Exercises — Unification — all three theorems as generalized Stokes

$k=1$. Yahan $M=[a,b]$ ek 1-dimensional interval hai, isliye $\omega$ ek $(k-1)=0$-form hona chahiye, yaani ek aam function $f$. Ek function ka exterior derivative $df=f^{\prime }(x)dx$ hota hai — ek $1$-form. Stokes kuch yun padhta hai: $\underset{{\int }_{\partial M}f}{\underbrace{f(b)-f(a)}}=\underset{{\int }_{M}df}{\underbrace{{\int }_a^b{f}^{\prime }(x)dx}}.$ Dekho Fundamental Theorem of Calculus.

$d\omega =(Q_x-P_y)dx\wedge dy.$ Yeh Green's Theorem hai ($k=2$, flat). Yahan "flat" ka matlab hai: $M$ ek aisa region hai jo ordinary plane ${\mathbb{R}}^2$ mein baitha hai, apne standard orientation ke saath (usual $x$–$y$ axes, positive area $dx\wedge dy$) — na ki 3D space mein curve karta hua surface (woh curved case Stokes' curl theorem hai). Ek bacha hua coefficient $Q_x-P_y$ "2D mein curl" hai.

$\mathbf{F}=(F_1,F_2,F_3)$ ka flux 2-form: $\omega =F_{1}dy\wedge dz+F_{2}dz\wedge dx+F_{3}dx\wedge dy,$ $d\omega =({\partial }_x{F}_1+{\partial }_y{F}_2+{\partial }_z{F}_3)dx\wedge dy\wedge dz=(\nabla \cdot \mathbf{F})dV.$ Dekho Divergence Theorem.

$d(fdx)=df\wedge dx$ use karo (Exterior Derivative se): $d\omega =d(x^{2}y)\wedge dx+d(x{y}^2)\wedge dy.$ Ab $d(x^{2}y)=2xydx+x^{2}dy$ aur $d(x{y}^2)=y^{2}dx+2xydy$. Har ek ko apne partner ke saath wedge karo: $d\omega =(2xydx+x^{2}dy)\wedge dx+(y^{2}dx+2xydy)\wedge dy.$ $dx\wedge dx=0$ aur $dy\wedge dy=0$ ko khatam karo, aur $dy\wedge dx=-dx\wedge dy$ flip karo: $d\omega =x^2(dy\wedge dx)+y^2(dx\wedge dy)=(y^2-x^2)dx\wedge dy.$ (Green se cross-check: $Q_x-P_y=y^2-x^2$. ✅)

Yahan $P=-y$, $Q=x$, isliye $Q_x-P_y=1-(-1)=2$. Tab ${\oint }_{\partial M}(-ydx+xdy)={\iint }_{M}2dA=2\cdot (\text{unit disk ki area})=2\pi .$ Geometrically (figure se): yeh line integral enclosed area ka do guna compute karta hai — ek classic trick. Do baar $\pi$ deta hai $2\pi$. Figure mein green caption yeh final identity record karta hai $\oint (-ydx+xdy)=2\cdot \text{area}=2\pi$.

$\nabla \cdot \mathbf{F}=1+1+1=3$. Isliye $\iint \mathbf{F}\cdot d\mathbf{S}={\iiint }_{M}3dV=3\cdot \underset{\text{ball ki volume}}{\underbrace{\frac{4}{3}\pi }}=4\pi .$

Andar wala route. $\nabla \cdot \mathbf{F}=2$, isliye ${\iint }_{M}2dA=2\cdot \pi =2\pi$. Boundary wala route. Circle ko $t\in [0,2\pi ]$ se parametrize karo: point $(\cos t,\sin t)$, outward normal $\hat{n}=(\cos t,\sin t)$. Tab $\mathbf{F}\cdot \hat{n}={\cos }^{2}t+{\sin }^{2}t=1$, aur arclength element $ds=dt$ (unit circle). Isliye ${\oint }_{\partial M}\mathbf{F}\cdot \hat{n}ds={\int }_0^{2\pi }1dt=2\pi .$ Dono $2\pi$ dete hain. Yeh agree karte hain kyunki yeh 2D Divergence theorem hai, yaani Generalized Stokes 1-form $\omega =-F_{2}dx+F_{1}dy$ ke liye (planar region $M$ par ek 1-form), jiska exterior derivative $d\omega =({\partial }_x{F}_1+{\partial }_y{F}_2)dx\wedge dy=(\nabla \cdot \mathbf{F})dx\wedge dy$ andar integrate hone wala 2-form hai. Boundary integrand ek 1-form, interior integrand ek 2-form — kabhi ulta nahi. ✅

Pehle $df=f_{x}dx+f_{y}dy$ jahan $f_x=3x^{2}y-y\cos (xy),f_y=x^3-x\cos (xy).$ Ab $d(df)=(f_{yx}-f_{xy})dx\wedge dy$. Dono mixed partials compute karo: $f_{xy}=3x^2-\cos (xy)+xy\sin (xy),f_{yx}=3x^2-\cos (xy)+xy\sin (xy).$ Dono equal hain, isliye $d(df)=0$. Yeh d squared equals zero hai, jo vector language mein $\nabla \times \nabla f=0$ hai: ek gradient field hamesha curl-free hota hai.

Green/Stokes ko chahiye ki $\omega$ poore $M$ par smooth ho. Lekin $\omega$ origin $(0,0)$ par blow up karta hai, jo disk ke andar hai — isliye origin ek valid $M$ ka hissa nahi hai. Jis region par $\omega$ smooth hai woh punctured disk hai, aur ek punctured disk ki boundary sirf outer circle nahi hai. Kyunki $d\omega =0$ sirf puncture se door hold karta hai, theorem ki hypothesis fail ho jaati hai; nonzero circulation $2\pi$ hole ko measure karta hai. (Direct check: $x=\cos t,y=\sin t$ ke saath, $\omega =({\sin }^{2}t+{\cos }^{2}t)dt=dt$, isliye $\oint ={\int }_0^{2\pi }dt=2\pi$.)

Notice karo $P=3x^{2}y={\partial }_x(x^{3}y)$ aur $Q=x^3={\partial }_y(x^{3}y)$, isliye $\omega =d(x^{3}y)$ exact hai. $d^2=0$ se, $d\omega =d(d(x^{3}y))=0$. Green tab deta hai ${\oint }_C\omega ={\iint }_{M}0dA=0.$ Exact form ka koi bhi closed-curve integral zero hota hai — interior "derivative of a derivative" zero hai. ✅ (Sanity check: $Q_x-P_y=3x^2-3x^2=0$.)

Work 1-form ke saath kaam karo; $d\omega \leftrightarrow$ curl. Compute karo $\nabla \times \mathbf{F}=\left({\partial }_{y}x-{\partial }_{z}z,{\partial }_{z}y-{\partial }_{x}x,{\partial }_{x}z-{\partial }_{y}y\right)=(-1,-1,-1).$ Triangle plane $x+y+z=1$ mein hai, jiska unit normal $\hat{n}=\frac{1}{\sqrt{3}}(1,1,1)$ hai. Isliye $(\nabla \times \mathbf{F})\cdot \hat{n}=\frac{-3}{\sqrt{3}}=-\sqrt{3}$, surface par constant. Triangle ki area $\frac{\sqrt{3}}{2}$ hai (side $\sqrt{2}$ ka equilateral triangle). Tab Stokes' Theorem (curl) se: ${\oint }_C\mathbf{F}\cdot d\mathbf{r}={\iint }_S(\nabla \times \mathbf{F})\cdot \hat{n}dS=(-\sqrt{3})\cdot \frac{\sqrt{3}}{2}=-\frac{3}{2}.$

$\omega =\frac{1}{2}(xdy-ydx)$ lo, ek 1-form. $d\omega$ compute karo: $P=-\frac{1}{2}y$, $Q=\frac{1}{2}x$ ke saath, $d\omega =(Q_x-P_y)dx\wedge dy=\left(\frac{1}{2}+\frac{1}{2}\right)dx\wedge dy=dx\wedge dy.$ Stokes se, ${\oint }_{\partial M}\omega ={\iint }_{M}dxdy=\text{Area}(M)$, jo formula prove karta hai. Ellipse: $xdy-ydx=(a\cos t)(b\cos t)dt-(b\sin t)(-a\sin t)dt=ab({\cos }^{2}t+{\sin }^{2}t)dt=abdt$. Tab $\text{Area}=\frac{1}{2}{\int }_0^{2\pi }abdt=\frac{1}{2}ab(2\pi )=\pi ab.✓$

$M$ ek $k=4$ manifold ke liye, $\omega$ ek $(k-1)=3$-form hai aur $d\omega$ ek $4$-form hai: ${\int }_{\partial M}\omega ={\int }_{M}d\omega ,\omega \in {\Lambda }^3,d\omega \in {\Lambda }^4.$ Yeh abhi bhi kyun hold karta hai: proof ne kabhi "2" ya "3" number use nahi kiya. $M$ ko chote 4-cells mein kaato; ${\int }_{\partial (\text{cell})}\omega$ sum karo. Har internal 3-dimensional wall do neighbouring cells ke beech share hoti hai opposite induced orientation ke saath, isliye woh contributions pairs mein cancel ho jaate hain — exactly wahi telescoping engine jo parent note mein hai. Sirf outer boundary $\partial M$ bachti hai, aur har cell per local leftover by definition $d\omega$ hai. Argument dimension-agnostic hai; yehi unification ki poori taaqat hai. ✅

$\omega$ ko $\mathbf{F}$ ka work 1-form maano. Tab $d\omega$ curl 2-form hai, aur $d(d\omega )$ divergence-of-curl 3-form hai. Lekin $d^2=0$ (d squared equals zero) force karta hai $d(d\omega )=0$, yaani 3-form $(\nabla \cdot (\nabla \times \mathbf{F}))dV=0$, isliye $\nabla \cdot (\nabla \times \mathbf{F})=0$. Dual reading: Stokes do baar apply karo. ${\int }_{M}d(d\omega )={\int }_{\partial M}d\omega ={\int }_{\partial (\partial M)}\omega =0$ kyunki $\partial (\partial M)=\varnothing$ (ek boundary ki koi boundary nahi hoti). Dono facts $d^2=0$ aur ${\partial }^2=\varnothing$ ek hi statement hain, integral sign ke dono taraf se dekho. ✅

"4D blob ko chote 4D blocks mein kaato. Har block ki 3D face ke across flow ka total nikaalo. Shared faces cancel ho jaati hain kyunki har ek ek block ke liye 'out' aur uske neighbour ke liye 'in' hai. Sirf outer skin bachti hai — aur har block ke andar bachne wala leftover by definition derivative $d\omega$ hai. Same trick, ek dimension zyada."