The proof of the theorem consists of 4 steps. We assume
Green's theorem, so what is of concern is how to boil down the three-dimensional complicated problem (Stokes' theorem) to a two-dimensional rudimentary problem (Green's theorem). When proving this theorem, mathematicians normally deduce it as a special case of a
more general result, which is stated in terms of
differential forms, and proved using more sophisticated machinery. While powerful, these techniques require substantial background, so the proof below avoids them, and does not presuppose any knowledge beyond a familiarity with basic vector calculus and linear algebra. Recognizing that the columns of are precisely the partial derivatives of at , we can expand the previous equation in coordinates as \begin{align} \oint_{\partial\Sigma}{\mathbf{F}(\mathbf{x})\cdot\,\mathrm{d}\mathbf{\Gamma}} &= \oint_{\gamma}{\mathbf{F}(\boldsymbol{\psi}(\mathbf{y}))\cdot J_{\mathbf{y}}(\boldsymbol{\psi})\mathbf{e}_u(\mathbf{e}_u\cdot\,\mathrm{d}\mathbf{y}) + \mathbf{F}(\boldsymbol{\psi}(\mathbf{y}))\cdot J_{\mathbf{y}}(\boldsymbol{\psi})\mathbf{e}_v(\mathbf{e}_v\cdot\,\mathrm{d}\mathbf{y})} \\ &=\oint_{\gamma}{\left(\left(\mathbf{F}(\boldsymbol{\psi}(\mathbf{y}))\cdot\frac{\partial\boldsymbol{\psi}}{\partial u}(\mathbf{y})\right)\mathbf{e}_u + \left(\mathbf{F}(\boldsymbol{\psi}(\mathbf{y}))\cdot\frac{\partial\boldsymbol{\psi}}{\partial v}(\mathbf{y})\right)\mathbf{e}_v\right)\cdot\,\mathrm{d}\mathbf{y}} \end{align}
Second step in the elementary proof (defining the pullback) The previous step suggests we define the function \mathbf{P}(u,v) = \left(\mathbf{F}(\boldsymbol{\psi}(u,v))\cdot\frac{\partial\boldsymbol{\psi}}{\partial u}(u,v)\right)\mathbf{e}_u + \left(\mathbf{F}(\boldsymbol{\psi}(u,v))\cdot\frac{\partial\boldsymbol{\psi}}{\partial v}(u,v) \right)\mathbf{e}_v Now, if the scalar value functions P_u and P_v are defined as follows, {P_u}(u,v) = \left(\mathbf{F}(\boldsymbol{\psi}(u,v))\cdot\frac{\partial\boldsymbol{\psi}}{\partial u}(u,v)\right) {P_v}(u,v) =\left(\mathbf{F}(\boldsymbol{\psi}(u,v))\cdot\frac{\partial\boldsymbol{\psi}}{\partial v}(u,v) \right) then, \mathbf{P}(u,v) = {P_u}(u,v) \mathbf{e}_u + {P_v}(u,v) \mathbf{e}_v . This is the
pullback of along , and, by the above, it satisfies \oint_{\partial\Sigma}{\mathbf{F}(\mathbf{x})\cdot\,\mathrm{d}\mathbf{l}}=\oint_{\gamma}{\mathbf{P}(\mathbf{y})\cdot\,\mathrm{d}\mathbf{l}} =\oint_{\gamma}{( {P_u}(u,v) \mathbf{e}_u + {P_v}(u,v) \mathbf{e}_v)\cdot\,\mathrm{d}\mathbf{l}} We have successfully reduced one side of Stokes' theorem to a 2-dimensional formula; we now turn to the other side.
Third step of the elementary proof (second equation) First, calculate the partial derivatives appearing in
Green's theorem, via the
product rule: \begin{align} \frac{\partial P_u}{\partial v} &= \frac{\partial (\mathbf{F}\circ \boldsymbol{\psi})}{\partial v}\cdot\frac{\partial \boldsymbol\psi}{\partial u} + (\mathbf{F}\circ \boldsymbol\psi) \cdot\frac{\partial^2 \boldsymbol\psi}{\partial v \, \partial u} \\[5pt] \frac{\partial P_v}{\partial u} &= \frac{\partial (\mathbf{F}\circ \boldsymbol{\psi})}{\partial u}\cdot\frac{\partial \boldsymbol\psi}{\partial v} + (\mathbf{F}\circ \boldsymbol\psi) \cdot\frac{\partial^2 \boldsymbol\psi}{\partial u \, \partial v} \end{align} Conveniently, the second term vanishes in the difference, by
equality of mixed partials. So, \begin{align} \frac{\partial P_v}{\partial u} - \frac{\partial P_u}{\partial v} &= \frac{\partial (\mathbf{F}\circ \boldsymbol\psi)}{\partial u}\cdot\frac{\partial \boldsymbol\psi}{\partial v} - \frac{\partial (\mathbf{F}\circ \boldsymbol\psi)}{\partial v}\cdot\frac{\partial \boldsymbol\psi}{\partial u} \\[5pt] &= \frac{\partial \boldsymbol\psi}{\partial v}\cdot(J_{\boldsymbol\psi(u,v)}\mathbf{F})\frac{\partial \boldsymbol\psi}{\partial u} - \frac{\partial \boldsymbol\psi}{\partial u}\cdot(J_{\boldsymbol\psi(u,v)}\mathbf{F})\frac{\partial \boldsymbol\psi}{\partial v} && \text{(chain rule)}\\[5pt] &= \frac{\partial \boldsymbol\psi}{\partial v}\cdot\left(J_{\boldsymbol\psi(u,v)}\mathbf{F}-{(J_{\boldsymbol\psi(u,v)}\mathbf{F})}^{\mathsf{T}}\right)\frac{\partial \boldsymbol\psi}{\partial u} \end{align} But now consider the matrix in that quadratic form—that is, J_{\boldsymbol\psi(u,v)}\mathbf{F}-(J_{\boldsymbol\psi(u,v)}\mathbf{F})^{\mathsf{T}}. We claim this matrix in fact describes a cross product. Here the superscript " {}^{\mathsf{T}} " represents the
transposition of matrices. To be precise, let A=(A_{ij})_{ij} be an arbitrary matrix and let \mathbf{a}= \begin{bmatrix}a_1 \\ a_2 \\ a_3\end{bmatrix} = \begin{bmatrix}A_{32}-A_{23} \\ A_{13}-A_{31} \\ A_{21}-A_{12}\end{bmatrix} Note that is linear, so it is determined by its action on basis elements. But by direct calculation \begin{align} \left(A-A^{\mathsf{T}}\right)\mathbf{e}_1 &= \begin{bmatrix} 0 \\ a_3 \\ -a_2 \end{bmatrix} = \mathbf{a}\times\mathbf{e}_1\\ \left(A-A^{\mathsf{T}}\right)\mathbf{e}_2 &= \begin{bmatrix} -a_3 \\ 0 \\ a_1 \end{bmatrix} = \mathbf{a}\times\mathbf{e}_2\\ \left(A-A^{\mathsf{T}}\right)\mathbf{e}_3 &= \begin{bmatrix} a_2 \\ -a_1 \\ 0 \end{bmatrix} = \mathbf{a}\times\mathbf{e}_3 \end{align} Here, {{math|{
e1,
e2,
e3}}} represents an orthonormal basis in the coordinate directions of \R^3. Thus for any . Substituting {(J_{\boldsymbol\psi(u,v)}\mathbf{F})} for , we obtain \left({(J_{\boldsymbol\psi(u,v)}\mathbf{F})} - {(J_{\boldsymbol\psi(u,v)}\mathbf{F})}^{\mathsf{T}} \right) \mathbf{x} =(\nabla\times\mathbf{F})\times \mathbf{x}, \quad \text{for all}\, \mathbf{x}\in\R^{3} We can now recognize the difference of partials as a
(scalar) triple product: \begin{align} \frac{\partial P_v}{\partial u} - \frac{\partial P_u}{\partial v} &= \frac{\partial \boldsymbol\psi}{\partial v}\cdot(\nabla\times\mathbf{F}) \times \frac{\partial \boldsymbol\psi}{\partial u} = (\nabla\times\mathbf{F})\cdot \frac{\partial \boldsymbol\psi}{\partial u} \times \frac{\partial \boldsymbol\psi}{\partial v} \end{align} On the other hand, the definition of a
surface integral also includes a triple product—the very same one! \begin{align} \iint_\Sigma (\nabla\times\mathbf{F})\cdot \, d\mathbf{\Sigma} &=\iint_D {(\nabla\times\mathbf{F})(\boldsymbol\psi(u,v))\cdot\frac{\partial \boldsymbol\psi}{\partial u}(u,v)\times \frac{\partial \boldsymbol\psi}{\partial v}(u,v)\,\mathrm{d}u\,\mathrm{d}v} \end{align} So, we obtain \iint_\Sigma (\nabla\times\mathbf{F})\cdot \,\mathrm{d}\mathbf{\Sigma } = \iint_D \left( \frac{\partial P_v}{\partial u} - \frac{\partial P_u}{\partial v} \right) \,\mathrm{d}u\,\mathrm{d}v
Fourth step of the elementary proof (reduction to Green's theorem) Combining the second and third steps and then applying
Green's theorem completes the proof. Green's theorem asserts the following: for any region D bounded by the Jordans closed curve γ and two scalar-valued smooth functions P_u(u,v), P_v(u,v) defined on D; \oint_{\gamma}{( {P_u}(u,v) \mathbf{e}_u + {P_v}(u,v) \mathbf{e}_v)\cdot\,\mathrm{d}\mathbf{l}} = \iint_D \left( \frac{\partial P_v}{\partial u} - \frac{\partial P_u}{\partial v} \right) \,\mathrm{d}u\,\mathrm{d}v We can substitute the conclusion of STEP2 into the left-hand side of Green's theorem above, and substitute the conclusion of STEP3 into the right-hand side.
Q.E.D. Proof via differential forms The functions \R^3\to\R^3 can be identified with the differential 1-forms on \R^3 via the map F_x\mathbf{e}_1+F_y\mathbf{e}_2+F_z\mathbf{e}_3 \mapsto F_x\,\mathrm{d}x + F_y\,\mathrm{d}y + F_z\,\mathrm{d}z . Write the differential 1-form associated to a function as . Then one can calculate that \star\omega_{\nabla\times\mathbf{F}}=\mathrm{d}\omega_{\mathbf{F}}, where is the
Hodge star and \mathrm{d} is the
exterior derivative. Thus, by generalized Stokes' theorem, \oint_{\partial\Sigma}{\mathbf{F}\cdot\,\mathrm{d}\mathbf{\gamma}} =\oint_{\partial\Sigma}{\omega_{\mathbf{F}}} =\int_{\Sigma}{\mathrm{d}\omega_{\mathbf{F}}} =\int_{\Sigma}{\star\omega_{\nabla\times\mathbf{F}}} =\iint_{\Sigma}{\nabla\times\mathbf{F}\cdot\,\mathrm{d}\mathbf{\Sigma}} == Applications ==