Unification — all three theorems as generalized Stokes: Worked examples

4.4.34 · D3 · Maths › Multivariable Calculus › Unification — all three theorems as generalized Stokes

Yeh page unification note ki practice floor hai. Parent note ne bataya kyun char classical theorems ek hi identity $\int_{\partial M} ω = \int_{M} d ω$ hain. Yahan hum identity ko haath se chalate hain — har dimension mein, har orientation sign ke saath, har degenerate input ke saath — aur hum hamesha dono sides compute karte hain taaki dekh sakein ke woh agree karte hain.

The scenario matrix

Compute karne se pehle, chalte hain har tarah ki situation ko naam dete hain jo yeh ek identity ko handle karni padti hai. Har row ek "cell" hai jahan reader land kar sakta hai; neeche har worked example us cell ke saath tagged hai jise woh cover karta hai.

Cell	Dimension $k$	Jo twist test ho rahi hai	Covered by
A	$k = 1$ (FTC)	endpoints par orientation sign, $f (b) - f (a)$	Example 1
B	$k = 2$ Green	curl term $Q_{x} - P_{y} \neq = 0$ (genuine circulation)	Example 2
C	$k = 2$ Green	degenerate: $Q_{x} - P_{y} = 0$ har jagah (conservative field) → answer $0$	Example 3
D	$k = 2$ Green as flux / 2D-divergence	expanding radial field, dono readings	Example 4
E	$k = 3$ Divergence	full 3D flux ek solid se bahar; boundary vs inside	Example 5
F	$k = 3$ Divergence	degenerate: $\nabla \cdot F = 0$ (incompressible) → net flux $0$	Example 6
G	any $k$	$d^{2} = 0$ trap: integrand already ek derivative ka derivative hai → $0$	Example 7
H	$k = 2$ Stokes (curl, surface in 3D)	non-flat surface, sirf boundary matters	Example 8
I	word problem	tank se fluid nikalna (real-world divergence)	Example 9
J	exam twist	galat orientation deliberately, sign flip hote dekho	Example 10

Hum poore waqt yeh background rules use karenge (from Exterior Derivative aur Differential Forms and the Wedge Product):

Cell A — Dimension 1: Fundamental Theorem Stokes ke costume mein

Worked example Example 1 (Cell A)

Maano $M = [1, 3]$ left-to-right oriented hai, aur $ω = f$ woh $0$ -form hai $f (x) = x^{2}$ . Verify karo $\int_{\partial M} ω = \int_{M} d ω$ .

Forecast: aage padhne se pehle guess karo ki dono sides kaunsa single number denge. (Hint: yeh $x^{2}$ ka change hai interval ke across.)

Step 1 — Boundary ko signed points ke roop mein dhundho. Oriented interval $[1, 3]$ ki boundary hai $\partial M = {3} - {1}$ : jis endpoint par tum pahuncho woh $+$ count hota hai, jise tum chhorte ho woh $-$ count hota hai. Yeh step kyun? Dimension 1 mein "boundary" koi edge nahin hai jis par chalo — woh do points hain, aur induced orientation exactly $\pm$ tags hain. Woh minus sign hi FTC ka poora content hai.

Step 2 — Un points par $0$ -form ko integrate karo (left side). Ek point par $0$ -form ko integrate karna sirf use evaluate karna hai: $\int_{\partial M} f = f (3) - f (1) = 9 - 1 = 8.$ Yeh step kyun? Ek point par integrate karne ke liye koi length nahin hai; values ka signed sum HI integral hai.

Step 3 — $d ω$ compute karo aur inside par integrate karo (right side). $d ω = f^{'} (x) d x = 2 x d x$ , toh $\int_{M} d ω = \int_{1}^{3} 2 x d x = [x^{2}]_{1}^{3} = 9 - 1 = 8.$ Yeh step kyun? Ek function par $d$ uska ordinary derivative-times- $d x$ hota hai; right side $f^{'}$ ka familiar Riemann integral hai.

Verify: left $= 8$ , right $= 8$ . ✅ Yeh literally $\int_{a}^{b} f^{'} = f (b) - f (a)$ hai — Fundamental Theorem of Calculus $k = 1$ ka costume hai.

Cell B — Green with genuine circulation

Worked example Example 2 (Cell B)

Maano $ω = - y d x + x d y$ unit disk $M$ par (boundary = unit circle, counterclockwise). Stokes verify karo.

Forecast: yeh field counterclockwise swirl karta hai. Kya tumhe expect hai ki circulation positive hogi, negative, ya zero?

Step 1 — $P$ aur $Q$ read karo. $ω = P d x + Q d y$ ke saath hamare paas $P = - y$ , $Q = x$ hai. Yeh step kyun? Green's Theorem ko dono component functions chahiye taaki curl term build kar sakein.

Step 2 — Antisymmetry use karke $d ω$ build karo. $d ω = (Q_{x} - P_{y}) d x \land d y = (1 - (- 1)) d x \land d y = 2 d x \land d y .$ Yeh step kyun? $d P \land d x + d Q \land d y$ ke chaar terms ek mein collapse hote hain kyunki $d x \land d x = d y \land d y = 0$ aur $d y \land d x = - d x \land d y$ . Jo bachta hai woh coefficient $Q_{x} - P_{y}$ 2D curl hai.

Step 3 — Inside integrate karo (right side). $\int_{M} 2 d x d y = 2 \cdot (area of unit disk) = 2 π .$ Yeh step kyun? Integrand constant $2$ hai, toh double integral sirf $2$ times region ki area hai; unit disk ki area $π$ hai. Yeh right-side "har patch par curl add karo" reading hai.

Step 4 — Boundary ke around integrate karo (left side). Circle ko parametrize karo $x = cos t, y = sin t$ , $t \in [0, 2 π]$ , toh $d x = - sin t d t$ , $d y = cos t d t$ : $\oint_{\partial M} (- y d x + x d y) = \int_{0}^{2 π} (sin^{2} t + cos^{2} t) d t = \int_{0}^{2 π} 1 d t = 2 π .$ Yeh step kyun? Boundary par directly chalna "left side" reading hai; yeh inside sum ke barabar hona chahiye.

Verify: dono sides $= 2 π$ . ✅ Positive, jaise forecast tha — field sach mein counterclockwise circulate karta hai. (Yeh field classic "twice the area" detector hai.)

Cell C — Green, degenerate: ek conservative field zero deta hai

Worked example Example 3 (Cell C)

Maano $ω = 2 x y d x + x^{2} d y$ kisi bhi closed loop $\partial M$ ke around (concreteness ke liye unit square lo). Verify karo ki answer $0$ hai.

Forecast: notice karo $ω = d (x^{2} y)$ . Ek exact form ka loop integral kya hona chahiye?

Step 1 — Curl term compute karo. $P = 2 x y, Q = x^{2}$ , toh $Q_{x} - P_{y} = 2 x - 2 x = 0$ . Yeh step kyun? Jab $Q_{x} - P_{y} \equiv 0$ toh field conservative (curl-free) hai; yeh degenerate cell hai jahan inside integrand identically vanish ho jaata hai.

Step 2 — Right side. $\int_{M} 0 d x d y = 0.$ Yeh step kyun? Inside integrand $Q_{x} - P_{y}$ identically zero hai, toh chahe loop konsa bhi region $M$ enclose kare, har patch par zero sum karne se zero aata hai. Answer loop ki shape par depend nahin karta.

Step 3 — Exactness view se confirm karo. Kyunki $ω = d (x^{2} y)$ exact hai, $d$ dobara lagane se $d ω = d (d (x^{2} y)) = 0$ milta hai $d^2=0$ se. Toh right side hai $\int_{M} d ω = \int_{M} d (d (x^{2} y)) = \int_{M} 0 = 0,$ jo Step 2 se match karta hai aur boundary integral $\oint_{\partial M} ω = 0$ bhi force karta hai. Yeh step kyun? Yeh dikhata hai ki zero structural hai, lucky nahin: ek exact form ka exterior derivative vanish ho jaata hai ( $d^{2} = 0$ ), toh uska inside integral — aur isliye uska loop integral — har closed loop ke liye zero hona chahiye.

Verify: dono readings $0$ dete hain. ✅ Degenerate cell handle ho gaya — kisi bhi closed curve ke around ek conservative field hamesha zero hota hai.

Cell D — 2D divergence: expanding radial field, dono readings

Figure — Unification — all three theorems as generalized Stokes

Worked example Example 4 (Cell D)

Maano $F = (x, y)$ unit disk par. Outward flux do tareekon se compute karo. (Yeh parent note ka unification example hai — yahan poora worked out.)

Forecast: arrows saare outward point karte hain aur radius ke saath badhte hain. Positive ya negative net flux?

Step 1 — Right side: inside par divergence. $\nabla \cdot F = \partial_{x} x + \partial_{y} y = 1 + 1 = 2$ , toh $\iint_{M} 2 d A = 2 \cdot π = 2 π .$ Yeh step kyun? 2D mein flux form ka exterior derivative $(\nabla \cdot F) d x \land d y$ hota hai; figure mein red outward arrows dekho — divergence measure karta hai ki woh kitna spread apart hote hain.

Step 2 — Left side: boundary circle ke across flux. Unit circle par outward unit normal $\overset{n}{^} = (cos t, sin t)$ hai, aur wahan $F = (cos t, sin t)$ hai, toh $F \cdot \overset{n}{^} = cos^{2} t + sin^{2} t = 1$ . Arclength element yahan $d s = d t$ hai aur total length $2 π$ hai: $\oint_{\partial M} F \cdot \overset{n}{^} d s = \int_{0}^{2 π} 1 d t = 2 π .$ Yeh step kyun? Figure mein teal boundary arrows fence cross karne wala flux hain — yeh "edge par chalo" reading hai, flux density $F \cdot \overset{n}{^}$ ko arclength $d s$ ke against integrate karte hue.

Verify: dono $= 2 π$ . ✅ Positive, jaise forecast tha: outward-spreading field ⇒ positive net outflow.

Cell E — Full 3D Divergence theorem

Worked example Example 5 (Cell E)

Maano $F = (x, y, z)$ aur $M$ solid unit ball. Verify karo $\iint_{\partial M} F \cdot d S = ∭_{M} \nabla \cdot F d V$ .

Forecast: divergence ek constant hai. Use ball ki volume se multiply karo — kya milta hai?

Step 1 — Flux 2-form aur uska $d$ build karo. $ω = x d y \land d z + y d z \land d x + z d x \land d y$ , aur parent ke rule se $d ω = (\partial_{x} x + \partial_{y} y + \partial_{z} z) d x \land d y \land d z = 3 d V .$ Yeh step kyun? $d$ ka har term sirf tab survive karta hai jab woh missing coordinate ko hit kare; baaki wedge karke zero ho jaate hain. Jo bachta hai woh divergence ka sum hai — Divergence Theorem dekho.

Step 2 — Right side: solid ball par integrate karo. Unit ball ka volume $\frac{4}{3} π$ hai, toh $∭_{M} 3 d V = 3 \cdot \frac{4}{3} π = 4 π .$ Yeh step kyun? Integrand constant divergence $3$ hai, toh triple integral sirf $3$ times ball ka volume hai; yeh right-side "solid ke har chhote chunk par source strength add karo" reading hai.

Step 3 — Left side: sphere se bahar flux. Unit sphere par outward normal $\overset{n}{^} = (x, y, z)$ (position vector) hai, toh $F \cdot \overset{n}{^} = x^{2} + y^{2} + z^{2} = 1$ . Unit sphere ki surface area $4 π$ hai, aur $d S = \overset{n}{^} d S$ : $\iint_{\partial M} F \cdot d S = \int_{sphere} 1 d S = 4 π .$ Yeh step kyun? Boundary par field exactly outward normal hai, toh flux density har jagah $1$ hai; total flux = surface area.

Verify: dono $= 4 π$ . ✅ Constant divergence $\times$ volume surface flux se match karta hai — $k = 3$ ka costume kaam karta hai.

Cell F — 3D, degenerate: incompressible field, net flux zero

Worked example Example 6 (Cell F)

Maano $F = (y, - x, 7)$ kisi bhi closed surface par (unit ball lo). Verify karo ki net flux $0$ hai divergence se.

Forecast: teesra component ek constant $7$ hai. Kya ek constant divergence mein contribute karta hai?

Step 1 — Divergence compute karo. $\nabla \cdot F = \partial_{x} (y) + \partial_{y} (- x) + \partial_{z} (7) = 0 + 0 + 0 = 0.$ Yeh step kyun? Zero divergence wala field incompressible hota hai — jo bhi kisi region mein flow karta hai woh bahar bhi nikalta hai. Yeh degenerate 3D cell hai.

Step 2 — Right side. $∭_{M} 0 d V = 0.$ Yeh step kyun? Inside integrand divergence hai, jo yahan identically zero hai, toh solid ke har chunk par use sum karne se zero aata hai chahe ball kitni bhi badi ho — right side surface ko touch karne se pehle hi $0$ ho jaata hai.

Step 3 — Boundary side interpret karo. Divergence Theorem se, closed sphere ke across total outward flux isliye $0$ bhi hona chahiye: kisi bhi patch ka outflow theek dusri jagah se inflow se cancel hota hai. Yeh step kyun? Yeh dikhata hai ki zero divergence zero net flux force karta hai chahe surface ki shape koi bhi ho — yeh ek strong structural fact hai, lucky cancellation nahin.

Verify: flux $= 0$ . ✅ Degenerate 3D case handle ho gaya. (Constant $7$ bhi kuch contribute nahin karta — uska derivative zero hai.)

Cell G — The $d^{2} = 0$ trap

Worked example Example 7 (Cell G)

Maano $f (x, y) = x^{3} y - sin (x y)$ . Dikhao ki $\int_{M} d (df) = 0$ kisi bhi region $M$ par, aur yeh $\oint_{\partial M} df$ ke barabar hai.

Forecast: $df$ already ek derivative hai. Parent note kya kehta hai $d$ ko do baar apply karne ke baare mein?

Step 1 — Pehla derivative. $df = f_{x} d x + f_{y} d y .$ Yeh step kyun? Dobara $d$ apply karne se pehle humein $1$ -form $df$ chahiye.

Step 2 — Doosra derivative, mixed partials dekho. $d (df) = (f_{y x} - f_{x y}) d x \land d y .$ Mixed partials ki equality se $f_{x y} = f_{y x}$ , toh $d (df) = 0$ identically. Yeh step kyun? Yeh $d^2=0$ hai: exterior derivative do baar apply karne se hamesha vanish hota hai. Form language mein yeh $\nabla \times \nabla f = 0$ hai.

Step 3 — Boundary side. Stokes kehta hai $\oint_{\partial M} df = \int_{M} d (df) = 0$ . Directly: $df$ exact hai, aur $\int_{\partial M} df = f (end) - f (start) = 0$ kisi bhi closed loop ke around. Yeh step kyun? Inside ( $d^{2} = 0$ ) aur boundary ( $\partial^{2} = 0$ , ek closed loop ke koi endpoints nahin) dono zero dete hain — yeh do facts Stokes ke through dual hain.

Verify: $d (df) = 0$ aur uska loop integral $= 0$ . ✅

Cell H — Stokes (curl) ek curved surface par: sirf boundary matters

Worked example Example 8 (Cell H)

Maano $F = (- y, x, 0)$ . $M$ = upper unit hemisphere ( $z \geq 0$ ) lo, jiska boundary $\partial M$ equator circle $x^{2} + y^{2} = 1$ hai plane $z = 0$ mein, counterclockwise oriented upar se dekha jaaye. Stokes' Theorem (curl) verify karo.

Forecast: surface curved hai aur upar bulge karti hai. Lekin boundary sirf flat equator hai. Kaunsa easier hai — aur kya unhe phir bhi agree karna chahiye?

Step 1 — Curl compute karo. $\nabla \times F = (\partial_{y} 0 - \partial_{z} x, \partial_{z} (- y) - \partial_{x} 0, \partial_{x} x - \partial_{y} (- y)) = (0, 0, 2) .$ Yeh step kyun? Stokes' curl theorem $(\nabla \times F) \cdot \overset{n}{^}$ ko surface par integrate karta hai; pehle curl vector chahiye. Woh seedha upar point karta hai magnitude $2$ ke saath.

Step 2 — Right side, flat-cap shortcut use karke. Yahan Stokes ka fayda hai: curl ka surface integral sirf boundary par depend karta hai, toh hum bulging hemisphere ko flat disk se replace kar sakte hain jise woh bound karta hai (same equator, Orientation and Induced Boundary Orientation se). Flat disk par $\overset{n}{^} = (0, 0, 1)$ , toh $(\nabla \times F) \cdot \overset{n}{^} = 2$ aur $d S = \overset{n}{^} d S$ : $\iint_{disk} (\nabla \times F) \cdot d S = \iint_{disk} 2 d S = 2 \cdot π = 2 π .$ Yeh step kyun? Figure mein plum boundary circle hi ek cheez hai jo integral "dekhta" hai; curved cap aur flat cap dono use share karte hain, toh dono same number dete hain.

Step 3 — Left side: equator ke around circulation. $x = cos t, y = sin t, z = 0$ ke saath: $F = (- sin t, cos t, 0)$ , $d r = (- sin t, cos t, 0) d t$ : $\oint_{\partial M} F \cdot d r = \int_{0}^{2 π} (sin^{2} t + cos^{2} t) d t = 2 π .$ Yeh step kyun? Yeh "boundary par chalo" left-side reading hai: hum field ka equator ke along push add karte hain, aur Stokes ke hisaab se yeh Step 2 mein compute ki gayi curl flux ke barabar hona chahiye.

Verify: dono $= 2 π$ . ✅ Curved surface aur flat disk agree karte hain kyunki Stokes sirf shared boundary ki parwah karta hai.

Cell I — Word problem: tank se fluid nikalna

Worked example Example 9 (Cell I)

Ek cubical tank $[0, 1]^{3}$ (metres) mein paani ka velocity field $v = (x, 2 y, 3 z)$ metres per second mein hai. Total kitne cubic metres per second tank se nikalte hain?

Forecast: field har direction mein speed up karta hai, toh paani "create" ho raha hai (ek source). Positive net outflow — lekin kitna?

Step 1 — Yeh ek flux question hai, yeh pehchano. Net outflow $= \iint_{\partial M} v \cdot d S$ , cube ki surface ke across $v$ ka flux, jahan $d S = \overset{n}{^} d S$ measure karta hai ki kitna paani har area patch $d S$ cross karta hai. Yeh step kyun? "Boundary ke across per second kitna nikalta hai" exactly Divergence Theorem ka left side hai.

Step 2 — Easy inside integral mein convert karo. $\nabla \cdot v = 1 + 2 + 3 = 6 s^{- 1}$ (ek constant), toh $∭_{M} 6 d V = 6 \cdot (1)^{3} = 6.$ Yeh step kyun? Chhe faces par sum karna tedious hai; divergence theorem surface ko volume se trade karta hai, jahan constant integrand use trivial bana deta hai — sirf $6$ times cube ka volume $1 m^{3}$ .

Verify — units ke saath: $6 s^{- 1} \times 1 m^{3} = 6 m^{3} / s$ . Positive jaise forecast tha. ✅ Tank $6$ cubic metres per second shed karta hai (ek source field). Units kaam karte hain: $[s^{- 1}] \cdot [m^{3}] = [m^{3} / s]$ , ek flow rate.

Cell J — Exam twist: galat orientation sign flip karta hai

Worked example Example 10 (Cell J)

Example 2 ( $ω = - y d x + x d y$ , unit disk) dobara consider karo, lekin ek student boundary clockwise walk karta hai (reversed orientation). Unhe kya milta hai, aur kyun yeh "galat" hai?

Forecast: field ki true circulation $+ 2 π$ hai. Walk reverse karne se kya hona chahiye?

Step 1 — Parametrization reverse karo. Clockwise: $x = cos t, y = - sin t$ , toh $d x = - sin t d t$ , $d y = - cos t d t$ : $\oint_{cw} (- y d x + x d y) = \int_{0}^{2 π} (- sin^{2} t - cos^{2} t) d t = - 2 π .$ Yeh step kyun? Travel ki direction reverse karne se har $d r$ negate ho jaata hai, integral ka sign flip ho jaata hai.

Step 2 — Inside integral se compare karo. Right side fixed hai: $\int_{M} 2 d A = + 2 π$ (region aur uska induced orientation nahin badla). Yeh step kyun? Stokes sirf tab hold karta hai jab $\partial M$ induced orientation carry kare. Clockwise walk opposite orientation use karta hai, toh $\int_{\partial M} ω = - 2 π \neq = + 2 π$ .

Step 3 — Fix. Plane mein ek region ke liye induced boundary orientation counterclockwise hai ("region on your left"). Use use karne se $+ 2 π = + 2 π$ restore ho jaata hai. Yeh step kyun? Yeh parent note se "orientation ignore karo" wali galti concrete bani hai: yeh magnitude nahin, sign ka error hai — aur yeh total hai, partial nahin.

Verify: clockwise $- 2 π$ deta hai; correct (ccw) $+ 2 π = \int_{M} d ω$ deta hai. ✅ Orientation decorative nahin hai — woh sign ka maalik hai.

Recall Kaunsa cell kaunsa hai — quick self-test

Field $(x, y)$ disk par, flux ::: Cell D, answer $2 π$ Field $(x, y, z)$ ball par, flux ::: Cell E, answer $4 π$ $Q_{x} - P_{y} = 0$ ek closed loop ke around ::: Cell C, answer $0$ $\nabla \cdot F = 0$ ek closed surface par ::: Cell F, net flux $0$ Boundary direction reverse karna ::: Cell J, $\int_{\partial M} ω$ ka sign flip karta hai $d (df)$ ko kisi bhi region par integrate karna ::: Cell G, hamesha $0$ by $d^{2} = 0$

Connections

Unification — all three theorems as generalized Stokes — parent identity jise yeh examples exercise karte hain
Fundamental Theorem of Calculus — Cell A
Green's Theorem — Cells B, C, D
Stokes' Theorem (curl) — Cell H
Divergence Theorem — Cells E, F, I
Exterior Derivative aur d squared equals zero — Cell G
Differential Forms and the Wedge Product — har $d ω$ computation
Orientation and Induced Boundary Orientation — Cell J