% (find-angg "LATEX/2009-2-C4-theorems.tex")
% (find-dn4ex "edrx08.sty")
% (find-angg ".emacs.templates" "s2008a")
% (defun c () (interactive) (find-zsh "cd ~/LATEX/ && ~/dednat4/dednat41 2009-2-C4-theorems.tex && latex    2009-2-C4-theorems.tex"))
% (defun c () (interactive) (find-zsh "cd ~/LATEX/ && ~/dednat4/dednat41 2009-2-C4-theorems.tex && pdflatex 2009-2-C4-theorems.tex"))
% (eev "cd ~/LATEX/ && Scp 2009-2-C4-theorems.{dvi,pdf} edrx@angg.twu.net:slow_html/LATEX/")
% (defun d () (interactive) (find-dvipage "~/LATEX/2009-2-C4-theorems.dvi"))
% (find-dvipage "~/LATEX/2009-2-C4-theorems.dvi")
% (find-pspage  "~/LATEX/2009-2-C4-theorems.pdf")
% (find-pspage  "~/LATEX/2009-2-C4-theorems.ps")
% (find-zsh0 "cd ~/LATEX/ && dvips -D 300 -o 2009-2-C4-theorems.ps 2009-2-C4-theorems.dvi")
% (find-zsh0 "cd ~/LATEX/ && dvips -D 600 -P pk -o 2009-2-C4-theorems.ps 2009-2-C4-theorems.dvi && ps2pdf 2009-2-C4-theorems.ps 2009-2-C4-theorems.pdf")
% (find-zsh0 "cd ~/LATEX/ && dvips -D 300 -o tmp.ps tmp.dvi")
% (find-pspage  "~/LATEX/tmp.ps")
% (ee-cp "~/LATEX/2009-2-C4-theorems.pdf" (ee-twupfile "LATEX/2009-2-C4-theorems.pdf") 'over)
% (ee-cp "~/LATEX/2009-2-C4-theorems.pdf" (ee-twusfile "LATEX/2009-2-C4-theorems.pdf") 'over)

\documentclass[oneside]{book}
\usepackage[latin1]{inputenc}
\usepackage{edrx08}       % (find-dn4ex "edrx08.sty")
%L process "edrx08.sty"  -- (find-dn4ex "edrx08.sty")
\input edrxheadfoot.tex   % (find-dn4ex "edrxheadfoot.tex")
\begin{document}

\input 2009-2-C4-theorems.dnt

%*
% (eedn4-51-bounded)

%Index of the slides:
%\msk
% To update the list of slides uncomment this line:
%\makelos{tmp.los}
% then rerun LaTeX on this file, and insert the contents of "tmp.los"
% below, by hand (i.e., with "insert-file"):
% (find-fline "tmp.los")
% (insert-file "tmp.los")

\def\na{\nabla}



Divergence theorem:
%
$$ \iint_S ¦F·\hat{¦n}\,dS = \iiint_V \na·¦F\,dV $$

Stokes' theorem:
%
$$ \oint_C ¦F·\hat{¦t}\,ds = \iint_S \hat{¦n}·\na×F\,dS $$

Identities involving the operator $\nabla$:
%
$$\begin{array}{l}
  \na(fg) = f\na g + g\na f \\
  \na(¦F·¦G) = (¦G·\na)¦F + (¦F·\na)¦G + ¦F×(\naצG) + ¦G×(\naצF) \\
  \na·(f¦F) = f\na· + \; ¦F·\na f \\
  \na·(¦FצG) = ¦G·(\naצF) - ¦F·(\naצG) \\
  \na·\naצF = 0 \\
  \na×(f¦F)=f\naצF + (\na f)צF \\
  \na×(¦FצG) = (¦G·\na)¦F - (¦F·\na)¦G + ¦F(\na·¦G) - ¦G(\na·¦F) \\
  \na×(\naצF) = \na(\na·¦F) - \na^2¦F \\
  \na×\na f = 0
  \end{array}
$$



%*

\end{document}

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