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\begin{document}

\input{213-Lecture-Macros}		%% 	macros, settings, et.c


\title{Math 213 - Change of Variables, Part I}			%%	title page
\author{Peter A. Perry}
\institute{University of Kentucky}
\date{November 1, 2019}
\maketitle

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%	
%	Due Date Reminders
%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\begin{frame}
\frametitle{Reminders}

\begin{itemize}

\item	Homework C4 on section 15.8 (triple integrals in spherical coordinates) is due 
tonight
\end{itemize}

\end{frame}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%
%	Unit Outline
%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\begin{frame}
\frametitle{Unit III: Multiple Integrals, Vector Fields}
\begin{itemize}

\item[{}]			Double Integrals in Polar Coordinates
\item[{}]			Triple Integrals (Part I)
\item[{}]			Triple Integrals (Part II)
\item[{}]			Triple Integrals in Cylindrical Coordinates
\item[{}]			Triple Integrals in Spherical Coordinates


\begingroup
\color{red}
\item[{}]			Change of Variables, Part I
\endgroup
\item[{}]			Change of Variables, Part II

\medskip

\item[{}]			Vector Fields
\item[{}]			Line Integrals (Scalar functions)
\item[{}]			Line Integrals (Vector functions)

\medskip

\item[{}]			Exam III Review
\end{itemize}
\end{frame}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%
%	Learning Goals
%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\section{Learning Goals}

\begin{frame}
\frametitle{Goals of the Day}
\begin{itemize}
\item	Understand what a transformation $T$ between two regions in the plane is
\item	Understand how to compute the \emph{Jacobian Matrix} and \emph{Jacobian determinant} of a  transformation
		and understand what the Jacobian determinant measures
\item	Understand how to compute double integrals using the change of variables formula
\end{itemize}
\end{frame}



%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%
%		New material begins here...
%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\section{Preview}

\begin{frame}
\frametitle{Preview: Calculus I versus Calculus III}
\footnotesize
If $x=g(u)$ maps $[c,d]$ to $[a,b]$, then
$$ \int_a^b f(x) \, dx = \int_c^d f(g(u)) \, g'(u) \, du $$

\medskip

In other words,
$$ \int_a^b f(x) \, dx = \int_c^d f(x(u)) \, \frac{dx}{du} \, du $$

\medskip

\hrulefill

\bigskip


If $x=g(u,v)$, $y=h(u,v)$, and if the region $S$ in the $uv$ plane is mapped to the region $R$ in the $xy$ plane,
then
$$
\iint_R f(x,y) \, dA = 
\iint_S f(x(u,v),y(u,v)) \, J(u,v)  \, du \, dv 
$$
The \emph{Jacobian determinant}
$$ J(u,v) =  \left| \frac{\dee(x,y)}{\dee(u,v)} \right| $$
measures how areas change under the map $(u,v) \mapsto (x,y)$. 


\end{frame}

\section{Transformations}

\begin{frame}
\frametitle{Transformations}
\begin{minipage}{0.4\textwidth}
\begin{tikzpicture}[scale=1]
\draw[thick,->,>=stealth]	(-0.5,0)	--	(2.5,0);
\draw[thick,->,>=stealth]	(0,-0.5)	--	(0,2.5);
\node[left]		at	(0,2.5)		{\tiny{$\theta$}};
\node[below]	at	(2.5,0)		{\tiny{$r$}};
\node[below]	at	(-0.25,0)	{\tiny{$0$}};
%%
\draw[orange,fill=orange!20]	(0.5,0.5)  rectangle	(2,1.5);
\node at	(1.25,1.0)				{\tiny{$S$}};
%%
\node[below]	at	(0.5,0)		{\tiny{$a$}};
\node[below]	at	(2,0)			{\tiny{$b$}};
%%
\node[left]	at	(0,0.5)			{\tiny{$\alpha$}};
\node[left]	at	(0,1.5)			{\tiny{$\beta$}};
%%
\draw[densely dashed]	(0.5,0)	--	(0.5,0.5);
\draw[densely dashed]	(2,0)		--	(2,0.5);
%%
\draw[densely dashed]	(0,0.5)	--	(0.5,0.5);
\draw[densely dashed]	(0,1.5)	--	(0.5,1.5);
%%
\only<2>
{
\draw[very thick,orange,->-]	(0.5,1.5)	--	(0.5,0.5);
}
\only<3>
{
\draw[very thick,orange,->-]	(0.5,0.5)	--	(2,0.5);
}
\only<4>
{
\draw[very thick,orange,->-]	(2,0.5)	--	(2,1.5);
}
\only<5>
{
\draw[very thick,orange,->-]	(2,1.5)	--	(0.5,1.5);
}
\end{tikzpicture}

\bigskip

\bigskip


\begin{tikzpicture}[scale=1]
\draw[thick,->,>=stealth]	(-0.2,0)	--	(2.5,0);
\draw[thick,->,>=stealth]	(0,-0.2)	--	(0,2.5);
\node[left]	at	(0,2.5)			{\scriptsize{$y$}};
\node[right] at	(2.5,0)			{\scriptsize{$x$}};
%%
\draw[cyan,fill=cyan!20]
	(20:1.0)	arc	(20:65:1.0)	--	++(65:1.5)	--	(65:2.5)	arc(65:20:2.5)	--	++(20:-1.5);
\node at (42.5:1.75)			{\scriptsize{$R$}};
%%
\node at	(10:2)					{\tiny{$\theta=\alpha$}};
\node at	 (77.5:2)				{\tiny{$\theta=\beta$}};
%%
\node at	(50:0.65)				{\tiny{$r=a$}};
\node at (42.5:3)					{\tiny{$r=b$}};
%%
\only<2>
{
\draw[very thick,cyan,->-]	(65:1.0)	arc	(65:20:1.0);
}
\only<3>
{
\draw[very thick,cyan,->-]	(20:1.0)	--	(20:2.5);
}
\only<4>
{
\draw[very thick,cyan,->-]	(20:2.5)	arc	(20:65:2.5);
}
\only<5>
{
\draw[very thick,cyan,->-]	(65:2.5)	--	(65:1.0);
}
\end{tikzpicture}
\end{minipage}
\quad
\begin{minipage}{0.5\textwidth}
\footnotesize
The polar coordinate map
$$ x = r \cos \theta, \quad  y = r\sin \theta $$
defines a transformation $T: S \to R$

\medskip

Before, we called  $R$  a \emph{polar rectangle}

\medskip

Here are the corresponding sides of the rectangle in the $r\theta$ plane and the polar rectangle in the $xy$ plane:

\medskip

\begin{itemize}
\uncover<2->
{
\item	The line $r=a$ 
}
\uncover<3->
{
\item	The line $\theta=\alpha$
}
\uncover<4->
{
\item	The line $r=b$
}
\uncover<5->
{
\item	The line $\theta=\beta$
}
\end{itemize}
\end{minipage}

\end{frame}

%% the same frame with r and theta renamed u and v


\begin{frame}
\frametitle{Transformations}
\begin{minipage}{0.4\textwidth}
\begin{tikzpicture}[scale=1]
\draw[thick,->,>=stealth]	(-0.5,0)	--	(2.5,0);
\draw[thick,->,>=stealth]	(0,-0.5)	--	(0,2.5);
\node[left]		at	(0,2.5)		{\tiny{$v$}};
\node[below]	at	(2.5,0)		{\tiny{$u$}};
\node[below]	at	(-0.25,0)	{\tiny{$0$}};
%%
\draw[orange,fill=orange!20]	(0.5,0.5)  rectangle	(2,1.5);
\node at	(1.25,1.0)				{\tiny{$S$}};
%%
\node[below]	at	(0.5,0)		{\tiny{$a$}};
\node[below]	at	(2,0)			{\tiny{$b$}};
%%
\node[left]	at	(0,0.5)			{\tiny{$\alpha$}};
\node[left]	at	(0,1.5)			{\tiny{$\beta$}};
%%
\draw[densely dashed]	(0.5,0)	--	(0.5,0.5);
\draw[densely dashed]	(2,0)		--	(2,0.5);
%%
\draw[densely dashed]	(0,0.5)	--	(0.5,0.5);
\draw[densely dashed]	(0,1.5)	--	(0.5,1.5);
%%
\only<2>
{
\draw[very thick,orange,->-]	(0.5,1.5)	--	(0.5,0.5);
}
\only<3>
{
\draw[very thick,orange,->-]	(0.5,0.5)	--	(2,0.5);
}
\only<4>
{
\draw[very thick,orange,->-]	(2,0.5)	--	(2,1.5);
}
\only<5>
{
\draw[very thick,orange,->-]	(2,1.5)	--	(0.5,1.5);
}
\end{tikzpicture}

\bigskip

\bigskip


\begin{tikzpicture}[scale=1]
\draw[thick,->,>=stealth]	(-0.2,0)	--	(2.5,0);
\draw[thick,->,>=stealth]	(0,-0.2)	--	(0,2.5);
\node[left]	at	(0,2.5)			{\tiny{$y$}};
\node[right] at	(2.5,0)			{\tiny{$x$}};
%%
\draw[cyan,fill=cyan!20]
	(20:1.0)	arc	(20:65:1.0)	--	++(65:1.5)	--	(65:2.5)	arc(65:20:2.5)	--	++(20:-1.5);
\node at (42.5:1.75)			{\tiny{$R$}};
%%
\node at	(10:2)					{\tiny{$v =\alpha$}};
\node at	 (77.5:2)				{\tiny{$v =\beta$}};
%%
\node at	(50:0.65)				{\tiny{$u=a$}};
\node at (42.5:3)					{\tiny{$u=b$}};
%%
\only<2>
{
\draw[very thick,cyan,->-]	(65:1.0)	arc	(65:20:1.0);
}
\only<3>
{
\draw[very thick,cyan,->-]	(20:1.0)	--	(20:2.5);
}
\only<4>
{
\draw[very thick,cyan,->-]	(20:2.5)	arc	(20:65:2.5);
}
\only<5>
{
\draw[very thick,cyan,->-]	(65:2.5)	--	(65:1.0);
}
\end{tikzpicture}
\end{minipage}
\quad
\begin{minipage}{0.5\textwidth}
\footnotesize
The polar coordinate map
$$ x = u \cos v , \quad  y = u\sin v  $$
defines a transformation $T: S \to R$

\medskip

Before, we called  $R$  a \emph{polar rectangle}

\medskip

Here are the corresponding sides of the rectangle in the $u\theta$ plane and the polar rectangle in the $xy$ plane:

\medskip

\begin{itemize}
\uncover<2->
{
\item	The line $u=a$ 
}
\uncover<3->
{
\item	The line $v =\alpha$
}
\uncover<4->
{
\item	The line $u=b$
}
\uncover<5->
{
\item	The line $v =\beta$
}
\end{itemize}
\end{minipage}

\end{frame}

\begin{frame}
\frametitle{Transformations}
\begin{minipage}{0.35\textwidth}
\begin{tikzpicture}[scale=1.5]
\draw[thick,->,>=stealth]	(-0.25,0)	--	(1.25,0);
\draw[thick,->,>=stealth]	(0,-0.25)	--	(0,1.25);
\node[left]		at	(0,1.25)		{\tiny{$v$}};
\node[below]	at	(1.25,0)		{\tiny{$u$}};
\node[below]	at	(-0.125,0)	{\tiny{$0$}};
%%
\draw[orange,fill=orange!20,opacity=0.5]
	(0,0)	rectangle	(1,1);
\node at (0.5,0.5)				{\tiny{$S$}};
%%
\node[below] at (1,0)			{\tiny{$(1,0)$}};
\node[left]	   at	(0,1)			{\tiny{$(0,1)$}};
%%
\only<2>
{
\draw[very thick,orange,->-]	(0,0)	--	(1,0);
}
\only<3>
{
\draw[very thick,orange,->-]	(1,0)	--	(1,1);
}
\only<4>
{
\draw[very thick,orange,->-]	(1,1)	--	(0,1);
}
\only<5>
{
\draw[very thick,orange,->-]	(0,1)	--	(0,0);
}
\end{tikzpicture}

\bigskip

\begin{tikzpicture}[scale=0.8]
\draw[thick,->,>=stealth]	(0,-0.5)	--	(0,2.5);
\draw[thick,->,>=stealth]	(-1.5,0)	--	(1.5,0);
\node[left] at	(0,2.5)			{\tiny{$y$}};
\node[below] at	(1.5,0)		{\tiny{$x$}};
%%
\draw[cyan,fill=cyan!20,opacity=0.5]
	(-1,0)	--	(1,0)	--	plot[domain=0:2,variable=\y]	({1-\y*\y/4},{\y})
						--  plot[domain=2:0,variable=\y]	({\y*\y/4 -1},{\y})
						-- cycle;
\node at (0.25,1.0)				{\tiny{$R$}};
%%
\draw[black,fill=black]	(0,2) circle(0.05cm);
\node[right]	at			(0,2)	{\tiny{$(0,2)$}};
\draw[black,fill=black]	(1,0)	circle(0.05cm);
\node[below] at			(1,0)	{\tiny{$(1,0)$}};
\draw[black,fill=black]	(-1,0)	circle(0.05cm);
\node[below] at			(-1,0)	{\tiny{$(1,0)$}};
%%
\node[right] at (0.75,1.5)	{\tiny{$x=1-y^2/4$}};
\node[left]	 at	(-0.75,1.5)	{\tiny{$x=y^2/4-1$}};
%%
\only<2>
{
\draw[very thick,cyan,->-]	(0,0)	--	(1,0);
}
\only<3>
{
\draw[very thick,cyan,->-]	plot[domain=0:2,variable=\y]	({1-\y*\y/4},{\y});
}
\only<4>
{
\draw[very thick,cyan,->-]	plot[domain=2:0,variable=\y]	({\y*\y/4 -1},{\y});
}
\only<5>
{
\draw[very thick,cyan,->-]	(-1,0)	--	(0,0);
}
\end{tikzpicture}
\end{minipage}
\quad
\begin{minipage}{0.55\textwidth}
\footnotesize
The equations
$$ x = u^2  -v^2, \quad y = 2uv $$
defines a transformation $T: S \to R$

\begin{itemize}
\uncover<2->
{
\item	$v=0$, $0 \leq u \leq 1$ maps to $0 \leq x \leq 1, \, y=0$
}
\smallskip
\uncover<3->
{
\item	$u=1$, $0\leq v \leq 1$ maps to the parametric curve $x=1-v^2, \, y = 2v$
}
\smallskip
\uncover<4->
{
\item	$v=1$, $0 \leq u \leq 1$ maps to the parametric curve $x=u^2-1, \, y = 2u$
}
\smallskip
\uncover<5->
{
\item	$u=0$, $0 \leq v \leq 1$ maps to $-1 \leq x \leq 0$, $y=0$
}
\end{itemize}
\end{minipage}
\end{frame}


\begin{frame}
\frametitle{Transformations}
\footnotesize
\begin{enumerate}
\item	Find the image of
		$$ S = \{ (u,v): 0 \leq u \leq 3, \, 0 \leq v \leq 2 \}$$
		under the transformation $x=2u+3v$, $y=u-v$
		
		\medskip
		
\item	Find the image of the disc $u^2 + v^2 \leq 1$ under the transformation
		$x=au$, $y=bv$
\end{enumerate}
\end{frame}


\section{The Jacobian}

\begin{frame}
\frametitle{The Jacobian}
\footnotesize
In \emph{one variable calculus}, the way a transformation $x=g(u)$ changes lengths of interals is measured by $g'(u)$:
$$ \Delta x = g'(u) \Delta u $$

\begin{center}
\begin{tikzpicture}
\draw[thick,->,>=stealth]	(0,0) 	--	(2,0);
\draw[black,fill=black]	(0.5,0)	circle(0.05cm);
\draw[black,fill=black]	(1.5,0)		circle(0.05cm);
\node[below] at (0.5,0)		{\tiny{$u \vphantom{+\Delta u}$}};
\node[below] at	(1.5,0)		{\tiny{$u+\Delta u$}};
\draw[decoration={brace,raise=0.1cm},decorate]	(0.5,0)	--	(1.5,0);
\node[above] at (1,0.2)		{\tiny{$\Delta u$}};
%%
\draw[thick,->,>=stealth]	(3.5,0)	--	(7,0);
\draw[black,fill=black]	(4,0)		circle(0.05cm);
\draw[black,fill=black]	(6.5,0)	circle(0.05cm);
\node[below]	at	(4,0)				{\tiny{$g(u)$}};
\node[below]  at	(6.5,0)			{\tiny{$g(u+\Delta u)$}};
\node[above] at	(5.25,0.2)			{\tiny{$\left| g'(u) \right| \Delta u$}};
\draw[decoration={brace,raise=0.1cm},decorate]	(4,0)	--	(6.5,0);
\end{tikzpicture}
\end{center}

\vfill

In \emph{two variable calculus}, the way a transformation $$x=g(u,v), \quad y=h(u,v)$$ changes areas is measured by the \emph{Jacobian determinant}
$$ J(u,v) = 
\begin{vmatrix}
				\dee x/\dee u		&&		\dee x/\dee v	\\[15pt]
				\dee y/\dee u		&&		\dee y/\dee v
\end{vmatrix},
\quad
\Delta A 	=	\left| J(u,v) \right| \, \Delta u \, \Delta v
$$
We'll now see why this is the case$\ldots$
\end{frame}

\begin{frame}
\frametitle{The Jacobian}
\footnotesize
A transformation $x=g(u,v)$, $y=h(u,v)$ maps a small rectangle $S$  into a distorted rectangle $R$ through 
the rule
$$ \bfr (u,v) = g(u,v) \ii + h(u,v) \jj $$

\medskip

\begin{minipage}{0.45\textwidth}
\begin{tikzpicture}[scale=1.2]
\draw[thick,->,>=stealth]	(-0.5,0)	--	(2,0);
\draw[thick,->,>=stealth]	(0,-0.5)	--	(0,2);
\node[left]	at	(0,2)				{\tiny{$v$}};
\node[below]	at	(2,0)			{\tiny{$u$}};
%%
\draw[orange,fill=orange!20]
	(0.5,1)	rectangle	(1.75,1.75);
\node at (1.125,1.375)			{\tiny{$S$}};
%%
\draw[thick,orange,fill=orange]	(0.5,1)	circle(0.05cm);
\node[below] at (0.45,1)		{\tiny{$(u_0,v_0)$}};
%%
\node[left]	at	(0.5,1.375)		{\tiny{$\Delta v$}};
\node[below]	at	(1.125,1)	{\tiny{$\Delta u$}};
%%
\end{tikzpicture}
\end{minipage}
\quad
\begin{minipage}{0.45\textwidth}
\begin{tikzpicture}[scale=1.2]
%%%%%%%%%%%%%%
%		x	=	u+ 0.5*v
%		y	=	-0.3*u + v
%%%%%%%%%%%%%%
\draw[thick,->,>=stealth]	(-0.5,0)	--	(2,0);
\draw[thick,->,>=stealth]	(0,-0.5)	--	(0,2);
\node[left]	at	(0,2)		{\tiny{$y$}};
\node[below]	at	(2,0)	{\tiny{$x$}};
%%
\draw[thick,cyan,fill=cyan!20]	
		plot[domain=1:1.75,variable=\v]			({0.5+0.5*\v},{-0.15+\v})	--
		plot[domain=0.5:1.75,variable=\u]		({\u+0.875},{-0.3*\u+1.75})	--
		plot[domain=1.75:1,variable=\v]			({1.75+0.5*\v},{-0.3*1.75+\v})	--
		plot[domain=1.75:0.5,variable=\u]		({\u+0.5},{-0.3*\u+1});
\node	at 	(1.8125,1.0375)	{\tiny{$R$}};
%%
\draw[cyan,fill=cyan]	(1,0.85)	circle(0.05cm);
\node[left]	at	(1,0.85)			{\tiny{$(x_0,y_0)$}};
\uncover<2->
{
\draw[very thick,->,>=stealth]	(1,0.85)	--	++(0.375,0.75);
\node[left]	at	(1.1875,1.225)	{\tiny{$\bfr_u \Delta u$}};
\draw[very thick,->,>=stealth]	(1,0.85)	--	++	(1.25,-0.375);
\node[left] at	 (1.6,0.5)	{\tiny{$\bfr_v \Delta v$}};
}
%%
\end{tikzpicture}
\end{minipage}

\medskip

\uncover<2->
{
$R$ has approximate area
$ \left| \bfr_u \times \bfr_v \right| \, \Delta u \, \Delta v $
where
$$ \bfr_u = \frac{\dee x}{\dee u} \ii + \frac{\dee y}{\dee u} \jj,
	\qquad
	\bfr_v	=\frac{\dee x}{\dee v} \ii + \frac{\dee y }{\dee v} \jj
$$
}
\end{frame}

\begin{frame}
\frametitle{The Jacobian}
\footnotesize
\begin{minipage}{0.45\textwidth}
\begin{tikzpicture}[scale=1.2]
\draw[thick,->,>=stealth]	(-0.5,0)	--	(2,0);
\draw[thick,->,>=stealth]	(0,-0.5)	--	(0,2);
\node[left]	at	(0,2)				{\tiny{$v$}};
\node[below]	at	(2,0)			{\tiny{$u$}};
%%
\draw[orange,fill=orange!20]
	(0.5,1)	rectangle	(1.75,1.75);
\node at (1.125,1.375)			{\tiny{$S$}};
%%
\draw[thick,orange,fill=orange]	(0.5,1)	circle(0.05cm);
\node[below] at (0.45,1)		{\tiny{$(u_0,v_0)$}};
%%
\node[left]	at	(0.5,1.375)		{\tiny{$\Delta v$}};
\node[below]	at	(1.125,1)	{\tiny{$\Delta u$}};
%%
\end{tikzpicture}
\end{minipage}
\quad
\begin{minipage}{0.45\textwidth}
\begin{tikzpicture}[scale=1.2]
%%%%%%%%%%%%%%
%		x	=	u+ 0.5*v
%		y	=	-0.3*u + v
%%%%%%%%%%%%%%
\draw[thick,->,>=stealth]	(-0.5,0)	--	(2,0);
\draw[thick,->,>=stealth]	(0,-0.5)	--	(0,2);
\node[left]	at	(0,2)		{\tiny{$y$}};
\node[below]	at	(2,0)	{\tiny{$x$}};
%%
\draw[thick,cyan,fill=cyan!20]	
		plot[domain=1:1.75,variable=\v]			({0.5+0.5*\v},{-0.15+\v})	--
		plot[domain=0.5:1.75,variable=\u]		({\u+0.875},{-0.3*\u+1.75})	--
		plot[domain=1.75:1,variable=\v]			({1.75+0.5*\v},{-0.3*1.75+\v})	--
		plot[domain=1.75:0.5,variable=\u]		({\u+0.5},{-0.3*\u+1});
\node	at 	(1.8125,1.0375)	{\tiny{$R$}};
%%
\draw[cyan,fill=cyan]	(1,0.85)	circle(0.05cm);
\node[left]	at	(1,0.85)			{\tiny{$(x_0,y_0)$}};
%%
\draw[very thick,->,>=stealth]	(1,0.85)	--	++(0.375,0.75);
\node[left]	at	(1.1875,1.225)	{\tiny{$\bfr_u \Delta u$}};
\draw[very thick,->,>=stealth]	(1,0.85)	--	++	(1.25,-0.375);
\node[left] at	 (1.6,0.5)	{\tiny{$\bfr_v \Delta v$}};
%%
%%
\end{tikzpicture}
\end{minipage}
\medskip

$R$ has approximate area
$ \left| \bfr_u \times \bfr_v \right| \, \Delta u \, \Delta v $
where
$$ \bfr_u = \frac{\dee x}{\dee u} \ii + \frac{\dee y}{\dee u} \jj,
	\qquad
	\bfr_v	=\frac{\dee x}{\dee v} \ii + \frac{\dee y }{\dee v} \jj
$$


Compute
$$  \bfr_u \times \bfr_v = 
\begin{vmatrix}
\ii								&&	\jj								&&	\kk	\\[5pt]
\dfrac{\dee x}{\dee u}	&&	\dfrac{\dee y}{\dee u}	&&	0		\\[10pt]
\dfrac{\dee x}{\dee v}	&&	\dfrac{\dee y}{\dee v}	&&	0
\end{vmatrix}
= \frac{\dee x}{\dee u} \frac{\dee y}{\dee v} - \frac{\dee x}{\dee v}  \frac {\dee y}{\dee u}
$$

\end{frame}

\begin{frame}
\frametitle{The Jacobian}
\begin{center}
\begin{minipage}{0.35\textwidth}
\begin{tikzpicture}[scale=1]
\draw[thick,->,>=stealth]	(-0.5,0)	--	(2,0);
\draw[thick,->,>=stealth]	(0,-0.5)	--	(0,2);
\node[left]	at	(0,2)				{\tiny{$v$}};
\node[below]	at	(2,0)			{\tiny{$u$}};
%%
\draw[orange,fill=orange!20]
	(0.5,1)	rectangle	(1.75,1.75);
\node at (1.125,1.375)			{\tiny{$S$}};
%%
\draw[thick,orange,fill=orange]	(0.5,1)	circle(0.05cm);
\node[below] at (0.45,1)		{\tiny{$(u_0,v_0)$}};
%%
\node[left]	at	(0.5,1.375)		{\tiny{$\Delta v$}};
\node[below]	at	(1.125,1)	{\tiny{$\Delta u$}};
%%
\end{tikzpicture}
\end{minipage}
\quad
\begin{minipage}{0.35\textwidth}
\begin{tikzpicture}[scale=1]
%%%%%%%%%%%%%%
%		x	=	u+ 0.5*v
%		y	=	-0.3*u + v
%%%%%%%%%%%%%%
\draw[thick,->,>=stealth]	(-0.5,0)	--	(2,0);
\draw[thick,->,>=stealth]	(0,-0.5)	--	(0,2);
\node[left]	at	(0,2)		{\tiny{$y$}};
\node[below]	at	(2,0)	{\tiny{$x$}};
%%
\draw[thick,cyan,fill=cyan!20]	
		plot[domain=1:1.75,variable=\v]			({0.5+0.5*\v},{-0.15+\v})	--
		plot[domain=0.5:1.75,variable=\u]		({\u+0.875},{-0.3*\u+1.75})	--
		plot[domain=1.75:1,variable=\v]			({1.75+0.5*\v},{-0.3*1.75+\v})	--
		plot[domain=1.75:0.5,variable=\u]		({\u+0.5},{-0.3*\u+1});
\node	at 	(1.8125,1.0375)	{\tiny{$R$}};
%%
\draw[cyan,fill=cyan]	(1,0.85)	circle(0.05cm);
\node[left]	at	(1,0.85)			{\tiny{$(x_0,y_0)$}};
%%
\draw[very thick,->,>=stealth]	(1,0.85)	--	++(0.375,0.75);
\node[left]	at	(1.1875,1.225)	{\tiny{$\bfr_u \Delta u$}};
\draw[very thick,->,>=stealth]	(1,0.85)	--	++	(1.25,-0.375);
\node[left] at	 (1.6,0.5)	{\tiny{$\bfr_v \Delta v$}};
%%
%%
\end{tikzpicture}
\end{minipage}
\end{center}

\smallskip

\footnotesize
The area of $R$ is approximately
$$ dA \simeq \left| \frac{\dee(x,y)}{\dee(u,v)} \right| \, \Delta u \, \Delta v $$
where
$$ \frac{\dee(x,y)}{\dee(u,v)} = 
\begin{vmatrix}
				\dee x/\dee u		&&		\dee x/\dee v	\\[10pt]
				\dee y/\dee u		&&		\dee y/\dee v
\end{vmatrix}
$$
is the \emph{Jacobian determinant} of the transformation
\end{frame}

\begin{frame}
\frametitle{The Jacobian}
\footnotesize
Find the Jacobian determinant of the following transformations.

\begin{enumerate}
\item	$x=2u+3v$, $y=u-v$
\item	$x=au$, $y=bv$
\item	$x=u^2-v^2$, $y=2uv$
\end{enumerate}
\end{frame}


\begin{frame}
\frametitle{Area Change in Polar Coordinates}
\footnotesize
Consider the transformation
$ x = u \cos v, \quad y = u \sin v $
\begin{center}

\begin{tikzpicture}[scale=1]
\draw[thick,->,>=stealth]	(-0.5,0)	--	(2.5,0);
\draw[thick,->,>=stealth]	(0,-0.5)	--	(0,2.5);
\node[left]		at	(0,2.5)		{\tiny{$v$}};
\node[below]	at	(2.5,0)		{\tiny{$u$}};
\node[below]	at	(-0.25,0)	{\tiny{$0$}};
%%
\draw[orange,fill=orange!20]	(0.5,0.5)  rectangle	(2,1.5);
\node at	(1.25,1.0)				{\tiny{$S$}};
%%
\node[below]	at	(0.5,0)		{\tiny{$a$}};
\node[below]	at	(2,0)			{\tiny{$b$}};
%%
\node[left]	at	(0,0.5)			{\tiny{$\alpha$}};
\node[left]	at	(0,1.5)			{\tiny{$\beta$}};
%%
\draw[densely dashed]	(0.5,0)	--	(0.5,0.5);
\draw[densely dashed]	(2,0)		--	(2,0.5);
%%
\draw[densely dashed]	(0,0.5)	--	(0.5,0.5);
\draw[densely dashed]	(0,1.5)	--	(0.5,1.5);
\end{tikzpicture}
\quad
\begin{tikzpicture}[scale=1]
\draw[thick,->,>=stealth]	(-0.2,0)	--	(2.5,0);
\draw[thick,->,>=stealth]	(0,-0.2)	--	(0,2.5);
\node[left]	at	(0,2.5)			{\tiny{$y$}};
\node[right] at	(2.5,0)			{\tiny{$x$}};
%%
\draw[cyan,fill=cyan!20]
	(20:1.0)	arc	(20:65:1.0)	--	++(65:1.5)	--	(65:2.5)	arc(65:20:2.5)	--	++(20:-1.5);
\node at (42.5:1.75)			{\tiny{$R$}};
%%
\node at	(10:2)					{\tiny{$v =\alpha$}};
\node at	 (77.5:2)				{\tiny{$v =\beta$}};
%%
\node at	(50:0.65)				{\tiny{$u=a$}};
\node at (42.5:3)					{\tiny{$u=b$}};
\end{tikzpicture}

\end{center}

$$
\begin{pmatrix}
				\dee x/\dee u		&&		\dee x/\dee v	\\[10pt]
				\dee y/\dee u		&&		\dee y/\dee v
\end{pmatrix}
=
\begin{pmatrix}
cos(v)		& -u \sin(v)	\\[10pt]
 \sin(v)			&	u \cos(v)
\end{pmatrix}
$$
so 
$$ 
J(u,v) =
\begin{vmatrix}
cos(v)		& \sin(v)			\\[5pt]
-u \sin (v)	&	u \cos(v)
\end{vmatrix}
=	u
$$


\end{frame}

\begin{frame}
\frametitle{Notation}
\footnotesize
The \emph{Jacobian Determinant} of a transformation $x=g(u,v)$, $y=h(u,v)$ is denoted
$$ \frac{\partial (x,y)}{\partial (u,v)}  =
\begin{vmatrix}
\dfrac{\dee x}{\dee u}	&	\dfrac{\dee x}{\dee v}	\\[10pt]
\dfrac{\dee y}{\dee u}	&	\dfrac{\dee y}{\dee v}
\end{vmatrix}
$$
The notation
$$
\left| \frac{\dee(x,y)}{\dee(u,v)} \right|
$$
denotes the \emph{absolute value} of this determinant.
\end{frame}



\section{Change of Variables Formula}

\begin{frame}
\frametitle{Change of Variables Formula}
\footnotesize
If the transformation $x=g(u,v)$, $y=h(u,v)$ maps a region $S$ in the $uv$-plane to a region $R$ in the $xy$ plane:
\begin{align*}
\iint_R f(x,y) \, dA  
	&	\simeq	\sum_{i=1}^n \,  \sum_{j=1}^n	f(x_i, y_j) \, \Delta A\\
	&	\simeq 
			\sum_{i=1}^n \, 
				\sum_{j=1}^n f(g(u_i,v_j),h(u_i,v_j)) 
							\left| \frac{\dee(x,y)}{\dee(u,v)} \right| 
						\, \Delta u \, \Delta v	\\
	&	\iint_S f(g(u,v),h(u,v)) \,
					 \left| \frac{\dee(x,y)}{\dee(u,v)} \right| \, du \, dv
\end{align*}

\medskip

\begin{center}
\fcolorbox{cyan}{white}
{
\begin{minipage}{0.9\textwidth}
{\color{cyan}\textbf{Change of Variables in a Double Integral}}
If $T$ is a one-to-one transformation with nonzero Jacobian and $T: S \to R$, then
$$ \iint_R f(x,y) \, dA = \iint_S f(x(u,v),y(u,v)) \, \left| \frac{\dee(x,y)}{\dee(u,v)} \right| \, du \, dv $$
\end{minipage}
}
\end{center}

\end{frame}

\begin{frame}
\frametitle{Change of Variables Formula}
\footnotesize
\begin{center}
\fcolorbox{cyan}{white}
{
\begin{minipage}{0.9\textwidth}
{\color{cyan}\textbf{Change of Variables in a Double Integral}}
If $T$ is a one-to-one transformation with nonzero Jacobian and $T: S \to R$, then
$$ \iint_R f(x,y) \, dA = \iint_S f(x(u,v),y(u,v)) \, \left| \frac{\dee(x,y)}{\dee(u,v)} \right| \, du \, dv $$
\end{minipage}
}
\end{center}


\medskip

\hrulefill

\medskip

\begin{enumerate}
\item	Use the transformation $x=2u+v$, $y=u+2v$ to find $\iint_R (x-3y) \, dA$ if $R$ is the triangular
region with vertices $(0,0)$, $(2,1)$ and $(1,2)$

\medskip

\item	Find $\iint_R (x+y) e^{x^2-y^2} \, dA$ if $R$ is the rectangle enclosed by $x-y=0$, $x-y=2$, $x+y=0$, and
$x+y=3$.
\end{enumerate}

\end{frame}



\end{document}
