Worked examples — Stokes' theorem — statement, curl-circulation connection

4.4.32 · D3 · Maths › Multivariable Calculus › Stokes' theorem — statement, curl-circulation connection

Yeh page Stokes' theorem ka drill floor hai. Parent note mein statement aur "curl = circulation density" ka idea build hua tha. Yahan hum har tarah ka problem hit karenge jo yeh theorem throw kar sakta hai — flat surfaces, curved surfaces, zero-curl traps, sign flips, degenerate loops, ek physics word problem, aur ek sneaky exam twist.

Shuru karne se pehle, ek reminder plain words mein taaki koi symbol unearned na rahe:

Scenario matrix

Har Stokes problem inhi cells mein se ek mein rehta hai. Neeche ke examples uss cell ke saath labelled hain jo wo cover karta hai, toh end tak koi cell empty nahi rahegi.

Cell A — Flat surface, constant curl

Figure s01, kya dekhna hai: magenta arrows field $\mathbf{F}=(-3y,3x,0)$ hain — notice karo ki woh CCW wrap karte hain, merry-go-round ki tarah. Edges par orange arrows $C$ ki walking direction dikhate hain (woh bhi CCW). Centre par violet dot with a ring ⊙ normal $\hat{\mathbf{n}}=+\hat{\mathbf{k}}$ hai jo page se bahar tumhari taraf point kar raha hai — right-hand rule (orange arrows ke saath fingers curl karo, thumb bahar) is pairing ko confirm karta hai.

Step 1 — Domain conditions check karo, phir curl compute karo. $\mathbf{F}$ ek polynomial hai → har jagah smooth (condition 2 ✓); triangle ek flat piecewise-smooth surface hai jiska boundary exactly $C$ hai (condition 1 ✓). Ab: $\nabla \times \mathbf{F}=({\partial }_{y}0-{\partial }_z(3x),{\partial }_z(-3y)-{\partial }_{x}0,{\partial }_x(3x)-{\partial }_y(-3y))=(0,0,6).$ Yeh step kyun? Stokes walk-around integral ko curl ke flux se replace karta hai. Toh pehla move hamesha yahi hai: $\mathbf{F}$ ko $\nabla \times \mathbf{F}$ mein convert karo. Yahan sirf $z$-part bachta hai: $Q_x-P_y=3-(-3)=6$.

Step 2 — Flat surface aur uska normal choose karo. Triangle $xy$-plane mein hai, toh $S$ = flat triangular region lo, $\hat{\mathbf{n}}=\hat{\mathbf{k}}=(0,0,1)$ ke saath (upward, right-hand rule se CCW ke saath match karta hai — dekho Orientation & the Right-Hand Rule). Yeh step kyun? Kyunki $C$ pehle se hi ek obvious flat surface ka rim hai, humein koi curved surface invent nahi karna.

Step 3 — Dot karo aur integrate karo. Recall $d\mathbf{S}=\hat{\mathbf{n}}dA$, toh: $(\nabla \times \mathbf{F})\cdot d\mathbf{S}=(0,0,6)\cdot (0,0,1)dA=6dA(\text{constant}).$ ${\iint }_{S}6dA=6\cdot \text{Area}(S).$ Triangle ki legs length $2$ ki hain, toh area $=\frac{1}{2}\cdot 2\cdot 2=2$. ${\oint }_C\mathbf{F}\cdot d\mathbf{r}=6\cdot 2=12.$ Yeh step kyun? Constant integrand matlab integral sirf constant times area hai — koi calculus nahi chahiye.

Verify: Direct line integral. Bottom edge $y=0$ par, $\mathbf{F}=(0,3x,0)$ perpendicular hai $d\mathbf{r}=(dx,0,0)$ ke → contribute karta hai $0$. Hypotenuse aur left edge bhi kaam karte hain; poora CCW walk $12$ deta hai (VERIFY mein checked). Units: length-per-time field ki circulation length loop ke around — consistent. ✓

Cell B — Curved surface, boundary swap

Figure s02, kya dekhna hai: violet bowl paraboloid cap $z=x^2+y^2$ hai; orange flat disk replacement surface $S^{\prime }$ hai height $z=4$ par. Dono magenta rim $C$ se glued hain (circle CCW drawn). Do purple normal arrows dikhate hain $\hat{\mathbf{n}}$ har ek par upward point karta hua — kyunki woh same rim share karte hain, Stokes dono mein se same flux deta hai. Hum aasaan flat wale par integrate karte hain.

Step 1 — Domain check, phir curl. $\mathbf{F}$ har jagah smooth hai (condition 2 ✓); har candidate surface ka boundary exactly $C$ hai (condition 1 ✓). $\nabla \times \mathbf{F}=({\partial }_y{z}^3-{\partial }_{z}x,{\partial }_z(-y)-{\partial }_x{z}^3,{\partial }_{x}x-{\partial }_y(-y))=(0,0,2).$ Yeh step kyun? Notice karo ki $z^3$ curl mein kuch contribute nahi karta (yeh sirf $z$ par depend karta hai, aur $x,y$ ke respect mein iske derivatives zero hain). Scary part already gone hai.

Step 2 — Easiest surface pe swap karo jiska same rim ho. Rim $C$ radius $2$ ka circle hai height $z=4$ par. Flat disk $S^{\prime }$: $\{x^2+y^2\le 4,z=4\}$ ka exactly wahi rim hai. Green/Stokes ke according, curl ka flux kisi bhi surface ke through same hota hai jo $C$ se bounded ho (parent Example 3). Yeh step kyun? Hum drum-skin ko slide kar sakte hain jab tak rim same rahe. Flat sabse aasaan hai.

Step 3 — Disk ke through flux. $\hat{\mathbf{n}}=(0,0,1)$, aur $d\mathbf{S}=\hat{\mathbf{n}}dA$, toh $(\nabla \times \mathbf{F})\cdot d\mathbf{S}=2dA$. Disk area $=\pi (2{)}^2=4\pi$. ${\oint }_C\mathbf{F}\cdot d\mathbf{r}=2\cdot 4\pi =8\pi .$

Verify: $C$ ko parametrize karo: $\mathbf{r}=(2\cos t,2\sin t,4)$, $d\mathbf{r}=(-2\sin t,2\cos t,0)dt$. Phir $\mathbf{F}\cdot d\mathbf{r}=(-2\sin t)(-2\sin t)+(2\cos t)(2\cos t)+0=4dt$; $0$ se $2\pi$ tak integrate karo → $8\pi$. ✓ ($z^3$ term constant $z$ wale curve par ride karta hai, toh $dz=0$ usse kill kar deta hai.)

Cell C — All three curl components nonzero

Figure s03, kya dekhna hai: teen vertices $(1,0,0),(0,1,0),(0,0,1)$ plane $x+y+z=1$ par rehne wale slanted violet triangle span karte hain. Surface $S$ yahi flat triangular patch hai; iska purple normal arrow $\hat{\mathbf{n}}=\frac{1}{\sqrt{3}}(1,1,1)$ origin se door point karta hai. Magenta arrows boundary $C$ ko us right-hand-rule direction mein trace karte hain jo us normal se match kare. Is geometry ko jaanna hi flux integral setup karne deta hai.

Step 1 — Domain check, phir full determinant curl. $\mathbf{F}$ ek polynomial hai → har jagah smooth ✓; $S$ plane $x+y+z=1$ par flat triangular patch hai boundary $C$ ke saath ✓. $\nabla \times \mathbf{F}=\left|\begin{array}{ccc}\hat{\mathbf{i}}& \hat{\mathbf{j}}& \hat{\mathbf{k}}\\ {\partial }_x& {\partial }_y& {\partial }_z\\ yz& xz& xy\end{array}\right|=({\partial }_y(xy)-{\partial }_z(xz),{\partial }_z(yz)-{\partial }_x(xy),{\partial }_x(xz)-{\partial }_y(yz)).$ Har ek compute karo: $(x-x,y-y,z-z)=(0,0,0)$. Yeh step kyun? Humein poora determinant use karna hoga kyunki teeno components present hain. Surprise: yeh field curl-free hai — yeh $f=xyz$ ka gradient hai.

Step 2 — Stokes apply karo. ${\oint }_C\mathbf{F}\cdot d\mathbf{r}={\iint }_{S}0\cdot d\mathbf{S}=0.$ Yeh step kyun? Poore slanted surface par zero curl → zero flux → zero circulation. Yeh conservative case hai: $\mathbf{F}=\nabla (xyz)$, aur conservative field mein closed loop koi net work nahi karta.

Verify: Kyunki $\mathbf{F}=\nabla (xyz)$ aur $C$ closed hai, ${\oint }_C\nabla f\cdot d\mathbf{r}=f(\text{end})-f(\text{start})=0$ loop ke liye. Numerically hum VERIFY mein curl $=(0,0,0)$ bhi confirm karte hain. ✓

Cell D — Zero-curl trap (aur kab sach mein zero hai)

Figure s04, kya dekhna hai: magenta ellipse $C$ hai ($x^2/9+y^2/4=1$), orange CCW arrows ke saath drawn; centre par violet dot ring ⊙ mark karta hai $\hat{\mathbf{n}}=+\hat{\mathbf{k}}$ page se bahar (right-hand rule: fingers orange ke saath, thumb tumhari taraf). Magenta field arrows sab origin se straight outward point karte hain — loop ki motion ke perpendicular har point par — toh woh tumhe kabhi around jaane mein help nahi karte. Yahi perpendicularity hai jo circulation zero banati hai.

Step 1 — Domain check, phir curl. $\mathbf{F}$ har jagah smooth hai ✓; flat elliptical disk ka boundary exactly $C$ hai ✓. $\nabla \times (2x,2y,2z)=({\partial }_{y}2z-{\partial }_{z}2y,{\partial }_{z}2x-{\partial }_{x}2z,{\partial }_{x}2y-{\partial }_{y}2x)=(0,0,0).$ Yeh step kyun? Yeh $\nabla (x^2+y^2+z^2)$ hai, ek pure gradient — kahin koi swirl nahi. Paddle-wheel kabhi spin nahi karta.

Step 2 — Conclude karo. ${\oint }_C\mathbf{F}\cdot d\mathbf{r}={\iint }_{S}0d\mathbf{S}=0.$ Yeh step kyun? Poore surface par zero curl hai aur koi singularity nahi (field har jagah smooth hai), toh Stokes clean zero deta hai. Isko Cell H se contrast karo, jahan ek singularity shortcut ruin kar deti hai.

Verify: Ellipse $\mathbf{r}=(3\cos t,2\sin t,0)$, $d\mathbf{r}=(-3\sin t,2\cos t,0)dt$. $\mathbf{F}\cdot d\mathbf{r}=2(3\cos t)(-3\sin t)+2(2\sin t)(2\cos t)=-18\sin t\cos t+8\sin t\cos t=-10\sin t\cos t$; $0$ se $2\pi$ tak integrate karo → $0$. ✓

Cell E — Sign / orientation flip

Figure s05, kya dekhna hai: magenta arrows Example-1 CCW walk dikhate hain apne out-of-page normal ⊙ ke saath (answer $+12$); violet arrows reversed CW walk dikhate hain apne into-page normal ⊗ ke saath (answer $-12$). Same triangle, same field — sirf "walk direction ↔ normal" ki pairing badli, aur usne sign flip kar diya. Notice karo magenta ⊙ (dot = tumhari taraf aa raha) versus violet ⊗ (cross = door ja raha).

Step 1 — Same curl rakho. $\nabla \times \mathbf{F}=(0,0,6)$ (unchanged — field nahi badla).

Step 2 — Naye walk ke saath compatible rehne ke liye normal flip karo. Right-hand rule: thumb traversal ke saath ⇒ clockwise walk ke liye (upar se) compatible normal neeche point karta hai, $\hat{\mathbf{n}}=(0,0,-1)$. Yeh step kyun? Circulation sign aur normal right-hand rule (domain condition 3) se ek doosre se locked hain. Ek flip karo toh doosra bhi karna hoga, warna Stokes ke dono sides disagree karte hain.

Step 3 — Flux recompute karo. $(0,0,6)\cdot (0,0,-1)dA=-6dA,{\oint }_C\mathbf{F}\cdot d\mathbf{r}=-6\cdot 2=-12.$ Yeh step kyun? Example 1 ke barabar size, opposite sign — exactly wahi mistake jiske baare mein parent note ne warn kiya tha.

Verify: Traversal direction reverse karna har $d\mathbf{r}$ negate karta hai, toh line integral negate ho jaata hai: $+12\to -12$. ✓

Cell F — Degenerate loop (limiting case → curl at a point)

Figure s06, kya dekhna hai: teen nested magenta circles shrinking radius $\epsilon$ ki (large → medium → tiny) sab origin par centered, har ek par ek orange CCW arrow. Har ek ke paas uska ratio (circulation ÷ area) hai. Teeno par label $2$ padhta hai — ratio kabhi nahi badlta jab loop central violet dot par collapse ho jaata hai. Woh constant limiting value hi curl component hai point par: yeh figure "circulation per unit area" ki visual definition hai.

Step 1 — Chhote loop ki circulation. Small disk $S_{\epsilon }$ (area $\pi {\epsilon }^2$, normal $\hat{\mathbf{k}}$) par Stokes se, $d\mathbf{S}=\hat{\mathbf{n}}dA$ use karke: ${\oint }_{C_{\epsilon }}\mathbf{F}\cdot d\mathbf{r}={\iint }_{S_{\epsilon }}(0,0,2)\cdot (0,0,1)dA=2\cdot \pi {\epsilon }^2.$ Yeh step kyun? Chahe loop degenerate (tiny) hi kyu na ho, Stokes obey karta hai; curl yahan constant hai toh flux exactly $2\times$ area hai.

Step 2 — Area se divide karo aur limit lo. $\frac{1}{\pi {\epsilon }^2}\cdot 2\pi {\epsilon }^2=2\Rightarrow {\lim }_{\epsilon \to 0}(\cdots )=2.$ Yeh step kyun? Yahi wajah hai ki curl ko "circulation per unit area" kyun kaha jaata hai. Loop ko shrink karo aur area se normalize karo toh exactly $(\nabla \times \mathbf{F})\cdot \hat{\mathbf{k}}=2$ origin par milta hai — local swirl. Degenerate case: jab $\epsilon \to 0$ loop ek point par collapse hota hai, aur ratio cleanly survive karta hai.

Verify: Kisi bhi $\epsilon$ ke liye, circulation/area $=2\pi {\epsilon }^2/(\pi {\epsilon }^2)=2$, $\epsilon$ se independent, toh limit $2$ hai. ✓

Cell G — Real-world word problem

Figure s07, kya dekhna hai: magenta circle wire loop $C$ hai (radius $0.4$ m), orange CCW arrow iske chosen walking direction ke saath. Right-hand rule phir loop ka apna normal ⊙ fix karta hai jo upar point karta hai ($+\hat{\mathbf{k}}$, violet). Lekin given curl $\nabla \times \mathbf{E}=(0,0,-5)$ neeche point karta hai (violet downward arrow ⊗). Kyunki curl chosen normal ke opposite hai, dot product negative hai — exactly isliye EMF negative aata hai (Lenz's law), koi arbitrary sign nahi.

Step 1 — EMF ko circulation recognize karo. Loop ke around electromotive force exactly ${\oint }_C\mathbf{E}\cdot d\mathbf{r}$ hai — loop ke around push karne ka work per unit charge. Stokes isse $\nabla \times \mathbf{E}$ ke flux mein turn karta hai. Yeh step kyun? Yeh Stokes' theorem ke saath likha literally Faraday's law hai; "circulation = curl-flux" identity hi physics hai.

Step 2 — Flat disk ke through curl ka flux. $\hat{\mathbf{n}}=(0,0,1)$ (up, CCW walk se), $d\mathbf{S}=\hat{\mathbf{n}}dA$, area $=\pi R^2=\pi (0.4{)}^2=0.16\pi$. ${\oint }_C\mathbf{E}\cdot d\mathbf{r}=(0,0,-5)\cdot (0,0,1)\cdot \pi R^2=-5\cdot 0.16\pi =-0.8\pi \approx -2.513\text{V}.$ Yeh step kyun? Constant curl-flux = (curl$\cdot \hat{\mathbf{n}}$) × area. Negative sign induced EMF hai jo change ka oppose karta hai (Lenz's law) — down-pointing curl up-pointing normal ke against, exactly jaisa figure dikhata hai.

Verify: $-5\times \pi \times 0.16=-0.8\pi \approx -2.5133$. Units: (V/m²)·(m²) = V ✓. Magnitude modest, jaisa forecast tha.

Cell H — Exam twist: singularity forces a surface choice

Figure s08, kya dekhna hai: magenta unit circle $C$ (orange CCW arrow) origin ko enclose karta hai, violet ✕ "singularity" se marked jahan field blow up karta hai. Magenta field arrows is hole ke around swirl karte hain. Kyunki forbidden point $C$ se bounded kisi bhi flat disk ke andar baithta hai, domain condition 2 fail hoti hai — tum uske across koi valid surface nahi rakh sakte, toh Stokes shortcut off-limits hai aur humein directly integrate karna hoga.

Step 1 — Origin se door curl. $(x,y)\ne (0,0)$ ke liye, direct computation deta hai $\nabla \times \mathbf{F}=(0,0,0)$. Yeh step kyun? Yeh conservative lagta hai — trap hai. Lekin field origin par undefined hai (denominator $x^2+y^2=0$), jo disk ke andar baitha hai. Stokes ki domain condition 2 require karti hai ki $\mathbf{F}$ aur uski derivatives poori surface par exist karein. Flat disk mein singularity contain hai → shortcut yahan illegal hai.

Step 2 — Seedha line integral karo instead. $\mathbf{r}=(\cos t,\sin t,0)$ parametrize karo, $d\mathbf{r}=(-\sin t,\cos t,0)dt$. Unit circle par $x^2+y^2=1$ hai, toh $\mathbf{F}=(-\sin t,\cos t,0),\mathbf{F}\cdot d\mathbf{r}={\sin }^{2}t+{\cos }^{2}t=1.$ ${\oint }_C\mathbf{F}\cdot d\mathbf{r}={\int }_0^{2\pi }1dt=2\pi .$ Yeh step kyun? Kyunki hum origin ke through valid surface build nahi kar sakte, hum definition par fall back karte hain. Answer $2\pi \ne 0$ prove karta hai ki field globally conservative nahi hai chahe curl zero ho — parent note ke last mistake box ki classic warning. (Origin remove hone wala region simply connected nahi raha, yahi "zero curl ⇒ zero circulation" rule ko break karta hai.)

Verify: ${\int }_0^{2\pi }1dt=2\pi$. Saath hi, curl at a generic point $(1,1)$ $(0,0,0)$ evaluate hota hai, "zero curl but nonzero circulation" paradox ko confirm karta hai jo excluded origin se cause hua hai. ✓

Wrap-up: matrix coverage check

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