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\begin{document}

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


\title{Math 213 - Semester Review - I}			%%	title page
\author{Peter A. Perry}
\institute{University of Kentucky}
\date{December 9, 2019}
\maketitle

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

\begin{frame}
\frametitle{Reminders}

\begin{itemize}


\item	Homework D5 (16.9, the Divergence Theorem) is due Wednesday night

\medskip

\item	There will be a drop-in review session for the final exam on Wednesday, December 18, 3:30-5:30 PM, CB 106. 

\item	Your final exam is Thursday, December 19 at 6:00 PM. Room assignments are the same as for Exams I - III

\item	On your final exam:

		\begin{itemize}
		\item	The multiple choice questions will be 50\% from Units I - III and 50\% from unit IV. 
		
		\smallskip
		
		\item All free response questions will be from unit IV. Since these questions typically involve
				integrals, they will also test material from unit III
		\end{itemize}

\end{itemize}

\end{frame}

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

\begin{frame}
\frametitle{Unit IV: Vector Calculus}
\begin{center}
\begin{itemize}
\item[{}]			Fundamental Theorem for Line Integrals
\item[{}]			Green's Theorem
\item[{}]			Curl and Divergence
\item[{}]			Parametric Surfaces and their Areas
\item[{}]			Surface Integrals
\item[{}]			Stokes' Theorem, I
\item[{}]			Stokes' Theorem, II 

\item[{}]			The Divergence Theorem
\item[{}]		
\begingroup
\color{red}	
\item[{}]			Review, I
\endgroup
\item[{}]			Review, II
\item[{}]			Review, III
\end{itemize}
\end{center}
\end{frame}

\section{Learning Goals}

\begin{frame}
\frametitle{Goals of the Day}
Calculus is about functions, derivatives, integrals, and ``fundamental theorems'' that relate them. Today we will review all of the 

\medskip

\begin{itemize}
\item	New functions
\item	New derivatives
\item	New integrals
\item New theorems
\end{itemize}

\medskip\noindent
that we've learned about in this course.
\end{frame}

\section{New Functions}

\begin{frame}
\frametitle{New Functions}
\footnotesize
\begin{itemize}
\item		\textbf{Vector functions}  $\bfr(t) = \langle x(t), y(t), z(t) \rangle$ for space curves, such as
			$$ \bfr(t) = \langle \cos t, \sin t, t \rangle $$
\item		\textbf{Vector functions} $ \bfr(u,v) = x(u,v) \ii + y(u,v) \jj + z(u,v) \kk$ for surfaces, such as
			$$\bfr(u,v) = \langle u \cos v, u \sin v, v \rangle$$
			
\item		\textbf{Functions of several variables} $f(x,y)$ and
			$g(x,y,z)$ such as
			$$ f(x,y) = x^2 + y^2, \quad g(x,y,z) = e^{xyz} $$
\item		\textbf{Transformations} $(x(u,v), y(u,v))$
			and $(x(u,v,w), y(u,v,w), z(u,v,w))$
			such as 
			$$ x(u,v) = u^2 - v^2, \quad y(u,v) = 2uv $$
\item		\textbf{Vector fields}
			$$
			\bfF(x,y,z) = P(x,y,z) \ii + Q(x,y,z) \jj
				+ R(x,y,z) \kk
			$$
\end{itemize}
\end{frame}

\section{New Derivatives}

\begin{frame}
\frametitle{New Derivatives - Vector Functions}
\footnotesize
\begin{itemize}
\item	The tangent vector to a space curve:
		$$ \bfr'(t) = \langle x'(t) , y'(t) , z'(t) \rangle $$
		$$ ds = \left| \bfr'(t) \right| \, dt, \quad d\bfr = \bfr'(t) \, dt $$
\item	The tangent vectors to a parameterized surface
		$$ \bfr_u(u,v) = \frac{\dee x}{\dee u} \ii +
				\frac{\dee y}{\dee u} \jj +
				\frac{\dee z}{\dee u} \kk, \quad
			\bfr_v(u,v) = \frac{\dee x}{\dee v} \ii +
				\frac{\dee y}{\dee v} \jj +
				\frac{\dee z}{\dee v} \kk
	   	$$
	   	and the element of area
	   	$$ dS = \left| \bfr_u \times \bfr_v \right| \, du \, dv, \quad
	   	     d\bfS = \left( \bfr_u \times \bfr_v \right) \, du \, dv
	   	 $$
\end{itemize}
\end{frame}

\begin{frame}
\frametitle{New Derivatives - Functions of Several Variables}
\footnotesize
\begin{itemize}
\item	The gradient of a function of a function of two variables
		$$ \nabla f (x,y) = \frac{\dee f}{\dee x} \ii + \frac{\dee f}{\dee y} \jj  $$
		(greatest change, directional derivatives, critical points)

\item	The Hessian of a function of two variables
		$$ \Hess(f)  = 
			\begin{pmatrix}
			\dfrac{\dee^2 f}{\dee x^2}	&	\dfrac{\dee^2 f}{\dee x \dee y} \\
			\\
			\dfrac{\dee^2 f}{\dee y \dee x} & \dfrac{\dee^2 f}{\dee y^2}
			\end{pmatrix}
		$$
		(determine whether critical points are local extrema or saddle points)
\end{itemize}
\end{frame}

\begin{frame}
\frametitle{New Derivatives - Transformations}
\footnotesize
\begin{itemize}
\item	The Jacobian matrix of a transformation $(x(u,v), y(u,v))$
		$$
		\begin{pmatrix}
		\dfrac{\dee x}{\dee u}	&	\dfrac{\dee x}{\dee v}	\\[10pt]
		\dfrac{\dee y}{\dee u}	&	\dfrac{\dee y}{\dee v}
		\end{pmatrix}
		$$
		(Area change from $uv$ plane to $xy$ plane)
\item	The Jacobian of a transformation $x(u,v,w), y(u,v,w), z(u,v,w)$
		$$
		\begin{pmatrix}
		\dfrac{\dee x}{\dee u}	&	\dfrac{\dee x}{\dee v}	& \dfrac{\dee x}{\dee w}\\[10pt]
		\dfrac{\dee y}{\dee u}	&	\dfrac{\dee y}{\dee v}	&	\dfrac{\dee y}{\dee w}\\[10pt]
		\dfrac{\dee z}{\dee u}	&	\dfrac{\dee z}{\dee v}	&	\dfrac{\dee z}{\dee w}
		\end{pmatrix}
		$$
		(Volume change from $uvw$ space to $xyz$ space)
\end{itemize}
\end{frame}

\begin{frame}
\frametitle{New Derivatives - Vector Fields}
\footnotesize
A \emph{vector field} is a function 
$$ \bfF (x,y,z) = P(x,y,z) \ii + Q(x,y,z) \jj + R(x,y,z) \kk $$
so there are \emph{nine} derivatives to choose from:
$$
\begin{pmatrix}
& \dfrac{\dee P}{\dee x}	&&	\dfrac{\dee P}{\dee y}	&&	\dfrac{\dee P}{\dee z}	& \\[10pt]
&\dfrac{\dee Q}{\dee x}	&&	\dfrac{\dee Q}{\dee y}	&&	\dfrac{\dee Q}{\dee z}	& \\[10pt]
&\dfrac{\dee R}{\dee x}	&&	\dfrac{\dee R}{\dee y}	&&	\dfrac{\dee R}{\dee z}	 & 
\end{pmatrix}
$$
Experience shows that there are two important ones, a scalar (the divergence)  and a vector (the curl).
\end{frame}

\begin{frame}
\frametitle{The Divergence}
\footnotesize
$$ \bfF (x,y,z) = P(x,y,z) \ii + Q(x,y,z) \jj + R(x,y,z) \kk $$

\vfill

$$
\begin{pmatrix}
& {\color{blue}\dfrac{\dee P}{\dee x}}	&&	{\color{gray}\dfrac{\dee P}{\dee y}}	&&	{\color{gray}\dfrac{\dee P}{\dee z}}	& \\[10pt]
&{\color{gray}\dfrac{\dee Q}{\dee x}}	&&	{\color{blue}\dfrac{\dee Q}{\dee y}}	&&	{\color{gray}\dfrac{\dee Q}{\dee z}}	& \\[10pt]
&{\color{gray}\dfrac{\dee R}{\dee x}}	&&	{\color{gray}\dfrac{\dee R}{\dee y}}	&&	{\color{blue}\dfrac{\dee R}{\dee z}}	 & 
\end{pmatrix}
$$

\vfill
$$ \diver \bfF = \nabla \cdot \bfF = \frac{\dee P}{\dee x} + \frac{\dee Q}{\dee y} + \frac{\dee R}{\dee z} $$

\vfill
The divergence is a \emph{scalar} which measures net flux of $\bfF$ per unit volume 
\end{frame}

\begin{frame}
\frametitle{The Curl}
\footnotesize
$$ \bfF (x,y,z) = P(x,y,z) \ii + Q(x,y,z) \jj + R(x,y,z) \kk $$

\vfill

$$
\begin{pmatrix}
& {\color{gray}\dfrac{\dee P}{\dee x}}	&&	{\color{red}\dfrac{\dee P}{\dee y}}	&&	{\color{blue}\dfrac{\dee P}{\dee z}}	& \\[10pt]
&{\color{red}\dfrac{\dee Q}{\dee x}}	&&	{\color{gray}\dfrac{\dee Q}{\dee y}}	&&	{\color{cyan} \dfrac{\dee Q}{\dee z}}	& \\[10pt]
&{\color{blue} \dfrac{\dee R}{\dee x}}	&&	{\color{cyan}\dfrac{\dee R}{\dee y}}	&&	{\color{gray}\dfrac{\dee R}{\dee z}}	 & 
\end{pmatrix}
$$

\vfill
$$
\curl \bfF = \nabla \times \bfF	=
\left( {\color{cyan} \frac{\dee R}{\dee y} - \frac{\dee Q}{\dee z} }\right)\ii +
\left( {\color{blue} \frac{\dee P}{\dee z} - \frac{\dee R}{\dee x} } \right)\jj +
\left( {\color{red} \frac{\dee Q}{\dee x} - \frac{\dee P}{\dee y}} \right)\kk
$$
\vfill
The curl is a \emph{vector}. The circulation of $\bfF$ around the boundary of an oriented area $d\bfS$ is $\curl \bfF \cdot \, d\bfS$

\end{frame}

\section{New Integrals}

\begin{frame}
\frametitle{New Integrals - Double Integrals}
\footnotesize
If $f(x,y)$ is a function of two variables defined on a region $D$ in the $xy$ plane, 
the double integral of $f$ over $D$ is $\iint_D f(x,y) \, dA$. It can be computed in 
the following ways:

\vfill
\begin{itemize}
\item	If $D=[a,b] \times [c,d]$
		$$ \iint_D f(x,y) \, dA = \int_a^b \, \int_c^d f(x,y) \, dy \, dx $$
\item	If $D = \{ (x,y): a \leq x \leq b, g_1(x) \leq y \leq g_2(y) \}$ then
		$$ \iint_D f(x,y) \, dA = \int_a^b \, \int_{g_1(x)}^{g_2(x)} f(x,y) \, dy \, dx $$
\item	If $D = \{ (r,\theta): \alpha \leq \theta \leq \beta, \, c \leq r \leq d \}$
		then
		$$ \iint_D f(x,y) \, dA = \int_{\alpha}^\beta \, \int_{c}^d f(r\cos\theta,r\sin\theta) \, r \, dr \, d\theta$$
\end{itemize}
\end{frame}


\begin{frame}
\frametitle{New Integrals - Triple Integrals}
\footnotesize
If $f(x,y,z)$ is a function of three variables defined on a region $E$ of $xyz$ space, the triple integral of $f$ over $E$ is $\iiint_E f(x,y,z) \, dV$. It can be computed in the following ways (among others!):

\begin{itemize}
\item	If $E=\{ (x,y,z): a \leq x \leq b, c \leq y \leq d, r \leq z \leq s \}$ then
	$$ \iiint_E f(x,y,z) \, dV = \int_a^b \, \int_c^d \, \int_r^s f(x,y,z) \, dz \, dy \, dx $$
\item	If $E= \{ (x,y,z): (x,y) \in D \text{ and } g_1(x,y) \leq z \leq g_2(x,y) \} $ then
	$$ \iiint_E f(x,y,z) \, dV 
	= \iint_D \, \left( \int_{g_1(x,y)}^{g_2(x,y)} f(x,y,z) \, dz \right) \, dA$$
\item	If $E = \{ (\rho,\theta,\phi): \alpha \leq \theta \leq \beta, \, \phi_1 \leq \phi \leq \phi_2, \, a \leq \rho \leq b \}$ then
	\begin{multline*} \iint_E f(x,y,z) \, dV =\\
		\int_\alpha^\beta \, \int_{\phi_1}^{\phi_2} \, \int_a^b
			f(\rho \sin \phi \cos \theta, \rho \sin \phi \sin \theta, \rho \cos \phi) \,
			\rho^2 \sin \phi \, d\rho \, d\phi \, d\theta 
	\end{multline*}
\end{itemize}
\end{frame}

\begin{frame}
\frametitle{New Integrals - Line Integrals}
\footnotesize
If the space curve $C$ is parameterized by $\bfr(t) = x(t) \ii + y(t) \jj + z(t) \kk$, $a \leq t \leq b$, then:
\vfill
\begin{itemize}
\item The line integral of a scalar function $f(x,y,z)$ over $C$, denoted $\int_C f \, ds$, is given by
		$$ \int_a^b f(x(t),y(t),z(t)) \, \left| \bfr'(t) \right| \, dt $$
\item	The line integral of a vector function $\bfF(x,y,z)$ over $C$, denotes $\int_C \bfF \cdot \, dr $, is given by
		$$ \int_a^b \bfF(x(t),y(t),z(t)) \cdot \bfr'(t) \, dt $$
\item	We also have
		\begin{align*}
		\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) \, dz &=  \int_a^b f(x(t),y(t),z(t)) \, z'(t) \, dt\
		\end{align*}
\end{itemize}
\end{frame}


\begin{frame}
\frametitle{New Integrals  - Surface Integrals}
\footnotesize
If $S$ is a surface parameterized by the vector function
$$\bfr(u,v) = x(u,v) \ii + y(u,v) \jj + z(u,v) \kk $$
where $u,v$ run over a domain $D$ in the $uv$ plane:

\vfill

\begin{itemize}
\item	The surface integral of a scalar function $f(x,y,z)$, denoted $\iint_S f \, d S$, is given by
$$\iint_D f(x(u,v), y(u,v), z(u,v)) \left| \bfr_u \times \bfr_v \right| \, du \, dv $$
\item	The surface integral of a vector function $\bfF(x,y,z)$, denoted 
$\iint_S \bfF \cdot \, d \bfS$, is given by
$$\iint_D \bfF(x(u,v), y(u,v), z(u,v)) \cdot \left( \bfr_u \times \bfr_v \right) \, du \, dv $$
\end{itemize}

\vfill

You can remember both of these formulas with the shorthand
\begin{align*}
dS &=	 \left| \bfr_u \times \bfr_v \right| \, du \, dv \\
d\bfS	&=	(\bfr_u \times \bfr_v) \, du \, dv
\end{align*}

\end{frame}


\section{New Theorems}

\begin{frame}
\frametitle{Lots of ``Fundamental Theorems''}
\footnotesize
In Calculus I you learned two versions of the Fundamental Theorem:

\vfill
\textbf{Fundamental Theorem of Calculus, Part I} \emph{Suppose that $f(x)$ is continuous on $[a,b]$ and let $F(x) = \int_a^x f(t) \, dt$. Then $F$ is differentiable on $(a,b)$ and}
$$ F'(x) = f(x)$$

\vfill
\textbf{Fundamental Theorem of Calculus, Part II} \emph{Suppose that $F$ is any antiderivative of $f$. Then}
$$ \int_a^b f(x) \, dx = F(b) - F(a)$$

\begin{center}
\begin{tikzpicture}
\draw[cyan,thick] (0,0)		--	(2,0);
\draw[red,fill=red] (0,0) circle(0.05cm);
\draw[red,fill=red]  (2,0) circle(0.05cm);
\node[below] at (0,0)	{$a$};
\node[below] at (2,0)	{$b$};
\end{tikzpicture}
\end{center}


\vfill
In this course we've seen \emph{four} theorems which reduce integrals ``by one dimension'':  the Fundamental Theorem for Line Integrals, Green's Theorem, Stokes' Theorem, and the Divergence Theorem

\end{frame}

\begin{frame}
\frametitle{The Fundamental Theorem for Line Integrals}
\footnotesize

Recall that a vector field $\bfF$ is called \emph{conservative} if there is a scalar function $\varphi$ so that $\bfF = \nabla \varphi$. 

\vfill

\textbf{Theorem} \emph{If $\bfF$ is a conservative vector field, and $C$ is a curve parameterized by $\bfr(t), a \leq t \leq b$, then}
$$ \int_C \bfF \cdot \, d\bfr = \varphi(\bfr(b)) - \varphi(\bfr(a)) $$

\vfill

\begin{center}
\begin{tikzpicture}[scale=0.75]
\draw[cyan,thick] plot[domain=0:4,variable=\t,smooth]
	({\t},{0.5*\t-1.4*sin(90*\t)});
\draw[red,fill=red] (0,0) circle(0.05cm);
\draw[red,fill=red]  (4,2) circle(0.05cm);
\node[above] at (0,0)	{$a$};
\node[below] at (4,2) {$b$};
\node[anchor=south east] at (2,1)		{$C$};
\end{tikzpicture}
\end{center}


\end{frame}

\begin{frame}
\frametitle{Green's Theorem}
\footnotesize
Recall that a domain $D$ is simply connected if it is connected (any two points of $D$ can be joined by a curve in $D$) and every simple closed curve in $D$ surrounds only points of $D$.

\vfill
\textbf{Theorem} \emph{Suppose that  $D$ is a simply connected domain and its boundary $C$ is a simple closed curve. Suppose that $\bfF(x,y) = P(x,y) \ii + Q(x,y) \jj$ is a vector field and that $P$ and $Q$ have continuous partial derivatives in a neighborhood of $D$. Then}
$$ \iint_D \left( \frac{\dee Q}{\dee x} - \frac{\dee P}{\dee y} \right) \, dA = 
     \oint_C P \, dx + Q \, dy 
$$

\vfill

\begin{center}
\begin{tikzpicture}[scale=0.8]
\draw[gray,thick,->,>=stealth]	(0,0)	--	(3,0)	node[anchor=west] {$x$};
\draw[gray,thick,->,>=stealth]	(0,0)	--	(0,3)	node[anchor=south] {$y$};
\draw[white,fill=cyan!25]
	plot[domain=0:360,variable=\t,smooth]
	({1.5+cos(\t)},{1.5+0.6*sin(\t)});
\draw[red,thick,->,>=stealth]	plot[domain=0:360,variable=\t,smooth]
	({1.5+cos(\t)},{1.5+0.6*sin(\t)});
\node at (1.5,1.5) 	{$D$};
\node at (2.5,1)		{$C$};
\end{tikzpicture}
\end{center}

\end{frame}

\begin{frame}
\frametitle{Stokes' Theorem}
\footnotesize
Recall that a surface $S$ is \emph{oriented} if there is a continuous choice of unit normal $\bfn$ at every point of $S$. The bounding curve $C$ has positive orientation if its direction is consistent with the direction of $\bfn$ via the right-hand rule.

\vfill
\textbf{Theorem}
\emph{Let $S$ be an oriented, piecewise smooth surface that is bounded by a simple closed curve $C$ with positive orientation. Let $\bfF$ be a vector field whose components have continuous partial derivatives in an open region on $\R^3$ that contains $S$. Then}
$$  \iint_S \curl \bfF \cdot \, d\bfS = \int_C \bfF \cdot \, d\bfr  $$

\vfill
\begin{center}
\tdplotsetmaincoords{80}{90}
\begin{tikzpicture}[tdplot_main_coords,scale=0.8]
%% cheap trick to give illusion of perspective
\draw[gray,thick,->,>=stealth]		(0,0,0)	--	(6,-0.5,0)	node[anchor=north east] {$x$};
%%
\draw[gray,thick,->,>=stealth]		(0,0,0)	--	(0,3,0)	node[anchor=west] {$y$};
\draw[gray,thick,->,>=stealth]		(0,0,0)	--	(0,0,2.5)	node[anchor=south] {$z$};
\draw[blue,left color=blue!80,right color=white] 
	plot[domain=270:450,variable=\t] ({2+cos(\t)},{2+sin(\t)},{1}) --
	plot[domain=3:1,variable=\y] ({2},{\y},{2-(\y-2)*(\y-2)});
\draw[thick,red,->,>=stealth] plot[domain=270:360,variable=\t] ({2+cos(\t)},{2+sin(\t)},{1});
\draw[thick,red]					 plot[domain=360:450,variable=\t] ({2+cos(\t)},{2+sin(\t)},{1});
\draw[red,dashed]				 plot[domain=90:270,variable=\t] ({2+cos(\t)},{2+sin(\t)},{1});
\node[below] at (3,2,1)		{$C$};
\node at (2,2,1.5)				{$S$};
\draw[thick,black,->,>=stealth]	(2,2,2)	--	++(0,0,0.5);
\node[anchor=south] at (2,2,2.5)	{$\mathbf{n}$};
\end{tikzpicture}
\end{center}
\end{frame}

\begin{frame}
\frametitle{Divergence Theorem}
\footnotesize
Recall that $E$ is a \emph{simple volume} if its boundary separates $\R^3$ into an ``inside'' and an ``outside.''

\vfill
\textbf{Theorem}
\emph{Let $E$ be a simple solid region and let $S$ be the boundary surface of $E$, given with positive (outward) orientation. Let $\bfF$ be a vector field whose component functions have continuous partial derivatives on an open region that contains $E$. Then}
$$ \iiint_E \diver \, \bfF \, dV = \iint_S \bfF \cdot d \bfS $$

\vfill

\begin{center}
\tdplotsetmaincoords{70}{125}
\begin{tikzpicture}[tdplot_main_coords,scale=0.6]
\draw[gray,thick,->,>=stealth]		(0,0,0)	--	(3,0,0)		node[anchor=north east] 	{$x$};
\draw[gray,thick,->,>=stealth]		(0,0,0)	--	(0,3,0)		node[anchor=north west] 	{$y$};
\draw[gray,thick,->,>=stealth]		(0,0,0)	--	(0,0,3)		node[anchor=south] 			{$z$};
%% face parallel to yz plane
\draw[red,left color=red!90,right color=red!20,opacity=0.5]	(2,0,0)	--	++(0,0,2)	--	++(0,2,0)	--	++(0,0,-2)	--	++(0,-2,0);
%% face parallel to xz plane
\draw[red,fill=red!20,opacity=0.5]	(2,2,0)	--	++(0,0,2)	--	++(-2,0,0)	--	++(0,0,-2)	--	++(2,0,0);
%% face parallel to xy plane
\draw[red,fill=red!20,opacity=0.5]	(2,2,2)	--	++(-2,0,0)	--	++(0,-2,0)	--	++(2,0,0)	--  ++(0,2,0);
\end{tikzpicture}
\end{center}

\end{frame}

\begin{frame}
\frametitle{The Unity of (Almost) All Mathematics}

\footnotesize
\renewcommand{\arraystretch}{2.5}
\begin{tabular}{llll}
\textbf{Theorem} & \textbf{Statement} & \textbf{Region} & \textbf{Boundary} \\
\textbf{FTC}	
		&	$\displaystyle{\int}_{\hspace{-1mm}a}^b \, F'(x) \, dx = F(b) - F(a)$		
		&	$[a,b]$	&	$\{a,b\}$	\\
\textbf{Green}	
		&	${\displaystyle{\iint}}_{\hspace{-1mm}D} \,
					\left(\dfrac{\dee Q}{\dee x}- \dfrac{\dee P}{\dee y}\right) \, dA = 
			{\displaystyle{\oint}}_{\hspace{-1mm}C}\,  \bfF \cdot d\bfr$  
		&	Domain $D$  & Curve $C$ \\
\textbf{Stokes}	
		&	${\displaystyle{\iint}}_{\hspace{-1mm}S} \, \curl \bfF \cdot \, d\bfS = 
			{\displaystyle{\oint}}_{\hspace{-1mm}C} \, \bfF \cdot \, d\bfr$ 
		& Surface $S$ & Curve $C$ \\
\textbf{Gauss}	
		&	$\displaystyle{\iiint}_{\hspace{-1mm}E} \, \diver \bfF \, \, dV = 
			{\displaystyle{\iint}}_{\hspace{-1mm}S} \, \bfF \cdot \, d \bfS$ 
		& Volume $E$ & Surface $S$ \\
\\
\textbf{Pattern}	
		&	${\displaystyle{\int}}_{\hspace{-1mm} \mathrm{region}} DF = 
			{\displaystyle{\int}}_{\hspace{-1mm} \mathrm{boundary}} F $
		&	Region & Boundary
\end{tabular}
\end{frame}





\end{document}

