classical Stokes’ theorem

Let $M$ be a compact, oriented two-dimensional differentiable manifold (surface) with boundary in ${ℝ}^3$,and $𝐅$ be a $C^2$-smooth vector field defined on an open set in ${ℝ}^3$ containing $M$.Then

$${\iint }_M(\nabla \times 𝐅)\cdot 𝑑𝐀={\int }_{\partial M}𝐅\cdot 𝑑𝐬.$$

Here, the boundary of $M$, $\partial M$ (which is a curve)is given the induced orientation from $M$. The symbol $\nabla \times 𝐅$denotes the curl of $𝐅$.The symbol $d𝐬$ denotes the line element $ds$ with a directionparallel to the unit tangent vector $𝐭$ to $\partial M$, while $d𝐀$ denotesthe area element $dA$ of the surface $M$ with a direction parallel to the unit outward normal $𝐧$to $M$. In precise terms:

$$d𝐀=𝐧dA,d𝐬=𝐭ds.$$

The classical Stokes’ theorem reduces to Green’s theorem on the plane if the surface $M$is taken to lie in the xy-plane.

The classical Stokes’ theorem, andthe other “Stokes’ type” theoremsare special cases of the general Stokes’ theorem involvingdifferential forms.In fact, in the proof we present below, we appeal to the general Stokes’ theorem.

Physical interpretation

(To be written.)

Proof using differential forms

The proof becomes a triviality once we express$(\nabla \times 𝐅)\cdot d𝐀$and $𝐅\cdot d𝐬$ in terms of differential forms.

Proof.

Define the differential forms $\eta$ and $\omega$ by

${\eta }_p(𝐮,𝐯)$$=⟨\mathrm{curl}𝐅(p),𝐮\times 𝐯⟩,$

${\omega }_p(𝐯)$$=⟨𝐅(p),𝐯⟩.$

for points $p\in {ℝ}^3$, and tangent vectors $𝐮,𝐯\in {ℝ}^3$.The symbol $⟨,⟩$ denotes the dot product in ${ℝ}^3$.Clearly, the functions ${\eta }_p$ and ${\omega }_p$ are linear and alternating in$𝐮$ and $𝐯$.

We claim

$\eta$$=\nabla \times 𝐅\cdot d𝐀$on $M$.(1)

$\omega$$=𝐅\cdot d𝐬$on $\partial M$.(2)

To prove (1), it suffices to checkit holds true when we evaluate the left- and right-hand sideson an orthonormal basis $𝐮,𝐯$ for the tangent space of $M$corresponding to the orientation of $M$,given by the unit outward normal $𝐧$.We calculate

$\nabla \times 𝐅\cdot d𝐀(𝐮,𝐯)$$=⟨\mathrm{curl}𝐅,𝐧⟩dA(𝐮,𝐯)$definition of $d𝐀=𝐧dA$

$=⟨\mathrm{curl}𝐅,𝐧⟩$definition of volume form $dA$

$=⟨\mathrm{curl}𝐅,𝐮\times 𝐯⟩$since $𝐮\times 𝐯=𝐧$

$=\eta (𝐮,𝐯).$

For equation (2), similarly, we only have to check that it holdswhen both sides are evaluated at $𝐯=𝐭$,the unit tangent vector of $\partial M$with the induced orientation of $\partial M$.We calculate again,

$𝐅\cdot d𝐬(𝐭)$$=⟨𝐅,𝐭⟩ds(𝐭)$definition of $d𝐬=𝐭ds$

$=⟨𝐅,𝐭⟩$definition of volume form $ds$

$=\omega (𝐭).$

Furthermore, $d\omega$ = $\eta$.(This can be checked by a calculationin Cartesian coordinates, but in fact this equationis one of the coordinate-free definitions of the curl.)

The classical Stokes’ Theorem now followsfrom the general Stokes’ Theorem,

$${\int }_M\eta ={\int }_M𝑑\omega ={\int }_{\partial M}\omega .\mathit{∎}$$