Question bank — Stokes' theorem — statement, curl-circulation connection

False. $S$ koi bhi piecewise-smooth oriented surface ho sakta hai — ek dome, ek saddle, ek crumpled sheet. Flatness sirf compute karne ke liye convenient hai; theorem kabhi iske liye nahi kehta.

True. Right side zero vector ko integrate karta hai, jo $0$ deta hai, isliye circulation bhi $0$ hona chahiye. Ye exactly "locally conservative" special case hai (dekho Conservative Fields & Potential Functions).

False. Ek vanishing loop coincidence ya symmetry se ho sakta hai. Conservative ka matlab hai har closed loop $0$ deta hai, jo force karta hai ki $\nabla \times \mathbf{F}=0$ everywhere ho (ek simply-connected region pe).

True, bas zaruri hai ki $\nabla \times \mathbf{F}$ unke beech har jagah defined ho. Dono same ${\oint }_C\mathbf{F}\cdot d\mathbf{r}$ ke barabar hain, isliye dono ek doosre ke barabar hone chahiye — ye "easy surface pe swap karo" wali trick hai.

False. Iske liye sirf continuous first partial derivatives ($C^1$) chahiye. Conservative fields wo boring case hai jahan dono sides zero hain.

False. 3D mein curl ek vector hai jisme teen components hain; Green ka scalar $Q_x-P_y$ sirf uska $\hat{\mathbf{k}}$-component hai (wo tukda jo plane se bahar point karta hai).

False. Traversal reverse karne se ${\oint }_C\mathbf{F}\cdot d\mathbf{r}$ ka sign flip ho jaata hai. Equality sahi rakhne ke liye tumhe $\hat{\mathbf{n}}$ bhi flip karna hoga, jo flux flip karta hai — right-hand rule unhe ek saath lock rakhta hai (dekho Orientation & the Right-Hand Rule).

True. $S$ ko $xy$-plane mein flat lo jahan $\hat{\mathbf{n}}=\hat{\mathbf{k}}$ ho; tab $(\nabla \times \mathbf{F})\cdot \hat{\mathbf{k}}=Q_x-P_y$ aur Stokes exactly Green's Theorem mein collapse ho jaata hai.

False. Theorem ko $\mathbf{F}$ ki zarurat hai ki wo saare $S$ pe $C^1$ ho. Ek singularity hypotheses mein ek hole bana deti hai; tumhe us point ko exclude karna hoga (jaise use cut out karo aur uska chhota boundary loop add karo).

True, lekin tab $C$ uski saari boundary hai — bahari rim aur andar ki rim dono, har ek right-hand rule se oriented. Circulation ka matlab hai har boundary curve ke upar sum karna.

$d\mathbf{S}$ ke saath dot missing hai. Curl ek vector hai; tumhe ise surface normal pe project karna hoga: $(\nabla \times \mathbf{F})\cdot d\mathbf{S}=(\nabla \times \mathbf{F})\cdot \hat{\mathbf{n}}dA$. Dot drop karna silently assume karta hai ki curl straight surface se hoke guzarta hai.

Sign determined hai, free nahi. Ek baar jab tum $C$ ki traversal fix kar lo, right-hand rule $\hat{\mathbf{n}}$ fix kar deta hai; doosra sign choose karna us orientation ko contradict karta hai jo tumne already choose ki thi aur galat answer deta hai.

Symmetry conservativeness nahi hai. Iska curl $(0,0,2)\ne 0$ hai, isliye ye circulate karta hai — unit-circle loop $2\pi$ deta hai, $0$ nahi.

Stokes circulation ko curl $\nabla \times \mathbf{F}$ ke flux se relate karta hai, $\mathbf{F}$ ke flux se nahi. $\mathbf{F}$ ka flux wo hai jise Divergence Theorem (Gauss) handle karta hai — ek alag theorem.

Sirf curl-flux equal hona forced hai. Do surfaces se hoke $\mathbf{F}$ ka ordinary flux alag ho sakta hai — jitna $\iiint \nabla \cdot \mathbf{F}dV$ unke beech hai (divergence theorem), jo zero hona zaroori nahi.

Vector area element $(1,1,1)dA$ mein already tilted graph ki $\sqrt{3}$ stretch baki hui hai; $(1,1,1)$ ki length $\sqrt{3}$ hai, $1$ nahi. Unit normal $\frac{1}{\sqrt{3}}(1,1,1)$ hai — dono ko confuse mat karo.

Har interior edge do neighboring patches se share hoti hai aur ek baar har taraf se chali jaati hai; wo opposite traversals cancel ho jaate hain. Sirf boundary edges ke paas koi neighbor nahi hota cancel karne ke liye, isliye wo $C$ ke roop mein bachti hain.

Normal $\hat{\mathbf{n}}$ wali area $\Delta A$ ki ek tiny loop ke liye, circulation $\approx (\nabla \times \mathbf{F})\cdot \hat{\mathbf{n}}\Delta A$ hoti hai. $\Delta A$ se divide karo aur loop ko shrink karo toh curl component milta hai — literally circulation density.

Sirf whi spin jo surface ke apne axis ke saath aligned hai woh surface se hoke guzarti hai; surface ke sideways spin ka us surface ke flux mein koi contribution nahi. Dot product ka $\cos \theta$ factor exactly aligned part rakhta hai.

Stokes ke dono sides ek reversal ke under sign flip karte hain (left ke liye traversal, right ke liye normal). Unhe ek consistent rule se lock karna equality ko equality rakhta hai na ki "ek sign tak" wali cheez.

Stokes ka right side $S$ pe sirf uske edge $C$ ke through depend karta hai; $C$ share karne wale do surfaces same ${\oint }_C$ share karte hain, isliye same curl-flux. Jo bhi surface integral ko simplest banaye use pick karo.

Ek conservative field $\mathbf{F}=\nabla f$ hota hai, aur iska curl $\nabla \times \nabla f=0$ identically hai. Tab Stokes har loop ki circulation ko $0$ ka flux banata hai, jo $0$ hai.

Kyunki ye $\mathbf{E}$ pe apply kiya hua Stokes hi hai: loop ke around $\mathbf{E}$ ki circulation, iske through $\nabla \times \mathbf{E}$ ke flux ke barabar hai, jise Faraday changing magnetic flux ke barabar set karta hai (dekho Maxwell's Equations).

Zero. Integrate karne ke liye koi length nahi hai, aur jo surface isse bound karta hai uski area zero hai, isliye dono sides consistently $0$ mein collapse ho jaate hain.

Iska boundary $C$ empty hai, isliye ${\oint }_C\mathbf{F}\cdot d\mathbf{r}=0$. Isliye ${\iint }_S(\nabla \times \mathbf{F})\cdot d\mathbf{S}=0$: kisi bhi closed surface se curl-flux hamesha zero hota hai.

Zero, kyunki $(\nabla \times \mathbf{F})\cdot \hat{\mathbf{n}}=0$ har point pe hai — spin surface ke parallel chalta hai aur iska kuch bhi surface se hoke nahi guzarta.

Nahi, sirf iska boundary matter karta hai (Stokes ke through) ya equivalently uska projected vector area. Same rim wale do dented surfaces ek constant curl ka same flux dete hain.

Tab $\nabla \times \mathbf{F}=0$ zero circulation force nahi karta, kyunki tum hole se bachte hue ek surface fill nahi kar sakte. Hole ko encircle karne wale loops nonzero circulation rakh sakte hain.

Zero. Ek constant field ka $\nabla \times \mathbf{F}=0$ hota hai, isliye iska curl-flux $0$ hai; equivalently, ek closed loop chalte hue ghar wapas aate ho aur pushes ka sum kuch nahi hota.