Y. Green, Gauss, Stokes formulae

Appendix Y. Green, Gauss, Stokes formulae


Let $D$ be a bounded domain in $\mathbb{R}^2$ and $L=\partial D$ be its boundary. Then \begin{equation} -\int_{L} \mathbf{A}\cdot \mathbf{n} \,ds= \iint_D (\nabla \cdot \mathbf{A})\,dS \tag{Gauss formula} \label{equ-Y.1} \end{equation} where the left-hand side expression is a linear integral, the right-hand side expression is an area integral and $\mathbf{n}$ is a unit inner normal to $L$.

Let $V$ be a bounded domain in $\mathbb{R}^3$ and $\Sigma=\partial V$ be its boundary. Then \begin{equation} -\iint_{\Sigma} \mathbf{A}\cdot \mathbf{n} \,dS= \iiint_D (\nabla \cdot \mathbf{A})\,dV \tag{Gauss formula} \label{equ-Y.2} \end{equation} where the left-hand side expression is a surface integral, the right-hand side expression is a volume integral and $\mathbf{n}$ is a unit inner normal to $\Sigma$.

Remark 1.

Let $D$ be a bounded domain in $\mathbb{R}^2$ and $L=\partial D$ be its boundary, counter–clockwise oriented (if $L$ has several components then inner components should be clockwise oriented). Then \begin{equation} \oint_{L} \mathbf{A}\cdot \, d\, \mathbf{r}= \iint_D (\nabla \times \mathbf{A})\cdot\mathbf{n}\,dS \tag{Green formula} \label{equ-Y.3} \end{equation} where the left-hand side expression is a line integral, the right-hand side expression is an area integral and $\mathbf{n}=\mathbf{k}$.

Let $\Sigma$ be a bounded piece of the surface in $\mathbb{R}^3$ and $L=\partial \Sigma$ be its boundary. Then \begin{equation} \oint_{L} \mathbf{A}\cdot \, d\, \mathbf{l}= \iint_\Sigma (\nabla \times \mathbf{A})\cdot\mathbf{n}\,dS \tag{Stokes formula} \label{equ-Y.4} \end{equation} where the left-hand side expression is a line integral, the right-hand side expression is a surface integral and $\mathbf{n}$ is a unit normal to $\Sigma$; orientation of $L$ should match to direction of $\mathbf{n}$.

Remark 2.