aapke right par hai, left par nahi, jo sign flip ka visual signature hai.

Example 4 — Region with a hole `[cell D]`

Worked example Annulus

Annulus $1 \leq x^{2} + y^{2} \leq 4$ (radii $1$ aur $2$ ) ki boundary par $\oint (- y) d x + x d y$ compute karo. Boundary mein do circles hain: outer $C_{2}$ counterclockwise, inner $C_{1}$ clockwise (taaki region har jagah aapke left par rahe).

Forecast: field phir se swirl wala hai. Double integral sirf ring ka area count karta hai, hole ka nahi. Expect karo $2 \times (ring area)$ .

Step 1. Sahi oriented boundary hai $C = C_{2} - C_{1}^{cc}$ jahan $C_{1}^{cc}$ inner circle counterclockwise hai. Yeh step kyun? Aapke left par region ke liye, outer curve counterclockwise chalti hai aur har inner hole clockwise chalti hai. Yeh parent proof ke Step 3 wala multiply-connected version hai.

Step 2. Curl integrand $2$ hai (Ex 1 ki tarah hi). Ring par integrate karo: $\iint_{D} 2 d A = 2 (π \cdot 2^{2} - π \cdot 1^{2}) = 2 (4 π - π) = 6 π .$ Yeh step kyun? Ring ka area hai (outer disk area) − (inner disk area) $= 4 π - π = 3 π$ .

Result: total oriented circulation $= 6 π$ .

Verify (by hand, dono circles): radius $R$ ki har circle counterclockwise $2 π R^{2}$ deti hai (Ex 1). Outer counterclockwise $= 2 π (4) = 8 π$ ; inner clockwise $= - 2 π (1) = - 2 π$ . Sum $= 8 π - 2 π = 6 π$ . ✅

Example 5 — Singularity: hypothesis breaks `[cell E]`

Worked example Trap field

$F = (\frac{- y}{x ^{2} + y ^{2}}, \frac{x}{x ^{2} + y ^{2}})$ ke liye, ek student $Q_{x} - P_{y} = 0$ compute karta hai aur conclude karta hai ki unit circle ke liye $\oint_{C} F \cdot d r = 0$ . Error kahan hai, aur sahi value kya hai?

Forecast: kuch galat hai — yeh field famous hai. Origin pe shak karo.

Step 1. Origin ke door curl integrand check karo. $Q_{x} - P_{y}$ likhna aur simplify karna $0$ deta hai jahan bhi field defined hai. Yeh step kyun? Student ki algebra ko blame karne se pehle hume confirm karna hoga.

Step 2. Flaw locate karo: $F$ $(0, 0)$ par undefined hai kyunki wahan $x^{2} + y^{2} = 0$ . Green's theorem demand karta hai ki $P, Q$ ke andar $D$ ke har jagah continuous partial derivatives hon — origin unit disk ke andar hai aur yeh violate karta hai. Yeh step kyun? Theorem ka conclusion void hai agar uska hypothesis fail ho. Koi hypothesis nahi, toh $\iint$ -to- $\oint$ equality nahi.

Step 3. Seedha line integral compute karo, $x = cos t, y = sin t$ (toh $x^{2} + y^{2} = 1$ ): $\oint_{C} F \cdot d r = \int_{0}^{2 π} [(- sin t) (- sin t) + (cos t) (cos t)] d t = \int_{0}^{2 π} 1 d t = 2 π .$ Yeh step kyun? Jab theorem available nahi hai, line integral ki definition par wapas jao.

Result: sahi value $2 π$ hai, $0$ nahi. Double-integral shortcut illegal tha.

Fix: origin ko ek tiny circle se exclude karo; resulting hole-punched region par theorem sach mein apply hota hai, aur woh sahi se $2 π$ report karta hai jo puncture par "live" karta hai. Yeh exactly wahi mistake hai jo parent note mein flag ki gayi thi.

Verify: $(- sin t) (- sin t) + (cos t) (cos t) = sin^{2} t + cos^{2} t = 1$ , integrate karke $2 π$ milta hai. ✅

Example 6 — Flux form `[cell F]`

Har red arrow fence ko outward cross karta hai — koi bhi inflow nahi, isliye total ek clean positive number hai.

Example 7 — Degenerate limit: curl as a shrinking loop `[cell G]`

Worked example Circulation per area

$F = (- y, x)$ ke liye, dikhao ki radius $ε$ ke small circle ke around circulation, uske area se divide karke, $ε \to 0$ par curl integrand approach karta hai.

Forecast: parent note ne $Q_{x} - P_{y}$ ko "local spin" kaha tha. Ek limit ko global loop integral se local value $2$ recover karni chahiye.

Step 1. Ex 1 se, radius $ε$ ke around circulation $2 π ε^{2}$ hai. Yeh step kyun? Hum woh closed-form reuse karte hain jis par humein already bharosa hai.

Step 2. Enclosed area $π ε^{2}$ se divide karo: $\frac{circulation}{area} = \frac{2 π ε ^{2}}{π ε ^{2}} = 2.$ Yeh step kyun? "Circulation per unit area" scalar curl ki physical definition hai.

Step 3. Limit lo: $lim_{ε \to 0} \frac{1}{π ε ^{2}} \oint_{C_{ε}} F \cdot d r = 2 = (Q_{x} - P_{y}) .$ Yeh step kyun? Yeh dikhata hai ki double-integral integrand sach mein loop integrals ki ek pointwise limit hai — Green's theorem ke peeche ka microscopic meaning, Fundamental Theorem of Calculus ke us idea ka echo ki boundary ek interior rate encode karti hai.

Result: ratio exactly $2$ hai har $ε$ ke liye, isliye limit $2$ hai. ✅ (Ek rare "no limit surprise" — field ka curl constant hai.)

Verify: $2 π ε^{2} / (π ε^{2}) = 2$ sabhi $ε > 0$ ke liye, isliye limit $2$ hai. ✅

Example 8 — Word problem: fluid across a channel gate `[cell H]`

Worked example Pond se kitna paani nikalta hai?

Ek shallow pond mein paani ka velocity field $F = (x, y)$ metres/second hai (origin par ek spring outward push kar raha hai). Pond boundary radius $3$ m ki circle hai. Boundary ke through net outward volume flow rate (per unit depth) kya hai?

Forecast: paani har jagah outward flow karta hai, isliye net outflow positive hai; Ex 6 se yeh $R = 3$ ke saath $2 π R^{2}$ hona chahiye.

Step 1. Net outflow rate $= \oint_{C} F \cdot n d s$ — flux form. Yeh step kyun? "Boundary ke across per second volume" precisely ek flux integral hai.

Step 2. $\nabla \cdot F = 1 + 1 = 2$ (per second), pond par integrate karo: $\iint_{D} 2 d A = 2 \cdot π \cdot 3^{2} = 18 π .$ Yeh step kyun? Divergence $= 2$ ka matlab hai har square metre "create" karta hai $2$ m³/s outflow; multiply karo area $9 π$ se.

Result: $18 π \approx 56.5 m^{3} / s$ per unit depth of pond.

Verify (units & value): $[divergence] = \frac{m/s}{m} = 1/ s$ ; times area $m^{2}$ deta hai $m^{2} / s$ (area flow per unit depth) ✅. Numerically $18 π = 56.548...$ , positive as forecast. ✅

Example 9 — Exam twist: triangle ka area uske edges se `[cell I]`

Worked example Bina double integral ke area

$Area = \frac{1}{2} \oint_{C} (x d y - y d x)$ use karke vertices $(0, 0)$ , $(4, 0)$ , $(0, 3)$ wale triangle ka area nikalo, counterclockwise chalte hue.

Forecast: classic $3$ - $4$ - $5$ -ish right triangle; area $= \frac{1}{2} \cdot 4 \cdot 3 = 6$ . Boundary formula ko yahi reproduce karna chahiye.

Step 1. Area formula $P = - y /2, Q = x /2$ choose karne se aata hai, jisse $Q_{x} - P_{y} = 1$ milta hai, isliye $\iint_{D} 1 d A = Area$ . Yeh step kyun? Hum integrand ko $1$ hone ke liye engineer karte hain taaki double integral literally area hi ho — plane mein koi antiderivative nahi chahiye, sirf edge sums.

Step 2. $(x_{1}, y_{1})$ se $(x_{2}, y_{2})$ tak ek straight segment ke liye, dikhaya ja sakta hai: $\frac{1}{2} \int (x d y - y d x) = \frac{1}{2} (x_{1} y_{2} - x_{2} y_{1}) .$ Yeh step kyun? Ek line segment par $x d y - y d x$ ka integral woh signed twice-area hai jo segment origin ke saath milkar triangle banata hai — "shoelace" building block.

Step 3. Teen edges $(0, 0) \to (4, 0) \to (0, 3) \to (0, 0)$ par sum karo: $\frac{1}{2} [(0 \cdot 0 - 4 \cdot 0) + (4 \cdot 3 - 0 \cdot 0) + (0 \cdot 0 - 0 \cdot 3)] = \frac{1}{2} [0 + 12 + 0] = 6.$ Yeh step kyun? Har bracket ek edge ka contribution hai; unhe add karna poori boundary ek baar walk karta hai.

Result: Area $= 6$ .

Verify: elementary geometry $\frac{1}{2} \cdot base \cdot height = \frac{1}{2} \cdot 4 \cdot 3 = 6$ . ✅

Har red edge origin se sweep kiye wedge ka signed area contribute karta hai; wedges add up ho jaate hain (cancellation ke saath) triangle ko. Yeh area measure karne ka line-integral tarika hai — boundary phir se interior jaanti hai.

Recall Kya matrix poora cover hua?

Positive curl ::: Ex 1 Zero by symmetry ::: Ex 2 Wrong orientation (sign flip) ::: Ex 3 Region with a hole (annulus) ::: Ex 4 Singularity breaks the hypothesis ::: Ex 5 Flux / divergence form ::: Ex 6 Degenerate shrinking loop (curl as a limit) ::: Ex 7 Real-world flux word problem ::: Ex 8 Exam twist (area from boundary) ::: Ex 9

Related deep structure: Green's theorem dono Stokes' Theorem (curl form) aur Divergence Theorem (flux form) ka flat-plane case hai; jab $Q_{x} - P_{y} = 0$ har jagah ho toh field aksar ek conservative field hoti hai jisme zero circulation hoti hai.

Flashcards

Circulation of

(- y, x)

around radius

R

(CCW)?

2 π R^{2}

Unit disk par

\oint y^{2} d x + 3 x y d y = 0

kyun?

Curl integrand

y

hai,

y = 0

ke baare mein symmetric disk par odd hai, isliye cancel ho jaata hai.

Clockwise loop — kya karte hain?

Sign flip karo; theorem ko CCW version par apply karo.

Annulus boundary orientation?

Outer CCW, inner CW (region aapke left par).

(- y, x) / (x^{2} + y^{2})

0

nahi

2 π

kyun deta hai?

Origin par singularity continuous-partials hypothesis tod deti hai.

Radius

R

ke across

(x, y)

ka flux?

2 π R^{2}

, kyunki divergence

= 2

Boundary se area?

Area = \frac{1}{2} \oint_{C} (x d y - y d x)