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\begin{document}

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


\title{Math 213 - Triple Integrals (Part II)}			%%	title page
\author{Peter A. Perry}
\institute{University of Kentucky}
\date{October 25, 2019}
\maketitle

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

\begin{frame}
\frametitle{Reminders}

\begin{itemize}
\item	Homework C1 on section 15.3 is due tonight
\item Homework C2 on section 15.6 is due \textbf{Monday night}!
\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)
\begingroup
\color{red}
\item[{}]			Triple Integrals (Part II)
\endgroup
\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

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

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

\section{Learning Goals}

\begin{frame}
\frametitle{Goals of the Day}
\begin{itemize}
\item	Review how to compute triple integrals as iterated integrals

\medskip

\item	Practice setting up triple integrals as iterated integrals of:

\smallskip

\begin{itemize}
\item	Type I  (over $xy$ plane), 

\smallskip

\item	Type II (over $yz$ plane), 

\medskip

\item	Type III (over $xz$ plane)
\end{itemize} 
\end{itemize}
\end{frame}


%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%
%		New material begins here...
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\section{Double}

\begin{frame}
\frametitle{Double Integrals}
\footnotesize


\begin{minipage}{0.3\textwidth}
\begin{tikzpicture}
\draw[thick,->,>=stealth]	(-0.5,0)	--	(2.5,0);
\draw[thick,->,>=stealth]	(0,-0.5)	--	(0,2.5);
\draw[black,thick,fill=orange!50]
	plot[domain=0.5:2.0]	
		({\x},{1+0.25*(\x-1.25)*(\x-1.25)})	
	--	(2,1.141)	
	--	(2,1.72)	
	--	plot[domain=2.0:0.5]	({\x},{2-0.5*(\x-1.25)*(\x-1.25)})	-- cycle;
\node[below] at (0.5,0)		{\scriptsize{$a$}};
\node[below]	at	(2.0,0)		{\scriptsize{$b$}};
\draw[dashed]	(0.5,0)	--	(0.5,1.14);
\draw[dashed]	(2,0)	--	(2,1.14);
\node[below] at	(1.25,0.9)	{\scriptsize{$g_1(x)$}};
\node[above] at	(1.25,2.0)	{\scriptsize{$g_2(x)$}};
\end{tikzpicture}

\bigskip

\begin{tikzpicture}
\draw[thick,->,>=stealth]	(-0.5,0)	--	(2.5,0);
\draw[thick,->,>=stealth]	(0,-0.5)	--	(0,2.5);
\draw[black,thick,fill=orange!50]
	plot[domain=0.5:2.0]	
		({1+0.25*(\x-1.25)*(\x-1.25)},{\x})	
	--	(1.141,2)	
	--	(1.72,2)
	--	plot[domain=2.0:0.5]	({2-0.5*(\x-1.25)*(\x-1.25)},{\x})	-- cycle;
\node[left] at (0,0.5)	{\scriptsize{$c$}};
\node[left]	at	(0,2)		{\scriptsize{$d$}};
\draw[dashed]	(0,0.5)	--	(1.14,0.5);
\draw[dashed]	(0,2)		--	(1.14,2);
\node[above] at		(0.5,1)	{\scriptsize{$h_1(y)$}};
\node[above] at		(2.5,1)	{\scriptsize{$h_2(y)$}};
\end{tikzpicture}
\end{minipage}
\qquad
\begin{minipage}{0.6\textwidth}
\footnotesize

\textbf{Type I}: $R$ lies between the graphs of two continuous functions of $x$
$$ D = \{(x,y): a \leq x \leq b, \, g_1(x) \leq y \leq g_2(x) \} $$
$$ \iint_D f(x,y) \, dA = \int_a^b \left( \int_{g_1(x)}^{g_2(x)} f(x,y) \, dy \right) \, dx $$

\bigskip

\bigskip

\textbf{Type II}: $R$ lies between the graphs of two continuous functions of $y$
$$ D = \{(x,y): c \leq y \leq d, \, h_1(y) \leq x \leq h_2(y) \} $$
$$ \iint_D f(x,y) \, dA = \int_c^d \left( \int_{h_1(y)}^{h_2(y)} f(x,y) \, dx \right) \, dy $$

\end{minipage}
\end{frame}

\begin{frame}
\frametitle{Integrals over Polar Regions}
\begin{minipage}{0.35\textwidth}
\begin{tikzpicture}[scale=0.9]
\draw[thick,->,>=stealth]	(-0.5,0)	--(3,0);
\draw[thick,->,>=stealth]	(0,-0.5)	--(0,3);
\draw[orange,thick,fill=orange!20]
	plot[domain=20:65,variable=\theta,samples=20,smooth]
		(	
			{(1+0.15*sin(4*(\theta-20)))*cos(\theta)},
			{(1+0.15*sin(4*\theta-20)))*sin(\theta)}		
		)	
	--	(65:2.5)
	--	plot[domain=65:20,variable=\theta,samples=20,smooth]
		(
			{(2.5 + 0.50*sin(4*(\theta-65)))*cos(\theta)},
			{(2.5 + 0.50*sin(4*(\theta-65)))*sin(\theta)}
		)
	--	(20:1);
\node[right] at (17:1.5)	{\tiny{$\theta=\alpha$}};
\node[left]	at	(65:2.1)	{\tiny{$\theta=\beta$}};
\node at (42.5:0.5)		{\tiny{$h_1(\theta)$}};
\node at (42.5:2.5)		{\tiny{$h_2(\theta)$}};
\end{tikzpicture}

\end{minipage}
\quad
\begin{minipage}{0.55\textwidth}
\footnotesize
If $f$ is continuous over a polar region of the form
$$ D = \{ (r,\theta): \alpha \leq \theta \leq \beta, \, h_1(\theta) \leq r \leq h_2(\theta) \} $$
then
\begin{multline*} \iint_D f(x,y) \, dA = \\ \int_\alpha^\beta \, \int_{h_1(\theta)}^{h_2(\theta)}	
	f(r\cos \theta, r\sin\theta) \, {\color{red} r} \, dr \, d\theta
\end{multline*}

\end{minipage}
\end{frame}

\section{Triple}

\begin{frame}
\frametitle{Riemann Sums}
\begin{minipage}{0.4\textwidth}
\begin{tikzpicture}[scale=0.5]
\draw[thick,->,>=stealth]	(0,0,0)	--	(5,0,0);
\draw[thick,->,>=stealth]	(0,0,0)	--	(0,3,0);
\draw[thick,->,>=stealth]	(0,0,0)	--	(0,0,3);
%%
\node[right]	at	(5,0,0)	{\scriptsize{$y$}};
\node[above]	at	(0,3,0)	{\scriptsize{$z$}};
\node[below]	at	(0,0,3)	{\scriptsize{$x$}};
%%
\foreach	\i	in	{1,2,3,4,5,6}	%%	x increment
	{
		\foreach		\j	in	{1,2,3,4}	%%	y increment
			{
				\draw[cyan,fill=cyan!20]	({1+0.5*(\i-1)},{0.5*(\j-1)},{3})	
					-- ++(0.5,0,0)	--	++(0,0.5,0)		--	++(-0.5,0,0)	--	++(0,-0.5,0);		%% front
				\draw[cyan,fill=cyan!50]	({1+0.5*(\i-1)},2,{1+0.5*(\j-1)})	
					--	++(0,0,0.5)		--	++(0.5,0,0)		--	++(0,0,-0.5)	--	++(-0.5,0,0);	%% top
			}
	};
\foreach	\i	in	{1,2,3,4}
		{	\foreach		\j	in {1,2,3,4}
			{
				\draw	[cyan,fill=cyan!80]	(4,{0.5*(\i-1)},{1+ 0.5*(\j-1)})	
					--	++(0,0,0.5)		--	++(0,0.5,0)	--	++(0,0,-0.5)	--	++(0,-0.5,0);	%% side
			}
		}
\end{tikzpicture}

\bigskip

\begin{center}
\begin{tikzpicture}[scale=0.5]
\draw[cyan,fill=cyan!30]	(0,0,0)	--	++(2,0,0)	--	++(0,2,0)	--	++(-2,0,0)	--	++(0,-2,0);		%% front
\draw[cyan,fill=cyan!50]	(0,2,0)	--	++(0,0,-2)	--	++(2,0,0)	--	++(0,0,2);							%% top
\draw[cyan,fill=cyan!80]	(2,0,0)	--	++(0,0,-2)	--	++(0,2,0)	--	++(0,0,2)	--	cycle;			%% side
%%
\draw[cyan,dashed]	(0,0,0)	--	(0,0,-2)	--	(2,0,-2);
\draw[cyan,dashed]	(0,0,-2)	--	(0,2,-2);
%%
\node[right]	at	(2,0,-0.6)	{\tiny{$\Delta x$}};
\node[below]	at	(1,0,0)		{\tiny{$\Delta y$}};
\node[right]	at	(2,1,-2)		{\tiny{$\Delta z$}};
%%
\end{tikzpicture}
\end{center}
\end{minipage}
\quad
\begin{minipage}{0.5\textwidth}
\footnotesize
Given a rectangular box
$$ B = [a,b] \times [c,d] \times [r,s] $$
and a function $f(x,y,z)$, we can divide the box into cubes of side
$\Delta x, \, \Delta y, \, \Delta z$ and volume
$$ \Delta V = \Delta x \, \Delta y \, \Delta z $$
The \emph{triple integral} of $f$ over the box $B$ is the limit of Riemann sums
$$ \sum_{i,j,k} f(x^*_{ijk},y^*_{ijk},z^*_{ijk}) \Delta V $$
and is denoted
$$ \iiint_B f(x,y,z) \, dV $$

\end{minipage}
\end{frame}

\section{Iterated}

\begin{frame}
\frametitle{Iterated Integrals Practice}

\footnotesize

Evaluate these integrals and sketch  the region of integration.

\vfill

\begin{enumerate}

\item Evaluate $\dint_0^1 \, \dint_0^1 \, \dint_0^{2-x^2-y^2} xye^z \, dz \, dy\, dx$

\smallskip

\item	Evaluate	$\dint_0^1 \, \dint_0^1 \, \dint_0^{1-z^2} 
\frac{z}{y+1} \, dx \, dz \, dy$

\smallskip

\item	Evaluate $\dint_1^2 \, \dint_0^{2z} \, \dint_0^{\ln x} xe^{-y} \, dy \, dx \, dz$

\end{enumerate}

\end{frame}

\begin{frame}
\frametitle{Triple Integrals as Iterated Integrals}

\footnotesize

\vfill


We have three ways of setting up a triple integral over a region 
$B$ as an iterated integral:

\begin{enumerate}
\item[\textbf{Type I}] $B$ lies over a region $D$ in the $xy$ plane so
$$ \iiint_B f(x,y,z) \, dV = \iint_D \left(\int_{u_1(x,y)}^{u_2(x,y)} f(x,y,z) \, dz \right) \, dA$$
\item[\textbf{Type II}] $B$ lies over a region $D$ in the $yz$ plane, so
$$ \iint_B f(x,y,z) \, dV = \iint_D \left(\int_{u_1(y,z)}^{u_2(y,z)} f(x,y,z) \, dx \right) \, dA $$
\item[\textbf{Type III}] $B$ lies over a region $D$ in the $xz$ plane, so
$$ \iiint_B f(x,y,z) \, dV = \iint_D \left(\int_{u_1(x,z)}^{u_2(x,z)} f(x,y,z) \, dy \right) \, dA
$$
\end{enumerate}


\end{frame}

\section{Type I}

\begin{frame}
\frametitle{Integrals Over Regions: Type I}
\footnotesize
Suppose that 
$$ E = \{ (x,y,z): (x,y) \in D, \, u_1(x,y) \leq z \leq u_2(x,y) \} . $$
\begin{minipage}{0.35\textwidth}
\begin{tikzpicture}[scale=0.6]
\begin{axis}[xlabel=$x$,ylabel=$y$,small,view={120}{30}]
%% curves which should partly be hidden by the graph
\addplot3+[color=red,thick,no markers,samples=5,samples y=0,variable=\t,domain=0:1]	({\t},{\t},0);
%% the surface
\addplot3[surf,domain=0:1,y domain=0:1] {x*y};
%% curves which should not be hidden by the graph
\addplot3+[color=red,thick,no markers,samples=5,samples y=0,variable=\t,domain=0:1]	({\t},0,0);
\addplot3+[color=red,thick,no markers,samples=5,samples y=0,variable=\t,domain=0:1]	(1,{\t},0);
\addplot3+[color=red,thick,no markers,samples=20,samples y=0,variable=\t,domain=0:1]	({\t},{\t},{\t*\t});
\addplot3+[color=red,dashed,thick,no markers,samples=5,samples y=0,variable=\t,domain=0:1]	(1,1,{\t});
\end{axis}
\end{tikzpicture}
\end{minipage}
\quad
\begin{minipage}{0.55\textwidth}
\footnotesize
\begin{multline*}
\iiint_E f(x,y,z) \, dV = \\
\iint_D \left[ \int_{u_1(x,y)}^{u_2(x,y)} f(x,y,z) \, dz \right] \, dA
\end{multline*}

\medskip

\hrulefill

\medskip
		Find $\diiint_E y \, dV$ if $E$ is the region 
			over $$D = \{ 0 \leq y \leq 1, \, \, y \leq x \leq 1\}$$
			where for each $(x,y)$, $$0\leq z \leq xy$$
\end{minipage}

\end{frame}

\section{Type II}

\begin{frame}
\frametitle{Integrals over Regions: Type II}
\footnotesize
If
$$ E = \{ (x,y,z): (y,z) \in D, \, u_1(y,z) \leq x \leq u_2(y,z) \} $$
then
$$ \iiint_E f(x,y,z) \, dV = \iint_D \left[ \int_{u_1(y,z)}^{u_2(y,z)} f(x,y,z) \, dx \right] \, dA $$

\medskip

\hrulefill

\bigskip

\begin{minipage}{0.35\textwidth}
\begin{tikzpicture}[scale=0.6]
\begin{axis}[xlabel=$x$,ylabel=$y$,small,view={120}{30}]
%% bounding curves of the domain in the yz plane
\addplot3+[color=red,thick,no markers,samples=5,samples y=0,variable=\t,domain=1:4]		
	(0,1,{\t});
\addplot3+[color=red,thick,no markers,samples=5,samples y=0,variable=\t,domain=1:4]		
	(0,{\t},4);
\addplot3+[color=red,thick,no markers,samples=5,samples y=0,variable=\t,domain=1:4]	
	(0,{\t},{\t});
%% the plane x=z, semi-opaque, parametric plot
\addplot3[surf,color=orange,opacity=0.4,samples=40,domain=1:4,y domain=1:4]
		(x,y,x);
\end{axis}
\end{tikzpicture}
\end{minipage}
\quad
\begin{minipage}{0.55\textwidth}
Find $\diiint_E \frac{z}{x^2+z^2} \, dV $ if
$$ E = \{ (x,y,z): 1 \leq y \leq 4, \, y \leq z \leq 4, \, 0 \leq x \leq z \}. $$
\end{minipage}


\end{frame}

\section{Type III}

\begin{frame}
\frametitle{Integrals over Regions: Type III}
\footnotesize

If 
$$ E = \{ (x,y,z): (x,z) \in D, \, u_1(x,z) \leq y \leq u_2(x,z) \} $$
then
$$ \diiint_E f(x,y,z) \, dV = \iint_D \left[ \int_{u_1(x,z)}^{u_2(x,z)} f(x,y,z) \, dy \right] \, dA $$

\medskip

\hrulefill

\bigskip

\begin{minipage}{0.35\textwidth}
\begin{tikzpicture}[scale=0.6]
\begin{axis}[xlabel=$x$,ylabel=$y$,small,view={120}{30}]
%% bounding curves of the domain in the yz plane
\addplot3+[color=red,dashed,fill=red!20,thick,no markers,samples=50,samples y=0,variable=\t,domain=0:360]		
	({2*cos(\t)},0,{2*sin(\t)});
%% the paraboloid, parametric plot, x means r, y means theta
\addplot3[surf,colormap/cool,z buffer=sort,samples=40,domain=0:2,y domain=0:360]
		({x*cos(y)},x^2,{x*sin(y)});
\end{axis}
\end{tikzpicture}
\end{minipage}
\quad
\begin{minipage}{0.55\textwidth}
Find $\diiint_E \sqrt{x^2+z^2} \, dV$ if $E$ is the region bounded by the paraboloid $y=x^2+z^2$ and the plane $y=4$

\medskip

\emph{Hint}: Remember that you can use polar coordinates to evaluate the double integral!
\end{minipage}

\end{frame}

\section{Volumes}

\begin{frame}
\frametitle{Volumes}
\footnotesize
\emph{Area} is computed by a double integral
$A(D) = \diint_D 1 \, dA $

\medskip

\emph{Volume} is computed by a triple integral
$ V(E) = \diiint_E 1 \, dV $

\vfill

\hrulefill

\vfill

\begin{minipage}{0.35\textwidth}
\begin{tikzpicture}
\begin{tikzpicture}[scale=0.45]
\begin{axis}[xlabel=$x$,ylabel=$y$,colormap/cool,view={120}{30}]]
%% bounding curves of the domain in the yz plane
\addplot3+[color=red,dashed,fill=red!20,thick,no markers,samples=50,samples y=0,variable=\t,domain=0:360]		
	({2*cos(\t)},0,{2*sin(\t)});
%% parameterized surface where x is r and y is theta
\addplot3[surf,z buffer=sort,samples=40,domain=0:2,y domain=0:360]
	({x*cos(y)},{x*x}, {x*sin(y)});
\addplot3[surf,z buffer=sort,samples=40,domain=0:2,y domain=0:360]
	({x*cos(y)},{8-x*x}, {x*sin(y)});
\end{axis}
\end{tikzpicture}
\end{tikzpicture}
\end{minipage}
\quad
\begin{minipage}{0.55\textwidth}
\footnotesize

Find the volume enclosed by the paraboloids $$y=x^2+z^2$$ and $$y=8-x^2 - z^2$$

\medskip

This is a Type III integral since we are given a range of $y$. 
\begin{itemize}
\item		Where do these surfaces intersect?
\item		What is the domain in the $xz$ plane?
\end{itemize}
\end{minipage}

\vfill

\end{frame}





\end{document}
