\documentclass[aspectratio=169]{beamer}
%\documentclass[11pt]{article} \usepackage{beamerarticle,hyperref}
\usepackage{pgfpages, wrapfig, mathtools, amsmath, amssymb, graphicx, qrcode, adjustbox, ../picins, %enumitem,
  stmaryrd, % for \sslash
  mathbbol % for \bbe.
}
\geometry{paperwidth=154.3mm,paperheight=90mm} % Best for PDF Annotator, full screen on 16x9.
%\pgfpagesuselayout{2 on 1}[letterpaper,border shrink=0mm]
%\pgfpagesuselayout{4 on 1}[letterpaper,border shrink=0mm,landscape]
\pgfpagesuselayout{8 on 1}[letterpaper,border shrink=0mm]

%\usepackage[breaklinks=true]{hyperref}
\hypersetup{colorlinks,
  linkcolor={green!50!black},
  citecolor={green!50!black},
  urlcolor=blue
}

\usepackage[all]{xy}

\newcommand{\bbR}{{\mathbb R}}
\newcommand{\ds}{\displaystyle}
\def\curl{{\operatorname{curl}}}
\def\divop{{\operatorname{div}}}
\def\grad{{\operatorname{grad}}}

\newenvironment{myitemize}{
  \begin{list}{$\bullet$}{\setlength{\leftmargin}{12pt}
  \setlength{\itemsep}{-4pt}
  \setlength{\labelwidth}{8pt}
  \setlength{\labelsep}{4pt}}
}{
  \end{list}
}

\begin{document} \beamertemplatenavigationsymbolsempty

\begin{frame}
\href{http://www.math.toronto.edu/~drorbn}{Dror Bar-Natan}:
\href{http://www.math.toronto.edu/~drorbn/classes}{Classes}:
\href{http://www.math.toronto.edu/~drorbn/classes/\#2122}{2021-22}:
\href{http://drorbn.net/2122-257}{MAT 257 Analysis II}:
\hfill \qrcode[height=3em,level=L,nolink]{drorbn.net/2122-257}
\newline{\large\bf The 3D Theorems}

\parbox{2in}{Hour 72 on April 8, 2022.
\vskip 5mm
There's also a handout at \url{http://drorbn.net/2122-257/ap/3DTheorems.pdf}}\hfill\adjustbox{valign=m}{\includegraphics[width=2.2in]{3DTheorems.pdf}}
\end{frame}

\begin{frame}
{\bf Some Small Print.}

{\tiny\begin{itemize}%[topsep=0pt,itemsep=0pt,leftmargin=8pt]
\input{Final-Details.tex}
\item More, more to date, and practice questions at {\footnotesize \url{http://drorbn.net/2122-257/ap/Final-Info.pdf}.}
\end{itemize}}
\end{frame}

\begin{frame}
{\bf Goal.} Gauss' and Stokes' theorems:
\[ \int_D\divop G\,dV = \int_{\partial D} G\cdot n\,dA
  \qquad
  \int_S(\curl F)\cdot n\, dA = \int_{\partial S}F\cdot T\, ds
\]
where
\begin{itemize}%[topsep=0pt,itemsep=0pt,leftmargin=12pt]
\item $D\subset\bbR^3$ is a domain (a compact oriented 3D manifold with boundary) and $S\subset\bbR^3$ is a surface (a compact oriented 2D manifold with boundary).
\item $F$ and $G$ are vector fields.
\item $n$ is the positive unit normal to $\partial D$ / $S$.
\item $T$ is the positive unit tangent to $\partial S$.
\item $dV$ is the 3D volume form on $D$, $dA$ the 2D volume form on $\partial D$ / $S$ (``Area''), and $ds$ is the 1D volume form on $\partial S$ (``arc length'').
\end{itemize}
\end{frame}

\begin{frame}
{\bf Volumes.} Recall, ``the volume form'' $dV$ (or $dA$ or $ds/d\ell$) on $M$ is the multiple of the orientation form which when fed with a positive orthonormal basis of $T_pM$ (at any $p\in M$) outputs 1. For a 3D $D\subset\bbR^3$, this is $dV=dx\wedge dy\wedge dz$. If $S\subset\bbR^3$ is 2D and $n$ is its positive unit normal, we have
\begin{multline*}
  dA(u,v) = \left|\begin{array}{ccc} -&u&- \\ -&v&- \\ -&n&- \end{array}\right| = (u\times v)\cdot n \\
  = (n_1 dy\wedge dz + n_2dz\wedge dx + n_3dx\wedge dy)(u,v)
\end{multline*}
so $dA=n_1 dy\wedge dz + n_2dz\wedge dx + n_3dx\wedge dy$.

{\bf Example.} On $S^2$, $dA=xdy\wedge dz + ydz\wedge dx + zdx\wedge dy$, as seen before.
\end{frame}

\begin{frame}
{\bf Reminder.} $\begin{pmatrix}u_1\\u_2\\u_3\end{pmatrix} \times \begin{pmatrix}v_1\\v_2\\v_3\end{pmatrix} = \begin{pmatrix}u_2v_3-u_3v_2\\u_3v_1-u_1v_3\\u_1v_2-u_2v_1\end{pmatrix}$ is the vector perpendicular to both $u$ and $v$ whose length is the area of the parallelogram generated by $u$ and $v$, and if that is not $0$, then $(u,v,u\times v)$ make a positive basis.

\parpic[r]{
  \def\u{{$|u|$}}
  \def\v{{$|v|$}}
  \def\vsin{{$|v|\sin\theta$}}
  \scalebox{1}{\input{Parallelogram.pdf_t}}
}
\vskip 12mm
{\bf Note.} The area of a parallelogram is given by
\[ A = |u|\cdot|v|\cdot\sin\theta \]
\[ = \pm |u|\cdot|v|\cdot\sqrt{1-\cos^2\theta} = \pm\sqrt{|u|^2|v|^2-\langle u,v\rangle^2}, \]
if you remember that $\cos\theta=\frac{\langle u,v\rangle}{|u|\cdot|v|}$.
\end{frame}

\begin{frame}
{\bf Aside.} $dA(u,v) = (u\times v)\cdot n = \pm|u\times v| = \pm\sqrt{|u|^2|v|^2-\langle u,v\rangle^2}$ so we can compute surface area as follows, if $S$ is the image of an orientation preserving 2-cube $c$:

\begin{multline*}
  \operatorname{Area}(S) = \int_S dA = \int_{I^2}c^\ast(dA) = \int_{I^2}dA(c_\ast e_1,c_\ast e_2) \\
  \int_{I^2}dA(\partial_1c,\partial_2c) = \int_{I^2}\sqrt{|\partial_1c|^2|\partial_2c|^2-\langle \partial_1c,\partial_2c\rangle^2}
\end{multline*}
\end{frame}

\begin{frame}
{\bf Recall} That in $\bbR^3$,
\[ \xymatrix{
  \Omega^0 \ar[r]^d & \Omega^1 \ar[r]^d & \Omega^2 \ar[r]^d & \Omega^3 \\
  \{\text{fncns }f\} \ar[r]^\grad \ar[u]_{\omega^0} &
  \{\text{V.F. }F\} \ar[r]^\curl \ar[u]_{\omega^1} &
  \{\text{V.F. }G\} \ar[r]^\divop \ar[u]_{\omega^2} &
  \{\text{fncns }g\} \ar[u]_{\omega^3}
} \]
via $\omega^0_f=f$, $\omega^1_F=F_1dx+F_2dy+F_3dz$, $\omega^2_G=G_1dy\wedge dz+G_2dz\wedge dx+G_3dx\wedge dy$, and $\omega^3_g=gdx\wedge dy\wedge dz$.
\end{frame}

\begin{frame}\null
{\bf Claim 1.} On an oriented curve in $\bbR^3$, $\omega^1_F=(T\cdot F)ds$.

{\em Proof.} Compute both sides on $T$ and get $T\cdot F$.

{\bf Claim 2.} On an oriented surface $S$ in $\bbR^3$, $\omega^2_G=(G\cdot n)dA$.

{\em Proof.} Compute both sides on $(u,v)$, a positive orthonormal basis of $T_pS$:
\[ \omega^2_G(u,v) = G\cdot(u\times v) = (G\cdot n)|u\times v| = (G\cdot n)dA(u,v). \]

{\bf Claim 3.} On an oriented domain $D$ in $\bbR^3$, $\omega^3_g=gdV$.

{\em Proof.} Of course.
\end{frame}

\begin{frame}
{\bf The 3D Theorems.} With all that,

\[ \int_Dd\omega^2_G = \int_{\partial D}\omega^2_G
  \quad\Longrightarrow\quad
  \int_D\divop G\,dV = \int_{\partial D} G\cdot n\,dA
\]

\[ \int_Sd\omega^1_F = \int_{\partial S}\omega^1_F
  \quad\Longrightarrow\quad
  \int_S(\curl F)\cdot n\, dA = \int_{\partial S}F\cdot T\, ds
\]
and that's all, for now!
\end{frame}

\begin{frame}\null
There's so much more, and I wish we had the time. Anyway,
\vskip 5mm
Good luck with the final!
\vskip 5mm
Have a wonderful summer!
\vskip 5mm
Have a wonderful year, next year!
\end{frame}

\end{document}
