Exercises — Unification — all three theorems as generalized Stokes

$k=1$. Here $M=[a,b]$ is a 1-dimensional interval, so $\omega$ must be a $(k-1)=0$-form, i.e. an ordinary function $f$. Its exterior derivative on a function is $df=f^{\prime }(x)dx$ — a $1$-form. Stokes reads $\underset{{\int }_{\partial M}f}{\underbrace{f(b)-f(a)}}=\underset{{\int }_{M}df}{\underbrace{{\int }_a^b{f}^{\prime }(x)dx}}.$ See Fundamental Theorem of Calculus.

$d\omega =(Q_x-P_y)dx\wedge dy.$ This is Green's Theorem ($k=2$, flat). Here "flat" means: $M$ is a region sitting inside the ordinary plane ${\mathbb{R}}^2$, with its standard orientation (the usual $x$–$y$ axes, positive area $dx\wedge dy$) — as opposed to a curved surface bending through 3D space (that curved case is Stokes' curl theorem). The single surviving coefficient $Q_x-P_y$ is the "curl in 2D."

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

Use $d(fdx)=df\wedge dx$ (from Exterior Derivative): $d\omega =d(x^{2}y)\wedge dx+d(x{y}^2)\wedge dy.$ Now $d(x^{2}y)=2xydx+x^{2}dy$ and $d(x{y}^2)=y^{2}dx+2xydy$. Wedge each against its partner: $d\omega =(2xydx+x^{2}dy)\wedge dx+(y^{2}dx+2xydy)\wedge dy.$ Kill $dx\wedge dx=0$ and $dy\wedge dy=0$, and flip $dy\wedge dx=-dx\wedge dy$: $d\omega =x^2(dy\wedge dx)+y^2(dx\wedge dy)=(y^2-x^2)dx\wedge dy.$ (Check against Green: $Q_x-P_y=y^2-x^2$. ✅)

Here $P=-y$, $Q=x$, so $Q_x-P_y=1-(-1)=2$. Then ${\oint }_{\partial M}(-ydx+xdy)={\iint }_{M}2dA=2\cdot (\text{area of unit disk})=2\pi .$ Geometrically (the figure): this line integral computes twice the enclosed area — a classic trick. Twice $\pi$ gives $2\pi$. The green caption in the figure records this final identity $\oint (-ydx+xdy)=2\cdot \text{area}=2\pi$.

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

Inside route. $\nabla \cdot \mathbf{F}=2$, so ${\iint }_{M}2dA=2\cdot \pi =2\pi$. Boundary route. Parametrize the circle by $t\in [0,2\pi ]$: point $(\cos t,\sin t)$, outward normal $\hat{n}=(\cos t,\sin t)$. Then $\mathbf{F}\cdot \hat{n}={\cos }^{2}t+{\sin }^{2}t=1$, and arclength element $ds=dt$ (unit circle). So ${\oint }_{\partial M}\mathbf{F}\cdot \hat{n}ds={\int }_0^{2\pi }1dt=2\pi .$ Both give $2\pi$. They agree because this is the 2D Divergence theorem, i.e. Generalized Stokes for the 1-form $\omega =-F_{2}dx+F_{1}dy$ (a 1-form on the planar region $M$), whose exterior derivative $d\omega =({\partial }_x{F}_1+{\partial }_y{F}_2)dx\wedge dy=(\nabla \cdot \mathbf{F})dx\wedge dy$ is the 2-form integrated inside. Boundary integrand a 1-form, interior integrand a 2-form — never the other way round. ✅

First $df=f_{x}dx+f_{y}dy$ where $f_x=3x^{2}y-y\cos (xy),f_y=x^3-x\cos (xy).$ Now $d(df)=(f_{yx}-f_{xy})dx\wedge dy$. Compute both mixed partials: $f_{xy}=3x^2-\cos (xy)+xy\sin (xy),f_{yx}=3x^2-\cos (xy)+xy\sin (xy).$ They are equal, so $d(df)=0$. This is d squared equals zero, which in vector language is $\nabla \times \nabla f=0$: a gradient field is always curl-free.

Green/Stokes needs $\omega$ smooth on all of $M$. But $\omega$ blows up at the origin $(0,0)$, which lies inside the disk — so the origin is not part of a valid $M$. The region where $\omega$ is smooth is the punctured disk, and a punctured disk's boundary is not just the outer circle. Because $d\omega =0$ holds only off the puncture, the theorem's hypothesis fails; the nonzero circulation $2\pi$ measures the hole. (Direct check: with $x=\cos t,y=\sin t$, $\omega =({\sin }^{2}t+{\cos }^{2}t)dt=dt$, so $\oint ={\int }_0^{2\pi }dt=2\pi$.)

Notice $P=3x^{2}y={\partial }_x(x^{3}y)$ and $Q=x^3={\partial }_y(x^{3}y)$, so $\omega =d(x^{3}y)$ is exact. By $d^2=0$, $d\omega =d(d(x^{3}y))=0$. Green then gives ${\oint }_C\omega ={\iint }_{M}0dA=0.$ Any closed-curve integral of an exact form vanishes — the interior "derivative of a derivative" is zero. ✅ (Sanity: $Q_x-P_y=3x^2-3x^2=0$.)

Work with the work 1-form; $d\omega \leftrightarrow$ curl. Compute $\nabla \times \mathbf{F}=\left({\partial }_{y}x-{\partial }_{z}z,{\partial }_{z}y-{\partial }_{x}x,{\partial }_{x}z-{\partial }_{y}y\right)=(-1,-1,-1).$ The triangle lies in the plane $x+y+z=1$, whose unit normal is $\hat{n}=\frac{1}{\sqrt{3}}(1,1,1)$. So $(\nabla \times \mathbf{F})\cdot \hat{n}=\frac{-3}{\sqrt{3}}=-\sqrt{3}$, constant over the surface. The triangle's area is $\frac{\sqrt{3}}{2}$ (an equilateral triangle of side $\sqrt{2}$). Then by Stokes' Theorem (curl): ${\oint }_C\mathbf{F}\cdot d\mathbf{r}={\iint }_S(\nabla \times \mathbf{F})\cdot \hat{n}dS=(-\sqrt{3})\cdot \frac{\sqrt{3}}{2}=-\frac{3}{2}.$

Take $\omega =\frac{1}{2}(xdy-ydx)$, a 1-form. Compute $d\omega$: with $P=-\frac{1}{2}y$, $Q=\frac{1}{2}x$, $d\omega =(Q_x-P_y)dx\wedge dy=\left(\frac{1}{2}+\frac{1}{2}\right)dx\wedge dy=dx\wedge dy.$ By Stokes, ${\oint }_{\partial M}\omega ={\iint }_{M}dxdy=\text{Area}(M)$, proving the formula. Ellipse: $xdy-ydx=(a\cos t)(b\cos t)dt-(b\sin t)(-a\sin t)dt=ab({\cos }^{2}t+{\sin }^{2}t)dt=abdt$. Then $\text{Area}=\frac{1}{2}{\int }_0^{2\pi }abdt=\frac{1}{2}ab(2\pi )=\pi ab.✓$

For $M$ a $k=4$ manifold, $\omega$ is a $(k-1)=3$-form and $d\omega$ is a $4$-form: ${\int }_{\partial M}\omega ={\int }_{M}d\omega ,\omega \in {\Lambda }^3,d\omega \in {\Lambda }^4.$ Why it still holds: the proof never used the number "2" or "3." Chop $M$ into tiny 4-cells; sum ${\int }_{\partial (\text{cell})}\omega$. Each internal 3-dimensional wall is shared by two neighbouring cells with opposite induced orientation, so those contributions cancel in pairs — exactly the telescoping engine from the parent note. Only the outer boundary $\partial M$ survives, and the local leftover per cell is by definition $d\omega$. The argument is dimension-agnostic; that is the whole force of the unification. ✅

Let $\omega$ be the work 1-form of $\mathbf{F}$. Then $d\omega$ is the curl 2-form, and $d(d\omega )$ is the divergence-of-curl 3-form. But $d^2=0$ (d squared equals zero) forces $d(d\omega )=0$, i.e. the 3-form $(\nabla \cdot (\nabla \times \mathbf{F}))dV=0$, hence $\nabla \cdot (\nabla \times \mathbf{F})=0$. Dual reading: apply Stokes twice. ${\int }_{M}d(d\omega )={\int }_{\partial M}d\omega ={\int }_{\partial (\partial M)}\omega =0$ because $\partial (\partial M)=\varnothing$ (a boundary has no boundary). The two facts $d^2=0$ and ${\partial }^2=\varnothing$ are the same statement seen from either side of the integral sign. ✅

"Cut the 4D blob into little 4D blocks. Add up the flow across every block's 3D face. Shared faces cancel because each is an 'out' for one block and an 'in' for its neighbour. Only the outer skin is left — and the leftover inside each block is, by definition, the derivative $d\omega$. Same trick, one more dimension."