Stokes' Theorem
Theorem: let \boldsymbol{\Sigma} be a smooth, oriented, two-dimensional manifold with a boundary, embedded in \mathbb{R}^3. The boundary, denoted \partial\boldsymbol{\Sigma}, is a piecewise smooth, closed curve whose orientation is induced by that of \boldsymbol{\Sigma} according to the right-hand rule.
Consider a vector field \mathbf{F}: U \to \mathbb{R}^3, where U is an open subset of \mathbb{R}^3 containing \boldsymbol{\Sigma}, and assume \mathbf{F} is of class C^1(U). Stokes’ theorem then states that the line integral of \mathbf{F} along the boundary \partial\boldsymbol{\Sigma} is equal to the surface integral of the curl of \mathbf{F} over \boldsymbol{\Sigma}:
\oint_{\partial\boldsymbol{\Sigma}} \mathbf{F} \cdot \mathrm{d}\boldsymbol{\Gamma} = \iint_{\boldsymbol{\Sigma}} (\nabla \times \mathbf{F}) \cdot \mathrm{d}^2\boldsymbol{\Sigma}
where, \mathrm{d}\boldsymbol{\Gamma} is the line element along the boundary curve, and \mathrm{d}^2\boldsymbol{\Sigma} = \mathbf{n} \, \mathrm{d}S is the vector surface element, with \mathbf{n} being the unit normal vector to the surface consistent with its orientation.
Proof: we relies on classical vector analysis and a parametrization of the surface, reducing the problem to Green’s theorem in the plane, transforming both sides of the theorem’s equation into double integrals over a parameter domain in \mathbb{R}^2 and then demonstrate their equality using Green’s theorem.
Let the surface \boldsymbol{\Sigma} be parametrized by a function \boldsymbol{\psi}: D \to \mathbb{R}^3, where D is a compact region in the uv-plane. We assume \boldsymbol{\psi}(u, v) is continuously differentiable. The boundary of D, denoted \gamma = \partial D, is a piecewise smooth Jordan curve. The parametrization maps this curve to the boundary of the surface in \mathbb{R}^3: \partial\boldsymbol{\Sigma} = \boldsymbol{\psi}(\gamma).
Let us first analyze the line integral on the left-hand side. The curve \partial\boldsymbol{\Sigma} is parametrized by \boldsymbol{\Gamma}(t) = \boldsymbol{\psi}(\gamma(t)), where \gamma(t) = (u(t), v(t)):
\oint_{\partial\boldsymbol{\Sigma}} \mathbf{F} \cdot \mathrm{d}\boldsymbol{\Gamma} = \oint_{\gamma} \mathbf{F}(\boldsymbol{\psi}(u,v)) \cdot \frac{\mathrm{d}\boldsymbol{\psi}}{\mathrm{d}t} \mathrm{d}t
By the chain rule, we have:
\frac{\mathrm{d}\boldsymbol{\psi}}{\mathrm{d}t} = \frac{\partial\boldsymbol{\psi}}{\partial u}\frac{\mathrm{d}u}{\mathrm{d}t} + \frac{\partial\boldsymbol{\psi}}{\partial v}\frac{\mathrm{d}v}{\mathrm{d}t}
Substituting this into the integral gives:
\begin{aligned} \oint_{\partial\boldsymbol{\Sigma}} \mathbf{F} \cdot \mathrm{d}\boldsymbol{\Gamma} &= \oint_{\gamma} \left( \mathbf{F}(\boldsymbol{\psi}) \cdot \frac{\partial\boldsymbol{\psi}}{\partial u} \frac{\mathrm{d}u}{\mathrm{d}t} + \mathbf{F}(\boldsymbol{\psi}) \cdot \frac{\partial\boldsymbol{\psi}}{\partial v} \frac{\mathrm{d}v}{\mathrm{d}t} \right) \mathrm{d}t \\ &= \oint_{\gamma} \left( P_u(u,v) \mathrm{d}u + P_v(u,v) \mathrm{d}v \right) \end{aligned}
where we have defined the components of a new vector field \mathbf{P}(u,v) = (P_u, P_v) in the uv-plane as follows:
\begin{aligned} P_u(u,v) &= \mathbf{F}(\boldsymbol{\psi}(u,v)) \cdot \frac{\partial\boldsymbol{\psi}}{\partial u}(u,v) \\ P_v(u,v) &= \mathbf{F}(\boldsymbol{\psi}(u,v)) \cdot \frac{\partial\boldsymbol{\psi}}{\partial v}(u,v) \end{aligned}
This has successfully transformed the line integral in \mathbb{R}^3 into a line integral in the uv-plane.
Now, by Green’s theorem, this line integral is equal to a double integral over the domain D:
\oint_{\gamma} (P_u \mathrm{d}u + P_v \mathrm{d}v) = \iint_D \left( \frac{\partial P_v}{\partial u} - \frac{\partial P_u}{\partial v} \right) \mathrm{d}u \mathrm{d}v
Using the product rule and chain rule for differentiation:
\begin{aligned} \frac{\partial P_v}{\partial u} & = \left(\frac{\partial \mathbf{F}}{\partial u}\right) \cdot \frac{\partial\boldsymbol{\psi}}{\partial v} + \mathbf{F} \cdot \frac{\partial^2\boldsymbol{\psi}}{\partial u \partial v} \\ \frac{\partial P_u}{\partial v} & = \left(\frac{\partial \mathbf{F}}{\partial v}\right) \cdot \frac{\partial\boldsymbol{\psi}}{\partial u} + \mathbf{F} \cdot \frac{\partial^2\boldsymbol{\psi}}{\partial v \partial u} \end{aligned}
The term \frac{\partial \mathbf{F}}{\partial u} denotes the derivative of the composite function \mathbf{F}(\boldsymbol{\psi}(u,v)) with respect to u. It can be expressed using the Jacobian matrix of \mathbf{F}, denoted J_{\mathbf{F}}, and the partial derivative of \boldsymbol{\psi}:
\frac{\partial \mathbf{F}}{\partial u} = J_{\mathbf{F}} \frac{\partial\boldsymbol{\psi}}{\partial u}
Assuming continuity of the second partial derivatives of \boldsymbol{\psi}, the terms involving second derivatives cancel upon subtraction:
\frac{\partial P_v}{\partial u} - \frac{\partial P_u}{\partial v} = \left( J_{\mathbf{F}} \frac{\partial\boldsymbol{\psi}}{\partial u} \right) \cdot \frac{\partial\boldsymbol{\psi}}{\partial v} - \left( J_{\mathbf{F}} \frac{\partial\boldsymbol{\psi}}{\partial v} \right) \cdot \frac{\partial\boldsymbol{\psi}}{\partial u}
This expression can be written using matrix notation. Let \mathbf{a} = \frac{\partial\boldsymbol{\psi}}{\partial u} and \mathbf{b} = \frac{\partial\boldsymbol{\psi}}{\partial v}.
The expression is:
(J_{\mathbf{F}}\mathbf{a})^{\mathsf{T}}\mathbf{b} - (J_{\mathbf{F}}\mathbf{b})^{\mathsf{T}}\mathbf{a} = \mathbf{a}^{\mathsf{T}}J_{\mathbf{F}}^{\mathsf{T}}\mathbf{b} - \mathbf{b}^{\mathsf{T}}J_{\mathbf{F}}^{\mathsf{T}}\mathbf{a}
A direct interpretation recognizes this as:
\frac{\partial P_v}{\partial u} - \frac{\partial P_u}{\partial v} = \frac{\partial\boldsymbol{\psi}}{\partial v} \cdot \left( J_{\mathbf{F}}^{\mathsf{T}} \frac{\partial\boldsymbol{\psi}}{\partial u} \right) - \frac{\partial\boldsymbol{\psi}}{\partial u} \cdot \left( J_{\mathbf{F}}^{\mathsf{T}} \frac{\partial\boldsymbol{\psi}}{\partial v} \right)
The components of the curl of \mathbf{F} are given by (\nabla \times \mathbf{F})_i = \epsilon_{ijk} \partial_j F_k. The Jacobian matrix of \mathbf{F} is (J_{\mathbf{F}})_{ij} = \partial_i F_j. The difference is:
\begin{aligned} \frac{\partial P_v}{\partial u} - \frac{\partial P_u}{\partial v} &= \sum_{i,j} \frac{\partial \psi_i}{\partial v} (\partial_j F_i) \frac{\partial \psi_j}{\partial u} - \sum_{i,j} \frac{\partial \psi_i}{\partial u} (\partial_j F_i) \frac{\partial \psi_j}{\partial v} \\ &= \sum_{i,j,k} \epsilon_{ijk} (\partial_j F_k) \left( \frac{\partial \psi_i}{\partial u} \frac{\partial \psi_j}{\partial v} - \frac{\partial \psi_i}{\partial v} \frac{\partial \psi_j}{\partial u} \right) \\ &= (\nabla \times \mathbf{F}) \cdot \left( \frac{\partial\boldsymbol{\psi}}{\partial u} \times \frac{\partial\boldsymbol{\psi}}{\partial v} \right) \end{aligned}
The vector \frac{\partial\boldsymbol{\psi}}{\partial u} \times \frac{\partial\boldsymbol{\psi}}{\partial v} is precisely the normal vector to the surface, and its magnitude is the Jacobian of the area transformation. Then:
\mathrm{d}^2\boldsymbol{\Sigma} = \left( \frac{\partial\boldsymbol{\psi}}{\partial u} \times \frac{\partial\boldsymbol{\psi}}{\partial v} \right) \mathrm{d}u \mathrm{d}v
Substituting this result back into the expression from Green’s theorem:
\iint_D \left( \frac{\partial P_v}{\partial u} - \frac{\partial P_u}{\partial v} \right) \mathrm{d}u \mathrm{d}v = \iint_D (\nabla \times \mathbf{F}) \cdot \left( \frac{\partial\boldsymbol{\psi}}{\partial u} \times \frac{\partial\boldsymbol{\psi}}{\partial v} \right) \mathrm{d}u \mathrm{d}v = \iint_{\boldsymbol{\Sigma}} (\nabla \times \mathbf{F}) \cdot \mathrm{d}^2\boldsymbol{\Sigma}
This establishes the equality and completes the proof.