Stokes' Curl Theorem

Let $\Sigma$ be a smooth oriented surface in ${\mathbb{R}}^3$ with boundary $\delta \Sigma =\Gamma$. If a vector field

$$\mathbf{F}(x,y,z)=(F_x(x,y,z),F_y(x,y,z),F_z(x,y,z))$$

is defined and has continuous first order partial derivatives in a region containing $\Sigma$, then

\iint_\Sigma (\nabla \times \boldsymbol F)\cdot \mathrm{d} \boldsymbol {\Sigma} = \int_{\delta \Sigma} \boldsymbol F\cdot \mathrm{d}(\boldsymbol \Gamma) $$where the '$\cdot$' operation stands for the [[Euclidean Space#v上的内积|inner product]] on $\mathbb{R}^3$ More explicitly, the equality says that

\begin{aligned} &\iint_\Sigma \left[ \left( \frac{ \partial F_z }{ \partial y } - \frac{ \partial F_y }{ \partial z } \right) \mathrm{d}y \mathrm{d}z + \left( \frac{ \partial F_x }{ \partial z } -\frac{ \partial F_z }{ \partial x } \right) \mathrm{d}z \mathrm{d}x + \left( \frac{ \partial F_y }{ \partial x } - \frac{ \partial F_x }{ \partial y } \right) \mathrm{d}x \mathrm{d}y \right] \
= &\int_{\delta \Sigma} F_x \mathrm{d}x + F_y \mathrm{d}y + F_z \mathrm{d}z \end{aligned}

This is actually an extension of [[Multiple Integral#green公式|Green's theorem]], and similarly if $\Omega$ is a [[Multiple Integral#单联通区域|simply connected]] space in $\mathbb{R}^{3}$, and $\boldsymbol F(x,y,z) = (P(x,y,z),Q(x,y,z),R(x,y,z))$ is a vector field with continuous first order partial derivatives, the following four statements are identical 1. For all smooth, closed curve $L$ in $\Omega$, we have

\oint_L \boldsymbol F\cdot \mathrm{d}\ell = 0

2. For all smooth curve $L$ in $\Omega$, the [[Curve Integral|curve integral]]

\int_L \boldsymbol F\cdot \mathrm{d}\ell

is independent from the integrating path 3. There exists a function $u(x,y,z)$ (or *potential function*) such that

\mathrm{d}u = P \mathrm{d}x + Q \mathrm{d}y + R \mathrm{d}z

$$whichcouldbeattainedby$$

u(A) = u(x,y,z) = \int_{A_0}^{A} \boldsymbol F \mathrm{d} \ell

where $A_0$ is the *reference point* 4. Every point in $\Sigma$ satisfies

\frac{ \partial P }{ \partial y } = \frac{ \partial Q }{ \partial x } ,;\frac{ \partial Q }{ \partial z } = \frac{ \partial R }{ \partial y },;\frac{ \partial R }{ \partial x } = \frac{ \partial Q }{ \partial z }