Differential Forms 06: Stokes, Cauchy-Green, and Area Forms

Differential forms / 06

Stokes, Cauchy-Green, and area forms

The generalized Stokes theorem says that exterior differentiation is the operation whose integral over a region is measured on the boundary.

Calculus root: the fundamental theorem with boundary terms

Stokes’ theorem is the fundamental theorem of calculus after the word “endpoint” is replaced by “boundary”:

\[\int_{[a,b]}df=f(b)-f(a)=\int_{\partial[a,b]}f.\]

In higher dimensions, adjacent boundary pieces cancel with opposite orientations. What remains is the outer boundary, and the derivative becomes the exterior derivative.

Example: rectangle cancellation

Partition a rectangle into two smaller rectangles. The shared edge appears once with upward orientation and once with downward orientation, so the line integrals on that edge cancel. Green’s theorem is the limiting form of this cancellation.

Example: integration by parts as Stokes

For functions \(u,v\) on \([a,b]\),

\[d(uv)=u\,dv+v\,du.\]

Integrating gives

\[\int_a^b u\,v'\,dx=[uv]_a^b-\int_a^b v\,u'\,dx.\]

The boundary term in integration by parts is the one-dimensional prototype of every Stokes boundary term.

For a compact oriented smooth \(n\)-manifold with boundary and an \((n-1)\)-form \(\omega\), Stokes’ theorem is

\[\int_{\partial M}\omega=\int_M d\omega.\]

Section 4.4 of the PDF uses the planar version to derive Green’s theorem, the Cauchy-Green formula, and the coordinate invariance of surface area.

Green’s theorem as Stokes

Let \(\Omega\subset\mathbb R^2\) have positively oriented boundary and let

\[\eta=P\,dx+Q\,dy.\]

Since

\[d\eta=(Q_x-P_y)\,dx\wedge dy,\]

Stokes becomes

\[\int_{\partial\Omega}P\,dx+Q\,dy =\iint_\Omega (Q_x-P_y)\,dx\,dy.\]

Example 1: area from a boundary integral

Take \(\eta=x\,dy\). Then \(d\eta=dx\wedge dy\), so

\[\int_{\partial\Omega}x\,dy=\operatorname{Area}(\Omega).\]

For the unit disk, parametrizing \(\partial\Omega\) by \((\cos t,\sin t)\) gives

\(\int_0^{2\pi}\cos^2 t\,dt=\pi.\)

Example 2: orientation reversal

If the same boundary is traversed clockwise, the induced orientation changes and

\[\int_{-\partial\Omega}\eta=-\int_{\partial\Omega}\eta.\]

This is why Stokes requires an oriented boundary convention, not only a geometric curve.

Cauchy-Green

Let \(\Omega\subset\mathbb C\) be a bounded smooth domain and \(f\in C^1(\overline\Omega)\). Fix \(z\in\Omega\). On \(\Omega\setminus\{z\}\) consider

\[\omega(w)={f(w)\over w-z}\,dw.\]

Using \(df=f_w\,dw+f_{\bar w}\,d\bar w\),

\[d\omega={f_{\bar w}(w)\over w-z}\,d\bar w\wedge dw.\]

Apply Stokes to \(\Omega\) with a small disk around \(z\) removed. The induced orientation on the inner circle is clockwise, so its boundary integral tends to \(-2\pi i f(z)\); moving that term to the other side gives

\[f(z)= {1\over 2\pi i} \left( \int_{\partial\Omega}{f(w)\over w-z}\,dw -\iint_\Omega {f_{\bar w}(w)\over w-z}\,d\bar w\wedge dw \right).\]

When \(f_{\bar w}=0\), this becomes the Cauchy integral formula.

Example 3: holomorphic input

If \(f(w)=w^2\) on the unit disk, then \(f_{\bar w}=0\) and

\[z^2={1\over 2\pi i}\int_{|w|=1}{w^2\over w-z}\,dw.\]

The area term vanishes because holomorphicity is precisely the disappearance of the \(d\bar w\) direction.

Example 4: nonholomorphic input

For \(f(w)=\bar w\), \(f_{\bar w}=1\). The boundary term and the area term must both be accounted for. On the unit circle \(\bar w=1/w\), so the boundary integral of \(\bar w/(w-z)\) may be computed by residues; the missing holomorphicity is exactly compensated by

\(\iint_{|w|<1}{1\over w-z}\,d\bar w\wedge dw.\)

Area as a 2-form

For a surface patch \(\sigma:U\to\mathbb R^3\) with coordinates \((u,v)\),

\[dA_{\mathrm{dens}}=\|\sigma_u\times\sigma_v\|\,du\,dv.\]

Equivalently, with

\[E=\sigma_u\cdot\sigma_u,\quad F=\sigma_u\cdot\sigma_v,\quad G=\sigma_v\cdot\sigma_v,\]

one has

\[dA_{\mathrm{dens}}=\sqrt{EG-F^2}\,du\,dv.\]

This is the positive area density. Once an orientation has been chosen, it is represented in an oriented chart by the 2-form

\[dA=\sqrt{EG-F^2}\,du\wedge dv.\]

If \(\Phi(s,t)=(u(s,t),v(s,t))\), then

\[\Phi^*(du\wedge dv)=\det(D\Phi)\,ds\wedge dt.\]

Thus the oriented area form changes sign under orientation reversal, while the unoriented area density uses $$

\det(D\Phi)

$$ and stays positive.