Exercises — Stokes' theorem — statement, curl-circulation connection

Left side circulation hai, loop $C$ ke around ek line integral: ${\oint }_C\mathbf{F}\cdot d\mathbf{r}.$ Right side curl ka flux hai disk $S$ ke through jo $C$ se bounded hai: ${\iint }_S(\nabla \times \mathbf{F})\cdot d\mathbf{S}.$ Abhi kuch compute nahi kiya — point sirf yeh pehchanna hai ki kaunsa object kahan rehta hai: left par ek loop, right par ek surface, aur curl sirf right par aata hai.

$P=2y,Q=-3x,R=z$ ke saath determinant layout use karke: $\nabla \times \mathbf{F}=({\partial }_{y}R-{\partial }_{z}Q,{\partial }_{z}P-{\partial }_{x}R,{\partial }_{x}Q-{\partial }_{y}P).$ Term by term:

• ${\partial }_{y}R-{\partial }_{z}Q={\partial }_y(z)-{\partial }_z(-3x)=0-0=0$
• ${\partial }_{z}P-{\partial }_{x}R={\partial }_z(2y)-{\partial }_x(z)=0-0=0$
• ${\partial }_{x}Q-{\partial }_{y}P={\partial }_x(-3x)-{\partial }_y(2y)=-3-2=-5$

To $\nabla \times \mathbf{F}=(0,0,-5)$. Yeh ek vector hai, constant, $-\hat{\mathbf{k}}$ mein point karta hua.

Apna right thumb $\hat{\mathbf{n}}=\hat{\mathbf{k}}$ ke along upar ki taraf point karo. Tumhari fingers upar se dekhe jaane par (yaani $+z$ side se) counter-clockwise curl karti hain. To $C$ ko $xy$-plane mein counter-clockwise traverse kiya jaata hai. Yahi figure mein amber arrow hai. Agar $\hat{\mathbf{n}}=-\hat{\mathbf{k}}$ hota, to sab kuch clockwise flip ho jaata.

L1.2 se, $\nabla \times \mathbf{F}=(0,0,-5)$. Surface $xy$-plane mein unit disk hai jiska $\hat{\mathbf{n}}=\hat{\mathbf{k}}$ hai, isliye $d\mathbf{S}=(0,0,1)dA$. Hum kya karte hain: curl ko normal se dot karo — yeh swirl ka woh piece extract karta hai jo actually surface ke through thread karta hai. $(\nabla \times \mathbf{F})\cdot \hat{\mathbf{n}}=(0,0,-5)\cdot (0,0,1)=-5.$ Integrand constant hai, isliye flux bas $-5\times (\text{disk ka area})$ hai: ${\iint }_S(\nabla \times \mathbf{F})\cdot d\mathbf{S}=-5\cdot \pi (1{)}^2=-5\pi .$ Isliye ${\oint }_C\mathbf{F}\cdot d\mathbf{r}=-5\pi$.

Unit circle parametrise karo: $\mathbf{r}(t)=(\cos t,\sin t,0)$, $t:0\to 2\pi$, isliye $d\mathbf{r}=(-\sin t,\cos t,0)dt$. Curve par $\mathbf{F}=(2\sin t,-3\cos t,0)$. Inhe dot karo: $\mathbf{F}\cdot d\mathbf{r}=\left(2\sin t\right)(-\sin t)+(-3\cos t)(\cos t)=-2{\sin }^{2}t-3{\cos }^{2}t.$ Simplify karo — ek algebra step. ${\sin }^{2}t=1-{\cos }^{2}t$ use karo, isliye $-2{\sin }^{2}t=-2(1-{\cos }^{2}t)=-2+2{\cos }^{2}t$. Phir $-2{\sin }^{2}t-3{\cos }^{2}t=(-2+2{\cos }^{2}t)-3{\cos }^{2}t=-2-{\cos }^{2}t.$ $0\to 2\pi$ integrate karo. Kyunki ${\int }_0^{2\pi }{\cos }^{2}tdt=\pi$: ${\int }_0^{2\pi }(-2-{\cos }^{2}t)dt=-2(2\pi )-\pi =-4\pi -\pi =-5\pi .✓$ L2.1 se exactly match karta hai — theorem hold karta hai.

Curl:

• ${\partial }_y(2z)-{\partial }_z(x)=0$
• ${\partial }_z(y)-{\partial }_x(2z)=0$
• ${\partial }_x(x)-{\partial }_y(y)=1-1=0$

To $\nabla \times \mathbf{F}=(0,0,0)$ har jagah. Stokes ke zariye, zero curl ka flux kisi bhi surface ke through $0$ hai, isliye ${\oint }_C\mathbf{F}\cdot d\mathbf{r}=0$ har closed loop ke liye. Yahi conservative-field situation hai: zero curl $\Rightarrow$ locally ek gradient $\Rightarrow$ closed-loop circulation vanish ho jaati hai.

Curl:

• ${\partial }_y(y^2)-{\partial }_z(x^2)=2y-0=2y$
• ${\partial }_z(z^2)-{\partial }_x(y^2)=2z-0=2z$
• ${\partial }_x(x^2)-{\partial }_y(z^2)=2x-0=2x$

To $\nabla \times \mathbf{F}=(2y,2z,2x)$.

Surface — $d\mathbf{S}=(-g_x,-g_y,1)d{A}_{xy}$ kahan se aata hai? Triangle ko ek graph $z=g(x,y)=1-x-y$ ke roop mein likho. Flat $xy$-shadow mein $dx$ ($x$ ke along) aur $dy$ ($y$ ke along) sides wala ek tiny rectangle socho. Uske do edges ko slanted surface par utha lo: $dx$-edge 3D vector $\mathbf{a}=(1,0,g_x)dx$ ban jaata hai ($x$ mein ek step lo, height $g_x$ se change hoti hai), aur $dy$-edge $\mathbf{b}=(0,1,g_y)dy$ ban jaata hai. Surface par actual patch woh parallelogram hai jo yeh dono span karte hain, aur woh vector jo patch ke perpendicular ho aur length patch ke area ke barabar ho unka cross product hai: $\mathbf{a}\times \mathbf{b}=(-g_x,-g_y,1)dxdy.$ Woh single object $(-g_x,-g_y,1)d{A}_{xy}$ hi $d\mathbf{S}$ hai: uski direction orientation deti hai (kaunsa side "upar/bahar" hai) aur uski magnitude tilted area deti hai. Yahan $g_x=g_y=-1$ hai, isliye $d\mathbf{S}=(1,1,1)d{A}_{xy}.$ Teesre slot mein $+1$ matlab yeh $+z$ side ki taraf point karta hai, yaani origin se door — required orientation. Acha.

Dot product: $(2y,2z,2x)\cdot (1,1,1)=2y+2z+2x=2(x+y+z).$ The collapse: plane $x+y+z=1$ par, yeh bas $2(1)=2$ hai — ek constant!

Shadow par integrate karo: $xy$-plane par project karne par triangle $x\ge 0,y\ge 0,x+y\le 1$ milta hai, jiska area $\frac{1}{2}$ hai. ${\oint }_C\mathbf{F}\cdot d\mathbf{r}={\iint }_{\text{tri}}2d{A}_{xy}=2\cdot \frac{1}{2}=1.$

Curl: ${\partial }_x(x)-{\partial }_y(-y)=1-(-1)=2$, baaki components $0$, isliye $\nabla \times \mathbf{F}=(0,0,2)$.

Pehle orientation — dono lids ko upward normal use karna chahiye. Boundary $C$ upar se dekhe jaane par CCW chala jaata hai. Right-hand rule se (thumb $\hat{\mathbf{n}}$ ke along, fingers $C$ ke along), compatible normal upar ki taraf point karta hai ($+z$ component positive) chahe kisi bhi lid ke saath cap karo. Isliye:

• flat disk → $\hat{\mathbf{n}}=+\hat{\mathbf{k}}$;
• hemisphere → outward-aur-upward normal $\hat{\mathbf{n}}=(x,y,z)$ (jiska $z\ge 0$ hai, isliye uska teesra slot $\ge 0$ hai: upward-facing, jaisi zaroorat hai).

Dono choices mein non-negative $\hat{\mathbf{k}}$-component hai, isliye dono CCW rim se induced same orientation hain — yahi humein compare karne deta hai.

(a) Flat disk: $\hat{\mathbf{n}}=\hat{\mathbf{k}}$, integrand $(0,0,2)\cdot (0,0,1)=2$, area $\pi$: ${\iint }_{\text{disk}}2dA=2\pi .$

(b) Hemisphere — geometrically same answer kyun, dekho. Curl constant vertical vector $(0,0,2)$ hai. Upar fix normal ke saath, $(0,0,2)\cdot d\mathbf{S}=2(d{S}_z)$ hai, jahan $d{S}_z$ patch ke horizontal shadow ka signed area hai — signed positive precisely isliye kyunki normal upar ki taraf point karta hai. Ab hemisphere ko tiny patches mein tile karo: equator ke paas ek steep patch bada hoga, lekin woh patla shadow cast karega; top ke paas ek flat patch chhota hoga lekin almost apne barabar shadow dega. Saare (positively-signed) shadows add karo aur tum exactly flat unit disk ko ek baar re-cover karte ho — na overlap, na gap. Isliye ${\iint }_{\text{hemisphere}}(0,0,2)\cdot d\mathbf{S}=2\cdot (\text{total shadow area})=2\cdot (\text{unit disk ka area})=2\pi .$ Agar hum (galti se) downward hemisphere normal lete, to har $d{S}_z$ sign flip ho jaata aur $-2\pi$ milta — rim se disagree karta, signal ki hum right-hand rule tod chuke hain.

Inhe agree karna chahiye — dono surfaces same boundary circle $C$ share karte hain same induced orientation ke saath, aur Stokes kehta hai curl ka flux $S$ par sirf uski boundary ke through depend karta hai. Yeh parent note ke Example 3 ka superpower hai: jo bhi lid aasan ho, woh chuno.

Curl (axis se door): $P=\frac{-y}{x^2+y^2},Q=\frac{x}{x^2+y^2}$ ke saath, $\hat{\mathbf{k}}$-component hai $Q_x-P_y=\frac{y^2-x^2}{(x^2+y^2{)}^2}-\frac{y^2-x^2}{(x^2+y^2{)}^2}=0,$ aur baaki components vanish ho jaate hain kyunki $R=0$ hai aur $P,Q$ $z$ par depend nahi karte. To $\nabla \times \mathbf{F}=0$ har jagah jahan yeh defined hai.

Line integral: $\mathbf{r}(t)=(\cos t,\sin t,0)$ par, $x^2+y^2=1$ hai, isliye $\mathbf{F}=(-\sin t,\cos t,0)$ aur $d\mathbf{r}=(-\sin t,\cos t,0)dt$: $\mathbf{F}\cdot d\mathbf{r}={\sin }^{2}t+{\cos }^{2}t=1,{\int }_0^{2\pi }1dt=2\pi .$

Resolution: Stokes require karta hai ki $\mathbf{F}$ surface par $C^1$ ho (continuous first derivatives — upar definition box dekho) jiska poora interior filled ho. Lekin $\mathbf{F}$ poori $z$-axis par undefined hai ($x=y=0$), aur unit circle ko cap karne wali har disk us axis se pass karti hai. Koi valid surface nahi hai, isliye Stokes simply apply nahi hota — koi contradiction nahi. Yahi parent note ki warning hai punctured surfaces ke baare mein.

Stokes ke zariye, ${\oint }_C\mathbf{F}\cdot d\mathbf{r}={\iint }_{\text{disk}}(0,0,x^2+y^2)\cdot (0,0,1)dA={\iint }_{\text{disk}}(x^2+y^2)dA.$

Polar coordinates mein switch karo ($x^2+y^2=r^2$, $dA=rdrd\theta$) kyunki integrand aur region dono circular hain — polar se limits clean ho jaati hain: ${\int }_0^{2\pi }{\int }_0^1{r}^2\cdot rdrd\theta ={\int }_0^{2\pi }d\theta {\int }_0^1{r}^{3}dr.$ Har piece evaluate karo: ${\int }_0^{2\pi }d\theta =2\pi$ aur ${\int }_0^1{r}^{3}dr={\left[\frac{r^4}{4}\right]}_0^1=\frac{1}{4}$. Multiply karo: ${\oint }_C\mathbf{F}\cdot d\mathbf{r}=2\pi \cdot \frac{1}{4}=\frac{\pi }{2}.$ To circulation $\frac{\pi }{2}$ hai.

Curl: pehle ki tarah, $\nabla \times \mathbf{F}=(0,0,2)$.

Wall flux: side wall ka outward normal $\hat{\mathbf{n}}=(x,y,0)$ purely horizontal hai, isliye $(\nabla \times \mathbf{F})\cdot \hat{\mathbf{n}}=(0,0,2)\cdot (x,y,0)=0.$ Wall ke through flux $0$ hai, isliye ${\oint }_{\partial S}\mathbf{F}\cdot d\mathbf{r}=0$.

Do rims opposite ways kyun chalti hain — right-hand rule se. Wall ke bahar khado taaki outward normal $\hat{\mathbf{n}}=(x,y,0)$ tumhare chest mein ghuse. Right-hand rule kehta hai: surface apni left par rakhte hue, boundary chalo us normal ko consistent rakhte hue. Wall par ek jagah yeh karo aur trace karo "forward" kahan lead karta hai:

• Top rim ($z=2$) ke along induced direction upar se dekhe jaane par clockwise nikalta hai.
• Bottom rim ($z=0$) ke along woh upar se dekhe jaane par counter-clockwise nikalta hai.

Woh opposite hain kyunki wall unke beech hai aur surface ko tumhare body ke same side par rehna hai jab tum bottom edge se top edge par slide karte ho — circulation ka sense flip ho jaata hai. Concretely, har single circle $\pm 2\pi$ deta hai (L3.2 se), isliye: ${\oint }_{\partial S}\mathbf{F}\cdot d\mathbf{r}=\underset{\text{bottom, CCW}}{\underbrace{(+2\pi )}}+\underset{\text{top, CW}}{\underbrace{(-2\pi )}}=0,$ zero wall-flux se agree karta hua. Vanishing flux precisely hume bata raha hai ki do rim circulations equal aur opposite hain.

Yahan $\nabla f=(f_x,f_y,f_z)$ hai, isliye hum $P=f_x,Q=f_y,R=f_z$ set karte hain aur curl compute karte hain:

• ${\partial }_{y}R-{\partial }_{z}Q=f_{zy}-f_{yz}$
• ${\partial }_{z}P-{\partial }_{x}R=f_{xz}-f_{zx}$
• ${\partial }_{x}Q-{\partial }_{y}P=f_{yx}-f_{xy}$

Clairaut's theorem ke zariye ($C^2$ function ke liye — continuous second partials — mixed partials equal hote hain, $f_{xy}=f_{yx}$ etc.), har bracket $0$ hai. Isliye $\nabla \times (\nabla f)=0$.

Stokes interpretation: ek gradient field $\mathbf{F}=\nabla f$ ko Stokes mein feed karo. Right-hand side $\nabla \times (\nabla f)=0$ ka flux ban jaata hai, jo $0$ hai kisi bhi surface ke through. Isliye left-hand side bhi vanish ho jaata hai: ${\oint }_C\nabla f\cdot d\mathbf{r}={\iint }_S(\nabla \times \nabla f)\cdot d\mathbf{S}={\iint }_{S}0\cdot d\mathbf{S}=0$ har closed loop $C$ ke around. Yahi statement hai ki potential fields kisi bhi closed path ke around zero net work karte hain — energy ka "jo upar jaata hai woh neeche aata hai." Stokes algebraic identity $\nabla \times \nabla f=0$ ko physical fact mein turn karta hai ki gradient fields conservative hote hain.

$\hat{\mathbf{n}}=\hat{\mathbf{k}}$ aur $\mathbf{F}=(P(x,y),Q(x,y),0)$ ke saath, curl ka sirf $\hat{\mathbf{k}}$-slot matter karta hai: $(\nabla \times \mathbf{F})\cdot \hat{\mathbf{k}}={\partial }_{x}Q-{\partial }_{y}P.$ Phir Stokes yeh likhta hai: ${\oint }_C(Pdx+Qdy)={\iint }_S(\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y})dA,$ jo verbatim Green's theorem hai. To Green, Stokes ka flat, 2D shadow hai.

Left side par Stokes apply karo, $\mathbf{E}$ ki circulation ko curl flux mein badal do: ${\oint }_C\mathbf{E}\cdot d\mathbf{r}={\iint }_S(\nabla \times \mathbf{E})\cdot d\mathbf{S}.$ Right side se equate karo, time derivative andar le jaao (fixed surface): ${\iint }_S(\nabla \times \mathbf{E})\cdot d\mathbf{S}=-{\iint }_S\frac{\partial \mathbf{B}}{\partial t}\cdot d\mathbf{S}.$ Kyunki yeh har surface $S$ ke liye hold karta hai, integrands pointwise match karne chahiye: $\boxed{\nabla \times \mathbf{E}=-\frac{\partial \mathbf{B}}{\partial t}}$ Faraday's law ka differential form — Maxwell's equations mein se ek. Stokes integral (loop/flux) aur local (curl) statements ke beech bridge hai.

L2.1 $:::-5\pi$ L2.2 $:::-5\pi$ (matches) L2.3 $:::0$ L3.1 $:::1$ L3.2 $:::2\pi$ dono lids ke liye L4.1 $:::2\pi$ (Stokes inapplicable — singular axis) L4.2 $:::\pi /2$ L4.3 $:::0$