Green's theorem — proof sketch, both forms

4.4.29 · Maths › Multivariable Calculus

YEH KEHTA KYA HAI

Yeh dono forms ek hi theorem hain jo do alag field arrangements ke liye likhi gayi hain — ek prove karo toh doosri free mil jaati hai (neeche dikhaya hai).

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

ORIENTATION KYUN MATTER KARTA HAI

PROVE KAISE KAREIN (derivation from scratch)

Hum circulation form prove karenge. Trick yeh hai: ise do hisson mein split karo aur dono alag prove karo. $\oint_{C} P d x = - \iint_{D} \frac{\partial P}{\partial y} d A (⋆)$ $\oint_{C} Q d y = + \iint_{D} \frac{\partial Q}{\partial x} d A (⋆ ⋆)$ Inhe add karo aur poora theorem mil jaata hai.

Step 1: $(⋆)$ ko ek "Type I" region pe prove karo

Type I region woh hai jise aise likha ja sake: $D = {(x, y) : a \leq x \leq b, g_{1} (x) \leq y \leq g_{2} (x)} .$ Yeh shape kyun? Kyunki phir har $x$ ke liye region ek vertical strip hai bottom curve $y = g_{1} (x)$ se top curve $y = g_{2} (x)$ tak — $y$ mein integrate karna aasaan hai.

Right-hand side. $- P_{y}$ ko $D$ ke upar integrate karo, pehle inner $y$ -integral karo: $- \iint_{D} \frac{\partial P}{\partial y} d A = - \int_{a}^{b} \int_{g_{1} (x)}^{g_{2} (x)} \frac{\partial P}{\partial y} d y d x .$ Yeh step kyun? Inner integral $y$ mein ek perfect derivative hai, toh Fundamental Theorem of Calculus use collapse kar deta hai: $\int_{g_{1} (x)}^{g_{2} (x)} \frac{\partial P}{\partial y} d y = P (x, g_{2} (x)) - P (x, g_{1} (x)) .$ Isliye $-\iint_D P_y\,dA = \int_a^b \big[P(x,g_1(x)) - P(x,g_2(x))\big]\,dx. \tag{RHS}$

Left-hand side. Boundary $C$ ko bottom curve $C_{1}$ ( $y = g_{1}$ , left→right) aur top curve $C_{2}$ ( $y = g_{2}$ , right→left) mein split karo, saath mein do vertical sides bhi. Split kyun? Kyunki $\oint P d x$ sirf horizontal motion "dekhta" hai — vertical sides pe $d x = 0$ , toh unka koi contribution nahi hota.

= \int_a^b P(x,g_1(x))\,dx + \int_b^a P(x,g_2(x))\,dx.$$ *$C_2$ pe limits flip kyun hain?* Counterclockwise orientation top curve ko **right se left** traverse karta hai, yaani $x$ $b$ se $a$ tak jaata hai. Standard order ke liye flip karo: $$\oint_C P\,dx = \int_a^b \big[P(x,g_1(x)) - P(x,g_2(x))\big]\,dx. \tag{LHS}$$ **(LHS) = (RHS).** Ho gaya: $(\star)$ sach hai. ✅ ### Step 2: $(\star\star)$ ko "Type II" region pe prove karo *Mirror argument* se — $D = \{c\le y\le d,\ h_1(y)\le x\le h_2(y)\}$ likho, pehle $x$ mein $Q_x$ integrate karo, FTC use karo, aur left/right boundary pieces match karo (ab vertical sides integral carry karti hain kyunki $dy\ne 0$). Wohi logic hai jisme $x,y$ ke roles swap hain. Isse $(\star\star)$ milta hai. ### Step 3: General regions ko decomposition se handle karo > [!intuition] Interior cuts cancel ho jaate hain > Jo region simultaneously Type I aur Type II nahi hai use **sub-pieces mein kaata jaata hai** jo hain. Theorem ko har piece pe apply karo. Jab line integrals add karo, toh har **internal cut do baar opposite directions mein traverse hoti hai**, isliye woh contributions pairwise cancel ho jaate hain. Sirf outer boundary bachti hai — poori region pe Green's theorem mil jaata hai. ### Step 4: Flux form free mein milti hai Circulation form lo aur use rotated field $\mathbf{G} = (-Q', P')$ pe apply karo... zyada clean: $\oint \mathbf{F}\cdot\mathbf{n}\,ds$ se shuru karo. Unit tangent $\mathbf{T}=(dx/ds, dy/ds)$ wali curve pe, outward normal hai $\mathbf{n}=(dy/ds, -dx/ds)$. *Kyun?* Tangent ko **clockwise 90°** rotate karna outward point karta hai jab region tumhari left pe ho. Toh $$\mathbf{F}\cdot\mathbf{n}\,ds = (P,Q)\cdot(dy,-dx) = P\,dy - Q\,dx.$$ Ab **circulation form** ko field $(\tilde P,\tilde Q)=(-Q,P)$ pe apply karo: $$\oint -Q\,dx + P\,dy = \iint_D\Big(\frac{\partial P}{\partial x} - \frac{\partial(-Q)}{\partial y}\Big)dA = \iint_D (P_x + Q_y)\,dA.$$ Left side exactly $\oint \mathbf{F}\cdot\mathbf{n}\,ds$ hai. Flux form proved. ✅ --- ## Worked Examples > [!example] 1 — Boundary se Area nikalna > Dikhao ki $\text{Area}(D) = \tfrac12\oint_C (x\,dy - y\,dx)$ aur ise unit disk pe use karo. > > $P=-y/2,\ Q=x/2$ choose karo. Toh $Q_x - P_y = \tfrac12-(-\tfrac12)=1$. > *Yeh choice kyun?* Hum integrand ko $1$ ke barabar engineer karte hain, toh double integral ban jaata hai $\iint_D 1\,dA = \text{Area}$. > $$\oint_C \tfrac12(x\,dy - y\,dx) = \iint_D 1\,dA = \text{Area}(D).$$ > Unit circle $x=\cos t,\ y=\sin t$: $x\,dy - y\,dx = \cos^2 t + \sin^2 t = 1$, toh > $$\text{Area}=\tfrac12\int_0^{2\pi}1\,dt = \pi.\ ✅$$ > [!example] 2 — Direct verification > $\mathbf{F}=(P,Q)=(y^2, 3xy)$ unit circle ke around. > $$Q_x - P_y = 3y - 2y = y.$$ > *Pehle yeh kyun compute karein?* Double integral usually line integral se aasaan hota hai, aur yeh humein expected answer bata deta hai. > $$\iint_D y\,dA = 0$$ > *Zero kyun?* Disk $y=0$ ke baare mein symmetric hai aur $y$ odd hai $y$ mein — positive aur negative halves cancel ho jaate hain. Toh $\oint_C \mathbf{F}\cdot d\mathbf{r}=0$ bina line integral compute kiye. > [!example] 3 — Flux form > $\mathbf{F}=(x,y)$ unit disk ki boundary ke across. > $\nabla\cdot\mathbf{F}=P_x+Q_y = 1+1 = 2$. > $$\oint_C \mathbf{F}\cdot\mathbf{n}\,ds = \iint_D 2\,dA = 2\pi.$$ > *Yeh intuitive kyun hai:* $\mathbf F$ radially outward point karta hai; circle ke through total outward flux area ke saath badhna chahiye — aur badhta hai, $2$ per unit area ki rate se. --- ## Common Mistakes > [!mistake] "Yeh kisi bhi closed curve pe kaam karta hai jo main chahoon." > **Kyun sahi lagta hai:** formula self-contained dikhta hai. **Dikkat:** $C$ **simple** honi chahiye (koi self-crossings nahi) aur $P,Q$ ke continuous partials $D$ ke andar *har jagah* hone chahiye. Origin pe $\mathbf{F}=(-y,x)/(x^2+y^2)$ jaisi singularity **hypothesis tod deti hai**, jisse $\oint = 2\pi$ milta hai lekin $Q_x-P_y=0$. **Fix:** singularities check karo; unhe ek chote hole se exclude karo. > [!mistake] Orientation bhool jaana. > **Kyun sahi lagta hai:** "ek loop toh loop hai." **Fix:** clockwise sign flip kar deta hai. Region apni **left** pe rakho; agar problem clockwise deta hai toh ek minus sign aage lagao. > [!mistake] Dono forms ke integrands confuse karna. > **Kyun sahi lagta hai:** dono partials ke sums/differences hain. **Fix:** circulation = $Q_x - P_y$ (**curl**); flux = $P_x + Q_y$ (**divergence**). Neeche mnemonic hai. --- > [!recall]- Feynman: ek 12-saal ke bache ko samjhao > Socho ek flat maidan hai jisme chhote chhote spinning tops bikre hue hain. Agar tum sab ka **total spin** jaanna chahte ho, toh tumhe har top count nahi karna — tum bas maidan ki boundary pe fence ke chaaro taraf chale jaao, measure karo ki hawa tumhe fence ke *saath saath* kitna dhakkelti hai. Boundary pe push exactly andar ke total spin ke barabar hoti hai! Aur agar tum measure karo ki hawa *fence ke through bahar* kitni nikal rahi hai har jagah, woh equals karta hai kitni "nayi hawa" andar create ho rahi hai. Edge hamesha jaanti hai andar kya ho raha hai. > [!mnemonic] Curl vs Divergence > **C**irculation mein **C**url hota hai → "**Q**uick **x** beats **P**lodding **y**": $Q_x - P_y$ (ek **difference**). > **D**ivergence **D**iagonal/straight-out hai → $P_x + Q_y$ (ek **sum**, dono "apne" derivatives). --- ## Flashcards #flashcards/maths Green's theorem (circulation form) kya hai? ::: $\oint_C P\,dx+Q\,dy=\iint_D (Q_x-P_y)\,dA$, $C$ positively oriented simple closed curve jo $D$ ko bound karti hai. Green's theorem (flux form) kya hai? ::: $\oint_C \mathbf F\cdot\mathbf n\,ds=\iint_D(P_x+Q_y)\,dA=\iint_D \nabla\cdot\mathbf F\,dA$. $Q_x-P_y$ kaunsa scalar represent karta hai? ::: $\mathbf F=(P,Q)$ ka scalar (z-component of the) curl. $C$ ki required orientation kya hai? ::: Counterclockwise (positive); region tumhari left pe. $\oint P\,dx$ prove karte waqt vertical sides kyun matter nahi karte? ::: Vertical segments pe $dx=0$ hota hai, toh unka $\int P\,dx$ mein koi contribution nahi hota. Proof mein inner integral kaunsa classical theorem collapse karta hai? ::: Fundamental Theorem of Calculus ($\int P_y\,dy$ collapse karta hai). General regions ko kaise handle kiya jaata hai? ::: Type I/II pieces mein kaat do; internal cuts do baar opposite directions mein traverse hoti hain aur cancel ho jaati hain. Green's theorem se area formula kya hai? ::: $\text{Area}=\tfrac12\oint_C(x\,dy-y\,dx)$. Arc length ke terms mein outward normal kya hai? ::: $\mathbf n=(dy/ds,\,-dx/ds)$ (tangent 90° clockwise rotate hua). $D$ ke andar singularity theorem kyun tod deti hai? ::: $P,Q$ ke continuous partials $D$ ke pore pe hone chahiye; singularity hypothesis violate karti hai (e.g. $(-y,x)/r^2$ deta hai $\oint=2\pi$ lekin integrand $=0$). --- ## Connections - [[Fundamental Theorem of Calculus]] — Green's theorem iska 2D generalization hai. - [[Stokes' Theorem]] — Green's, Stokes ka plane mein ek flat region tak restricted version hai. - [[Divergence Theorem]] — flux form 2D divergence theorem hai. - [[Curl and Divergence]] — dono integrands provide karte hain. - [[Line Integrals]] aur [[Double Integrals]] — woh dono sides jo equate ho rahi hain. - [[Conservative Vector Fields]] — agar $Q_x=P_y$ har jagah ho, toh saare loops $0$ dete hain. ## 🖼️ Concept Map ```mermaid flowchart TD GT[Green's theorem] FTC[Fundamental Thm of Calculus] Circ[Circulation tangential form] Flux[Divergence normal form] Orient[Counterclockwise orientation] Curl[Scalar curl Qx minus Py] Split[Split into two halves] TypeI[Type I region] InnerY[Inner y-integral] Bound[Boundary curves C1 and C2] FTC -->|2D ancestor of| GT GT -->|form A| Circ GT -->|form B| Flux Circ -->|same theorem as| Flux Circ -->|integrand is| Curl Curl -->|requires| Orient Orient -->|region on left| GT Circ -->|proof splits into| Split Split -->|proven on| TypeI TypeI -->|do| InnerY InnerY -->|collapsed by| FTC TypeI -->|split boundary into| Bound Bound -->|matches RHS| InnerY ``` ## 🔬 Deep Dive > [!intuition] Aur gehrai mein jao — visual, zero se > Is topic ke step-by-step 3Blue1Brown-style breakdowns. - [[4.4.29 D1 Foundations|D1 · Foundations — har symbol zero se]] - [[4.4.29 D2 Visual Walkthrough|D2 · Visual walkthrough — derivation pictures mein]] - [[4.4.29 D3 Worked Examples|D3 · Worked examples — har scenario]] - [[4.4.29 D4 Exercises|D4 · Exercises — graded, full solutions]] - [[4.4.29 D5 Question Bank|D5 · Question bank — concept traps]]