Question bank — Green's theorem — proof sketch, both forms

False. $C$ simple bhi honi chahiye (koi self-crossings nahi) aur $P,Q$ ke continuous partials $D$ ke poore region par hone chahiye; andar koi singularity ho toh hypothesis toot jaati hai.

False. Ek loop ka zero dena sirf ek data point hai; conservative ka matlab hai ki har closed loop zero deti hai, jiske liye $Q_x=P_y$ throughout ek simply-connected region mein hona zaroori hai.

False. Green's theorem ek 2D Fundamental Theorem hai: proof mein inner $y$-integral FTC se boundary values par collapse ho jaata hai, isliye poora result ek dimension upar jaake "boundary remembers the interior" hai.

False. Ye ek hi theorem hai do field arrangements ke liye; circulation ko $(\tilde{P},\tilde{Q})=(-Q,P)$ par apply karne se bina kisi naye kaam ke $(P,Q)$ ke liye flux form milta hai.

True — bas partials $D$ ke poore region par continuous honi chahiye. Agar koi hole/singularity hai, toh double integral ki hypothesis fail ho jaati hai aur loop integral nonzero ho sakta hai.

False. Yeh sirf line integral side ko negate karta hai; fixed region $D$ par double integral unchanged rehta hai. Toh ek clockwise loop $-{\iint }_D(Q_x-P_y)dA$ ke barabar hota hai.

True. Stokes' ${\oint }_C\mathbf{F}\cdot d\mathbf{r}$ ko ek surface ke through curl ke flux se relate karta hai; surface ko $xy$-plane mein flat lene par $\text{curl}\cdot \hat{\mathbf{k}}$ reduce ho jaata hai $Q_x-P_y$ mein, aur circulation form wapas mil jaata hai.

True. Dono kehte hain ki boundary ke across total outward flux andar ke divergence ke integral ke barabar hai; Green's flux form plane mein wahi statement hai.

Error: punctured disk simply connected nahi hoti — hole ke around ek loop ek point tak shrink nahi ho sakta, isliye curl-free wahan conservative force nahi karta.

Error: field origin par singular hai, toh Green's ki hypothesis kisi bhi $D$ par fail hoti hai jo usse contain kare; actual loop integral $2\pi$ hai, $0$ nahi.

Error: yeh divergence hai, flux form mein use hota hai. Circulation integrand difference $Q_x-P_y$ hai, yaani scalar curl.

Yeh asar mein sahi hi hai — $P_y-Q_x$ par swap karna reversed orientation ke liye minus lagane jaisa hi hai. Bas dhyan raho ki ek baar flip karo, do baar nahi.

Error: vertical segment par $x$ constant hota hai isliye $dx=0$; woh pieces $\oint Pdx$ mein kuch contribute nahi karti.

Error: kuch regions (jaise annulus, ya galat taraf se padhi gayi L-shape) poori tarah Type I nahi hoti; tumhe Type I/II pieces mein kaat na padta hai jahan internal cuts cancel ho jaate hain.

Error: har cut do baar opposite directions mein traverse hota hai, isliye woh Line Integrals cancel ho jaate hain, sirf outer boundary bachti hai.

Kyunki $Q_x-P_y$ counterclockwise (positive) scalar curl hai; uske sign convention se match karne ke liye tumhe aise chalna chahiye ki region tumhare left par rahe.

Agar $Q_x-P_y$ ek symmetric region par odd function hai (jaise $y$ mein odd integrand ek disk par), toh double integral cancellation se $0$ hota hai, isliye koi computation kiye bina loop integral $0$ hai.

Unit tangent $\mathbf{T}=(dx/ds,dy/ds)$ ko 90° clockwise rotate karna outward point karta hai jab region tumhare left par ho; woh rotation $(a,b)\mapsto (b,-a)$ bhejta hai.

Taaki har piece ek single-valued graph $y=g(x)$ ho; counterclockwise top curve right-to-left chalti hai, jo woh minus sign deti hai jo FTC ke boundary evaluation se match karta hai.

Theorem ko poore enclosed region par continuous partials chahiye; ek chote loop se singularity hata dene par hypothesis wapas aati hai, aur woh inner loop (oriented taaki region tumhare left par rahe, yaani hole ke around clockwise) boundary sum mein enter karta hai.

Circulation loop ke around net spin (curl) measure karta hai; flux uske through net outflow (divergence) measure karta hai — Green's dono scalar quantities ko boundary-versus-interior statements ke roop mein package karta hai.

Kyunki $Q_x-P_y=\frac{1}{2}-(-\frac{1}{2})=1$, toh double integral ${\iint }_{D}1dA=\text{Area}(D)$ ban jaata hai; field ko engineer kiya gaya hai taaki integrand exactly $1$ ho.

Dono sides trivially $0$ hain; yeh consistent hai lekin koi information nahi deta — ek degenerate sanity check hai.

Double integral $0$ hai aur boundary ki length zero hai, toh dono sides vanish ho jaati hain; theorem limit mein vacuously hold karta hai.

Yeh simply connected nahi hai, isliye uski boundary do loops hai; outer ko counterclockwise orient karo aur inner ko clockwise (region throughout tumhare left par), aur dono line integrals ka sum karo.

Theorem fail ho sakta hai — proof ko $D$ contain karne wale open set par continuous partial derivatives chahiye, sirf $P,Q$ ki continuity nahi.

Koi baat nahi — Green's theorem sirf piecewise-smooth maangta hai, isliye finite corners allowed hain; tum smooth arcs par integrate karte ho aur add kar dete ho.

Theorem directly apply nahi hota kyunki $C$ simple nahi hai; isko simple sub-loops mein split karo, har ek ke saath apna consistent orientation, aur unhe alag treat karo.

Positive: $\nabla \cdot \mathbf{F}=2$, toh flux $={\iint }_{D}2dA=2A$ — radially-outward field hamesha bahar ki taraf push karta hai.