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\begin{document}

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


\title{Math 213 - Exam III Review}			%%	title page
\author{Peter A. Perry}
\institute{University of Kentucky}
\date{November 13, 2019}
\maketitle

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

\begin{frame}
\frametitle{Reminders}

\begin{itemize}

\item	Exam III is tonight, 5:00-7:00 PM
\item	Homework C7 is due on Friday Night
\item Thanksgiving is coming!

\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



\item[{}]			Change of Variables, Part I

\item[{}]			Change of Variables, Part II

\medskip


\item[{}]			Vector Fields

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


\medskip

\begingroup
\color{red}
\item[{}]			Exam III Review
\endgroup
\end{itemize}
\end{frame}

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

\section{Learning Goals}

\begin{frame}
\frametitle{Goals of the Day}
\begin{itemize}
\item	Learn how to get your best score yet on a Calculus Exam
\end{itemize}
\end{frame}

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

\section{Overview}

\begin{frame}
\frametitle{Fantastic Integrals and How to Compute Them}
\footnotesize


\begin{minipage}{0.8\textwidth}
{\renewcommand{\arraystretch}{1.2}
\begin{tabular}{lll}
Double Integral	$\iint_D f(x,y) \, dA$	&&	Iterated integral (Cartesian coordinates)\\
														&&	\quad Type I Region\\
														&&	\quad Type II Region\\ 
														&&	Iterated integral (polar coordinates)\\
														&&	Change of variables theorem\\
														\\
Triple Integral $\iiint_E f(x,y,z) \, dV$		&&	Iterated integral in $xyz$\\
														&&	\quad over $xy$, $xz$, or $yz$ plane\\
														&&	Iterated integral (cylindrical coordinates)) \\
														&&	Iterated integral (spherical coordinates)	\\
														&&	Change of variables theorem\\
														\\
Line integral $\int_C f(x,y) \, ds$			&&	$ds=\sqrt{(x')^2+(y')^2} \,dt$\\
Line integral $\int_C f(x,y) \, dx$			&&	$dx=x'(t) dt$\\
Line integral $\int_C f(x,y) \, dy$			&& 	$dy =y'(t) dt$\\
Line integral $\int_C f(x,y,z) \, ds$		&& 	$ds = \sqrt{(x'(t))^2+(y'(t))^2+(z'(t))^2} \, dt$			
\end{tabular}
}
\end{minipage}

\end{frame}

\begin{frame}
\frametitle{Transformations: More Than Meets the Eye}
\footnotesize
\begin{minipage}{0.55\textwidth}
In two dimensions, a \emph{transformation} is a map $$(u,v) \to (x(u,v),y(u,v))$$ that transforms a region $S$ of the $uv$ plane to a region $R$ of the $xy$ plane. The `volume change' factor is the absolute value of
the \emph{Jacobian}
$$ \frac{\dee(x,y)}{\dee(u,v)} = 
\begin{vmatrix}
\dfrac{\dee x}{\dee u}	&	 \dfrac{\dee x}{\dee v}\\[10pt]
\dfrac{\dee y}{\dee u}	&	\dfrac{\dee y}{\dee v}
\end{vmatrix}
$$

\medskip

Example:

\medskip
\begin{align*}
x(u,v)	&=	u \cos v\\
y(u,v)	&=	u \sin v
\end{align*}
\end{minipage}
\qquad
\begin{minipage}{0.35\textwidth}
\begin{tikzpicture}[scale=1.2]
\draw[thick,->,>=stealth]	(-1,0) -- (1.5,0) 	node[anchor=west]  {$u$};
\draw[thick,->,>=stealth]  (0,-0.25) -- (0,2) 	node[anchor=south] {$v$};
\draw[dashed,red,opacity=0.5]		(0.5,0)			--	(0.5,0.78);
\draw[dashed,red,opacity=0.5]		(1,0)			--	(1,0.78);
\draw[dashed,red,opacity=0.5]		(0,0.78)		--	(1,0.78);
\draw[dashed,red,opacity=0.5]		(0,1.57)	--	(1,1.57);
\node[below] at (0.5,0)		{\footnotesize{$1$}};
\node[below] at (1,0)			{\footnotesize{$2$}};
\node[left] 		at (0,0.78)		{\footnotesize{$\pi/2$}};
\node[left]		at	(0,1.57)		{\footnotesize{$\pi$}};
\draw[red,fill=red!20,opacity=0.5]	(0.5,0.78) rectangle (1,1.57);
\node at	(0.75,1.2)				{$S$};
\end{tikzpicture}

\bigskip

\begin{tikzpicture}[scale=0.6]
\draw[help lines] (-2,-2) grid(2,2);
\draw[thick,->,>=stealth]	(-2.5,0) -- (2.5,0) 	node[anchor=west] {$x$};
\draw[thick,->,>=stealth]  (0,-2.5) -- (0,2.5) 	node[anchor=south] {$y$};
\draw[red,fill=red!20,opacity=0.5] (0,1) arc(90:180:1) -- (-2,0) -- (-2,0) arc(180:90:2) -- (0,2);
\node at (-1,1)	{$R$};
\end{tikzpicture}

\end{minipage}

\end{frame}

\begin{frame}
\frametitle{Transformations: More Than Meets the Eye}
\footnotesize
In three dimensions, a \emph{transformation} is a map $$(u,v,w) \to (x(u,v,w),y(u,v,w),z(u,v,w))$$ that transforms a region $S$ of $uvw$ space to a region $R$ of $xyz$ space. The `volume change factor' is the absolute value of the  \emph{Jacobian}
$$ \dfrac{\dee(x,y,z)}{\dee(u,v,w)} = 
\begin{vmatrix}
\dfrac{\dee x}{\dee u}	&	\dfrac{\dee x}{\dee v}	&	\dfrac{\dee x}{\dee w}	\\[10pt]
\dfrac{\dee y}{\dee u}	&	\dfrac{\dee y}{\dee v}	&	\dfrac{\dee y}{\dee w}	\\[10pt]
\dfrac{\dee z}{\dee u}	&	\dfrac{\dee z}{\dee v}	&	\dfrac{\dee z}{\dee w}	\\
\end{vmatrix}
$$
Example:
$$ (u,v,w) \to (u \cos v, u \sin v, w) $$
\end{frame}

\begin{frame}
\frametitle{Transformations: More Than Meets the Eye}
\footnotesize
Example:
$$ (u,v,w) \to (u \cos v, u \sin v, w) $$

\vfill

\begin{minipage}{0.45\textwidth}
\tdplotsetmaincoords{70}{110} 
\begin{tikzpicture}[tdplot_main_coords]
\draw[thick,gray,->,>=stealth] (0,0,0) -- (4,0,0) node[anchor=north east]{$u$}; 
\draw[thick,gray,->,>=stealth] (0,0,0) -- (0,3,0) node[anchor=north west]{$v$}; 
\draw[thick,gray,->,>=stealth] (0,0,0) -- (0,0,3.5) node[anchor=south]{$w$};
%% top
\draw[blue,fill=blue!20,opacity=0.5]	(1,0,3)	--	(2,0,3)	--	(2,1.57,3)	--(1,1.57,3)	--	cycle;
%% face parallel to vw plane
\draw[blue,fill=blue!40,opacity=0.5]	(2,0,3)	--	(2,0,1)	--	(2,1.57,1)	--	(2,1.57,3)	-- cycle;
%% face parallel to uw plane
\draw[blue,fill=blue!80,opacity=0.5]	(2,1.57,1)	--	(1,1.57,1)	--	(1,1.57,3)	--	(2,1.57,3)	--	cycle;
\end{tikzpicture}
\end{minipage}
\qquad
\begin{minipage}{0.35\textwidth}
\tdplotsetmaincoords{70}{110} 
\begin{tikzpicture}[tdplot_main_coords]
\draw[thick,gray,->,>=stealth] (0,0,0) -- (4,0,0) node[anchor=north east]	{$x$}; 
\draw[thick,gray,->,>=stealth] (0,0,0) -- (0,3,0) node[anchor=north west]	{$y$}; 
\draw[thick,gray,->,>=stealth] (0,0,0) -- (0,0,3) node[anchor=south]			{$z$};
%%	top
\draw[blue,left color=blue!10,right color=blue!50,opacity=0.5]	
	(1,0,3)	--	(2,0,3)	--	(2,0,3) arc (0:90:2)	--	(0,1,3)	--	(0,1,3) arc(90:0:1)	-- cycle;
%% front face
\draw[blue,left color=blue!20,right color=blue,opacity=0.5]  
	(2,0,3)	--	(2,0,1)	--	(2,0,1) arc (0:90:2)	--	(0,2,3) -- (0,2,3) arc (90:0:2);
\end{tikzpicture}
\end{minipage}
\end{frame}

\section{Double Integrals}

\begin{frame}
\frametitle{Double Integrals}
\footnotesize
\begin{minipage}{0.45\textwidth}
Find the $ \iint_D xy^2 \, dA$ if $D$ is the region 
bounded by the curves $y=x^2$ and $x=y^2$ in the 
first quadrant.
\end{minipage}
\qquad
\begin{minipage}{0.45\textwidth}
\begin{tikzpicture}[scale=1.5]
\draw[help lines,step=0.2,opacity=0.4]	(0,0)	grid (1,1);
\draw[thick,gray,->,>=stealth]	(-0.5,0)	--	(1.5,0)	node[anchor=west]	{$x$};
\draw[thick,gray,->,>=stealth]	(0,-0.5)	--	(0,1.5)	node[anchor=south] {$y$};
\draw[thick,red,fill=red!30,opacity=0.5]	plot[domain=0:1]	(\x,{\x*\x}) -- plot[domain=1:0] ({\x*\x},{\x});
\node[anchor=south east] at (0.42,0.6) 		{\scriptsize{$x=y^2$}};
\node[anchor=north west] at (0.45,0.3) 		{\scriptsize{$y=x^2$}};
\end{tikzpicture}
\end{minipage}

\vfill

\begin{minipage}{0.45\textwidth}
Find the $ \iint_D xy \, dA$ if $D$ is the region 
shown.
\end{minipage}
\qquad
\begin{minipage}{0.45\textwidth}
\begin{tikzpicture}[scale=0.55]
\draw[thick,gray,->,>=stealth]	(-3,0)	--	(3,0)	node[anchor=west]	{$x$};
\draw[thick,gray,->,>=stealth]	(0,-3)	--	(0,3)	node[anchor=south] {$y$};
%%
\draw[thick,red,fill=red!30,opacity=0.5]
	(0,1)	--	(0,1) arc (90:360:1) -- (2,0)	--	(2,0) arc(360:90:2) -- (0,1);
\node[anchor=south east] 	at (-1.42,1.42) 	{\scriptsize{$x^2+y^2=4$}};
\node[anchor=west]			at (-0.65,0.4)		{\scriptsize{$x^2+y^2=1$}};
\end{tikzpicture}
\end{minipage}
\end{frame}


\section{Triple Integrals}

\begin{frame}
\frametitle{Triple Integrals, Cartesian Coordinates}
\footnotesize
Set up $\iiint_E f(x,y,z) \, dV$ six  different ways if
$$ E = \{ (x,y,z): 0 \leq y \leq 1, \, y \leq x \leq 1, \, 0 \leq z \leq y \}$$ 

\medskip
\begin{align*}
{\color{purple} 
	\int_0^1 \, \int_y^1 }
			\,   \int_0^y f(x,y,z) \, dz  {\color{purple} \, dx  \, dy }
	&=		{\color{purple} \int_0^1 \, \int_0^x} \, \int_0^y  f(x,y,z) \, dz  {\color{purple} \, dy \, dx  }\\
\int_0^1 \, {\color{purple} \int_0^x \, \int_0^y}  f(x,y,z) {\color{purple} \, dz  \, dy } \, dx  
	&=	\int_0^1 \, {\color{purple} \int_0^x \, \int_z^x } f(x,y,z) {\color{purple} \, dy \, dz } \, dx \\
{\color{purple} \int_0^1 \, \int_0^x}  \, \int_0^y  f(x,y,z) \, dz  {\color{purple} \, dy  \, dx  }
	&=	{\color{purple} \int_0^1 \,   \int_y^1}   \, \int_z^x  f(x,y,z)\, dy  \,  {\color{purple}dx  \, dz}
\end{align*}
etc.
\end{frame}

\begin{frame}
\frametitle{Triple Integrals, Cylindrical Coordinates}
\footnotesize

For cylindrical coordinates
$$ dV = {\color{red} r}  \, dr \, d\theta \, dz $$

\bigskip

Find $\iiint_E y \, dV$ if $E$ lies below the plane $z=x+2$, above the $xy$ plane, and between the cylinders 
$x^2+y^2=1$ and $x^2+y^2=4$.

\tdplotsetmaincoords{60}{110} 
\begin{tikzpicture}[tdplot_main_coords,scale=0.7]
%%	plane
\draw[top color=blue!10,bottom color=blue!40]
	(-2,-3,0)	-- (-2,4,0)	-- (2,4,4)	--	(2,-3,4);
%% cylinder base
\draw[dashed,color=blue!70]
	(1,0,0)	arc(0:360:1)	--	(2,0,0) arc(0:360:2);
%% cylinder top
\draw[dashed,color=blue]
	plot[domain=0:360,variable=\t]
		({cos(\t)},{sin(\t)},{cos(\t)+2});
\draw[dashed,color=blue]
	plot[domain=0:360,variable=\t]
		({2*cos(\t)},{2*sin(\t)},{2*cos(\t)+2});
%% axes
\draw[thick,gray,->,>=stealth] (0,0,0) -- (5,0,0) node[anchor=north east]	{$x$}; 
\draw[thick,gray,->,>=stealth] (0,0,0) -- (0,4,0) node[anchor=north west]	{$y$}; 
\draw[thick,gray,->,>=stealth] (0,0,0) -- (0,0,4) node[anchor=south]			{$z$};
%% verticals for cylinder
\foreach \theta in {10,55,100,145,190,235,280,335}
	{	
		\draw[dashed,blue,opacity=0.5] ({2*cos(\theta)},{2*sin(\theta)},0)	--	++(0,0,{2*cos(\theta)+2});
	}
\end{tikzpicture}

\end{frame}

\begin{frame}
\frametitle{Triple Integrals, Spherical Coordinates}
\footnotesize

For spherical coordinates, 
$$ dV = {\color{red} \rho^2 \sin \phi } \, d \rho \, d\theta \, d\phi $$

\medskip

Find the volume of the region contained inside the sphere $x^2+y^2+z^2 = 4$ but outside the cylinder $x^2+y^2=1$

\medskip

\tdplotsetmaincoords{60}{135} 
\begin{tikzpicture}[tdplot_main_coords,scale=0.5]
%% sphere boundary
\draw[blue,shading=ball,opacity=0.5]	
	(0,0,0) circle(2cm);
%% take off top
\draw[dashed,blue,left color=white,right color=blue!50]
	(1,0,1.76) arc(0:360:1);
%% axes
\draw[thick,gray,->,>=stealth] (0,0,0) -- (4,0,0) node[anchor=north east]	{$x$}; 
\draw[thick,gray,->,>=stealth] (0,0,0) -- (0,4,0) node[anchor=north west]	{$y$}; 
\draw[thick,gray,->,>=stealth] (0,0,0) -- (0,0,4) node[anchor=south]			{$z$};
%% cylinder top and bottom
\draw[blue,opacity=0.5]
	(1,0,1.76) arc(0:360:1);
\draw[dashed,blue,opacity=0.5]
	(1,0,-1.76) arc(0:360:1);
%% cylinder sides
\draw[dashed,blue,opacity=0.5]
	(0.707,-0.707,-1.76) -- ++(0,0,3.52);
\draw[dashed,blue,opacity=0.5]
	(-0.707,0.707,-1.76) -- ++(0,0,3.52);
\end{tikzpicture}
\end{frame}

\section{Change of Variables}

\begin{frame}
\frametitle{Change of Variables Theorem}
\footnotesize
$$ \iint_R f(x,y) \, dA = \iint_S f(x(u,v),y(u,v)) \left| \frac{\dee(x,y)}{\dee(u,v)} \right| \, du \, dv $$

\bigskip

Using the change of variables 
$$x=\sqrt{2}u -  \sqrt{\frac{2}{3}} v, \quad y = \sqrt{2} u + \sqrt{\frac{2}{3}}v$$
find $\iint_D (x^2 - xy + y^2) \, dA$ if $D$ is the region $x^2 - xy + y^2 \leq 2$. 

\bigskip

\begin{minipage}{0.45\textwidth}
\begin{center}
\begin{tikzpicture}[scale=0.6]
\draw[help lines]	(-2,-2) grid (2,2);
\draw[thick,->,>=stealth]	(-2.5,0)		--	(2.5,0)	node[anchor=west] 	{$u$};
\draw[thick,->,>=stealth]	(0,-2.5)		--	(0,2.5)	node[anchor=south]	{$v$};
\draw[red,fill=red!20,opacity=0.5]		(0,0)	circle(1cm);
\node[anchor=south west] at (0.05,0.05)	{$S$};
\end{tikzpicture}
\end{center}
\end{minipage}
\qquad
\begin{minipage}{0.45\textwidth}
\begin{tikzpicture}[scale=0.6]
\draw[help lines]	(-2,-2) grid (2,2);
\draw[thick,->,>=stealth]	(-2.5,0)		--	(2.5,0)	node[anchor=west] 	{$x$};
\draw[thick,->,>=stealth]	(0,-2.5)		--	(0,2.5)	node[anchor=south]	{$y$};
\draw[red,fill=red!20,opacity=0.5,smooth]		plot[domain=0:360,variable=\t]
			({1.414*cos(\t)-0.816*sin(\t)},{1.414*cos(\t)+0.816*sin(\t)});
\node at (0.5,0.5)	{$R$};
\end{tikzpicture}
\end{minipage}
\end{frame}


\section{Line Integrals}

\begin{frame}
\frametitle{Line Integral Formulas}

\scriptsize

Arc length differential: 
$ds							=	\sqrt{x'(t)^2+y'(t)^2+z'(t)^2} \, dt
$

\vfill
\vfill

Line integral with respect to arc length
\begin{align*}
\int_C f(x,y,z)  \,ds	&=	\int_a^b f(x(t),y(t),z(t)) \, \sqrt{x'(t)^2+y'(t)^2+z'(t)^2} \, dt\\
\intertext{Line integral with respect to $x$, $y$, $z$}
\int_C f(x,y,z) \, dx	&=	\int_a^b f(x(t),y(t),z(t))  x'(t) \, dt\\
\int_C f(x,y,z) \, dy	&=	\int_a^b f(x(t),y(t),z(t))  y'(t) \, dt\\
\int_C f(x,y,z) \, dy	&=	\int_a^b f(x(t),y(t),z(t))  z'(t) \, dt\\
\intertext{Line integral of a vector field}
\int_C \bfF \cdot d\bfr	&=	\int_a^b \bfF(x(t),y(t),z(t)) \cdot \bfr'(t) \, dt 
\end{align*}

\end{frame}

\begin{frame}
\frametitle{Line Integral Problems}

\footnotesize
Find $\int_C 4x^3 \, ds$ if $C$ is a line segment from $(-2,1)$ to $(1,2)$

\vfill
Find $\int_C xy \, ds$ if $C$ is a circle of radius $2$.

\vfill
Find $\int_C xy \, ds$ if $C$ is the helix $\langle \cos(t), \sin(t), 2t \rangle$ from $t=0$ to $t=\pi$. 

\vfill
Find $\int_C y \, dx$ if $C$ is the quarter-circle of radius $3$ in the first quadrant with center at the origin.


\vfill

\end{frame}



\end{document}
