\input{preamblem273}

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\begin{center}
	\textsc{M273 Final Overview}
\end{center}
	
	\bigskip
	
	\begin{center}
		The formulas in the box will be provided on the exam.\\{\small
			\begin{tabular}{|c|c|c|}
				\hline
				& & \\
				$\ds dA= r\, dr\, d \theta$ & $\ds dV= r\, dz\,dr\,d\theta $ & $\ds dV = \rho^2 \, \sin \phi \,d \rho \, d\phi \, d \theta$ \\
				& & \\ \hline\hline
				& & \\
				$\ds \oint\limits_{\partial\mathcal{D}} P\,dx - Q\,dy = \iint\limits_{\mathcal{D}}\Sp{Q_x - P_y}\,dA$ & $\ds \oint\limits_{\partial\mathcal{S}}\vec{F}\cdot \,d\vec{r} = \oiint\limits_{\mathcal{S}}(\nabla \times \vec{F})\cdot \,d\vec{S}$ & $\ds \iint\limits_{\partial\mathcal{E}} \vec{F}\cdot\,d\vec{S} = \iiint\limits_{\mathcal{E}} \nabla\cdot\vec{F}\,dV$ \\
				& & \\
				
				\hline
		\end{tabular}}
	\end{center}

%\begin{center}
%	\begin{tabular}{|c|c|c|}
%		\hline
%		% after \\: \hline or \cline{col1-col2} \cline{col3-col4} ...
%		& & \\
%		$\ds \mathrm{d}A= r\, \mathrm{d}r\, \mathrm{d} \theta$ & $\ds \mathrm{d}V= r\, \mathrm{d}z\,\mathrm{d}r\,\mathrm{d}\theta $ & $\ds \mathrm{d}V = \rho^2  \sin (\phi) \,\mathrm{d} \rho \, \mathrm{d}\phi \, \mathrm{d} \theta$ \\
%		& & \\
%		\hline
%		& & \\
%		$\ds \oint_{\partial \mathcal{D}} P \D x + Q \D y = \iint\limits_\mathcal{D} (Q_x - P_y) \D A$ & $\ds \oint_{\partial \mathcal{S}} \vec{F} \cdot \D \vec{r} = \iint\limits_\mathcal{S} (\nabla \times \vec{F}) \cdot \D \vec{S}$ & $\ds \oiint\limits_{\partial \mathcal{E}} \vec{F} \cdot \D \vec{S} = \iiint\limits_{\mathcal{E}} (\nabla \cdot \vec{F}) \D V$\\
%		&&\\
%		\hline
%	\end{tabular}
%\end{center}

\bigskip

\textsc{\S12.1 Vectors in the Plane}

	\begin{itemize}
		\item[-] $\vec{v} = \V{ v_1, v_2 }$
		\item[-] Magnitude of a vector: $\norm{\vec{v}} = \sqrt{v_1^2 + v_2^2}$
		\item[-] Geometric vector addition, "tip-to-tail"
		\item[-] $\ds \hat{e}_\vec{v} = \frac{\vec{v}}{\norm{\vec{v}}}$ a unit vector which points in the direction of $\vec{v}$
	\end{itemize}

\medskip

\textsc{\S12.2 Vectors in Three Dimensions}

	\begin{itemize}
		\item[-] $\vec{v} = \V{ v_1, v_2, v_3 }$
		\item[-] Parameterization of a line through the point $(x_0, y_0, z_0)$ with direction vector $\vec{v}$
		
			$$ \vec{r}(t) = \V{x_0, y_0, z_0} + t\vec{v} $$
			
			\medskip
			
	\end{itemize}

\medskip

\textsc{\S12.3 Dot Product and the Angle Between Two Vectors}

	\begin{itemize}
		\item[-] Dot product between two vectors $\vec{v} = \V{ v_1, v_2, v_3}$ and $\vec{w} = \V{ w_1, w_2, w_3}$
		
			$$ \vec{v} \cdot \vec{w} = v_1w_1 + v_2w_2 + v_3w_3 $$
			
			\medskip
	
		\item[-] $\ds \vec{v}\cdot\vec{w} = \norm{\vec{v}}\norm{\vec{w}}\cos\theta$ where $0 \leq \theta \leq \pi$ is the angle between the two vectors.
		\item[-] If $\vec{v}\cdot\vec{w} = 0$ then $\vec{v}$ and $\vec{w}$ are orthogonal
		\item[-] $\ds \vec{v} \cdot \vec{v} = \norm{\vec{v}}^2$
		\item[-] Projection: given two vectors $\vec{u}$ and $\vec{v} \neq \vec{0}$, then projection of $\vec{u}$ onto $\vec{v}$ is given by
		
			$$ \vec{u}_{||\vec{v}} = \Sp{ \frac{\vec{u}\cdot\vec{v}}{\norm{\vec{v}}^2} } \vec{v} $$
			
			\medskip
			
		\item[-] Orthogonal decomposition: given two vectors $\vec{u}$ and $\vec{v} \neq \vec{0}$ then $\vec{u}$ can be written as the sum of two orthogonal vectors
		
			$$ \vec{u} = \vec{u}_{||\vec{v}} + \vec{u}_{\perp\vec{v}} $$
			
			\medskip
			
	\end{itemize}

	\medskip
	
\textsc{\S12.4 The Cross Product}

	\begin{itemize}
		\item[-] Given two vectors $\vec{v}$ and $\vec{w}$ the cross product is the vector
		
			$$ \vec{v} \times \vec{w} =  \begin{vmatrix} \ih & \jh & \kh \\ v_1 & v_2 & v_3 \\ w_1 & w_2 & w_3 \end{vmatrix} $$
			
			\medskip
			
		\item[-] $\ds \norm{\vec{v} \times \vec{w}} = \norm{\vec{v}}\norm{\vec{w}}\sin\theta$
		\item[-] Anticommutative: $\ds \vec{w} \times \vec{v} = -\vec{v} \times \vec{w}$
		\item[-] $\vec{v} \times \vec{w}$ is orthogonal to both $\vec{v}$ and $\vec{w}$, forming a right-handed system
	\end{itemize}

	\medskip
	
\textsc{\S12.5 Planes in 3-Space}

	\begin{itemize}
		\item[-] A plane through a point $(x_0, y_0, z_0)$ with normal vector $\vec{n} = \V{ a,b,c}$ has equation
		
			$$ a(x-x_0)+b(y-y_0)+c(z-z_0) = 0 $$
			
			\medskip
			
	\end{itemize}

	\medskip
	
\textsc{\S12.6 A Survey of Quadric Surfaces}

	\begin{itemize}
		\item[-] Ellipsoid
		
			$$ \Sp{\frac{x}{a}}^2 + \Sp{\frac{y}{b}}^2 + \Sp{\frac{z}{c}}^2 = 1 $$
			
			\begin{center}
				\begin{tikzpicture}
				\begin{axis}[
				axis lines = none,
				view={105}{40},
				]
				\addplot3[surf,fill=white,opacity=0.6,z buffer=sort,domain=-0.5*pi:0.5*pi,y domain=-pi:pi]
				( { cos(deg(x)) * cos(deg(y)) }, { cos(deg(x)) * sin(deg(y)) }, { 5*sin(deg(x)) });
				\end{axis}
				\end{tikzpicture}
			\end{center}
		
		\item[-] Elliptic Cone
		
			$$ \Sp{\frac{x}{a}}^2 + \Sp{\frac{y}{b}}^2 = \Sp{\frac{z}{c}}^2 $$
			
			\Null
			
			\begin{center}
				\begin{tikzpicture}
				\begin{axis}[
				axis lines = none,
				%view={105}{40},
				]
				\addplot3[surf,fill=white,opacity=0.6,z buffer=sort,domain=0:1,y domain=0:2*pi]
				( { (1-x)*cos(deg(y)) }, { (1-x)*sin(deg(y)) }, { x });
				\addplot3[surf,fill=white,opacity=0.6,z buffer=sort,domain=0:1,y domain=0:2*pi]
				( { (x)*cos(deg(y)) }, { (x)*sin(deg(y)) }, { 1+x });
				\end{axis}
				\end{tikzpicture}
			\end{center}
		
		\item[-] Hyperboloids
		
			$$ \text{One Sheet: } \Sp{\frac{x}{a}}^2 + \Sp{\frac{y}{b}}^2 = \Sp{\frac{z}{c}}^2 + 1 \hspace{3em} \text{and} \hspace{3em} \text{Two Sheet: } \Sp{\frac{x}{a}}^2 + \Sp{\frac{y}{b}}^2 = \Sp{\frac{z}{c}}^2 - 1 $$
			
			\begin{multicols}{2}
				\begin{center}
					\begin{tikzpicture}
					\begin{axis}[
					axis lines = none,
					%view={105}{40},
					]
					\addplot3[surf,fill=white,opacity=0.6,z buffer=sort,domain=-5:5,y domain=0:2*pi]
					( { 1*sqrt(1+x^2)*cos(deg(y)) }, { 1*sqrt(1+x^2)*sin(deg(y)) }, { 1*x });
					\end{axis}
					\end{tikzpicture}
				\end{center}
				
				\begin{center}
					\begin{tikzpicture}
					\begin{axis}[
					axis lines = none,
					%view={105}{40},
					]
					\addplot3[surf,fill=white,opacity=0.6,z buffer=sort,domain=0:0.05,y domain=0:2*pi]
					( { sinh(deg(x))*cos(deg(y)) }, { sinh(deg(x))*sin(deg(y)) }, { 2*cosh(deg(x)) });
					\addplot3[surf,fill=white,opacity=0.6,z buffer=sort,domain=0:0.05,y domain=0:2*pi]
					( { sinh(deg(x))*cos(deg(y)) }, { sinh(deg(x))*sin(deg(y)) }, { -2*cosh(deg(x)) });
					\end{axis}
					\end{tikzpicture}
				\end{center}
			\end{multicols}
		
		\item[-] Paraboliods
		
			$$ \text{Elliptic Paraboloid: } z = \Sp{\frac{x}{a}}^2 + \Sp{\frac{y}{b}}^2 \hspace{3em} \text{and} \hspace{3em} \text{Hyperbolic Paraboloid: } z = \Sp{\frac{x}{a}}^2 - \Sp{\frac{y}{b}}^2 $$
			
			\begin{multicols}{2}
				\begin{center}
					\begin{tikzpicture}
					\begin{axis}[
					axis lines = none,
					%view={105}{40},
					]
					\addplot3[surf,fill=white,opacity=0.6,z buffer=sort,domain=0:2*pi,y domain=0:2*pi]
					( { 7*sqrt(x)*cos(deg(y)) }, { 2*sqrt(x)*sin(deg(y)) }, { x });
					\end{axis}
					\end{tikzpicture}
				\end{center}
				
				\medskip
				
				\begin{center}
					\begin{tikzpicture}
					\begin{axis}[
					axis lines = none,
					%view={105}{40},
					]
					\addplot3[surf,fill=white,opacity=0.6,z buffer=sort,domain=-5:5,y domain=-5:5]
					( { x }, { y }, { (x/2)^2 - (y/3)^2 });
					\end{axis}
					\end{tikzpicture}
				\end{center}
			\end{multicols}

	\end{itemize}

	\medskip
	
\textsc{\S13.1 Vector-Valued Functions}

	\begin{itemize}
		\item[-] A vector-valued function is another name for a parameterization
		\item[-] $\vec{r}(t) = \V{ x(t), y(t), z(t)}$ is a vector-valued function/parameterization
	\end{itemize}

\medskip

\textsc{\S13.2 Calculus of Vector-Valued Functions}

	\begin{itemize}
		\item[-] You can perform limits, differentiation, and integration componentwise. e.g. the integral of a vector-valued function is the vector-valued function consisting of the integral of its components.
		\item[-] Tangent line to $\vec{r}(t)$ at $t = t_0$
		
			$$ \vec{\mathscr{L}}(t) = \vec{r}(t_0) + t \vec{r}'(t_0) $$
	\end{itemize}

\medskip

\textsc{\S13.3 Arc Length and Speed}

	\begin{itemize}
		\item[-] Speed: given a position/path parameterization $\vec{r}(t)$ then the speed is given by $\norm{\vec{r}'(t)}$
		\item[-] Arc Length $s$ is the actual distance traveled on a path. The arc length of $\vec{r}(t)$ for $t \in [a,b]$ is
		
			$$ s = \int_a^b \norm{\vec{r}'(t)}\,dt $$
			
			\medskip
			
		\item[-] An arc length parameterization of a given parameterization $\vec{r}(t)$ is a re-parameterization with the parameter $s$ via the substitution $t = g^{-1}(s)$ which is found by evaluating the arc length integral and finding the inverse (i.e. solving for $t$).
		
			$$ s = g(t) = \int_a^t \norm{\vec{r}'(u)}\,du $$
			
			\medskip
			
	\end{itemize}

	\medskip
	
\textsc{\S13.4 Curvature}

	\begin{itemize}
		\item[-] As stated at the beginning of this overview, if curvature is on the final, you'll be given all of the necessary formulas---you don't have to memorize them. So they are not listed here.
		\item[-] The unit tangent vector $\ds \vec{T} = \frac{\vec{r}'(t)}{\norm{\vec{r}'(t)}}$ points in the direction of motion (it's parallel to the velocity vector $\vec{r}'(t)$)
		\item[-] The unit normal vector $\ds \vec{N} = \frac{\vec{T}'(t)}{\norm{\vec{T}'(t)}}$ points in the direction of bending
	\end{itemize}

\medskip

\textsc{\S13.5 Motion in 3-Space}

	\begin{itemize}
		\item[-] No new content is presented in this section. It simply applies what has already been taught to more physically-relevant application problems
		\item[-] The major thing in this section is that the velocity $\vec{v}(t) = \vec{r}'(t)$ and the acceleration $\vec{a}(t) = \vec{r}''(t)$
	\end{itemize}

\medskip

\newpage

\textsc{\S14.1 Functions of Two or More Variables}

	\begin{itemize}
		\item[-] This is an introductory section on functions of two or more variables
		\item[-] A function $f(x,y)$ has a domain which is a subset of $\mathbb{R}^2$
		\item[-] A vertical trace is obtained by intersecting a surface $z = f(x,y)$ with a plane $x = a$ or $y = b$
		\item[-] A horizontal trace (or level curve) is obtained by intersecting a surface $z = f(x,y)$ with a plane $z = c$
		\item[-] A contour plot/map is obtained by considering many horizontal traces of a surface $z = f(x,y)$ and projecting all of them onto the $xy$-plane
	\end{itemize}

\medskip

\textsc{\S14.2 Limits and Continuity}

	\begin{itemize}
		\item[-] If $f(x,y)$ is continuous at $(a,b)$ then $\ds \lim_{(x,y) \to (a,b)} f(x,y) = f(a,b)$ i.e. "just plug it in"
		\item[-] Limits can be shown to not exist by considering different paths to $(a,b)$
		\item[-] Common paths to try: $x$-axis, $y$-axis, $y = mx$, and $y = x^2$
		\item[-] Unlike in Calc I, if any finite number of paths yields the same limiting value, this does not prove the limit exists and equals this value (\textit{all} paths most agree, nearly impossible to check explicitly)
	\end{itemize}

\textsc{\S14.3 Partial Derivatives}

	\begin{itemize}
		\item[-] $f_x(x,y)$ and $f_y(x,y)$ are the first order $x$- and $y$-partials of $f(x,y)$, respectively
		\item[-] $f_{xy}$ is obtained by first differentiating $f$ with respect to $x$ then with respect to $y$
		\item[-] $f_{xy} = f_{yx}$ (provided they exist and are continuous)---this generalizes to higher order derivatives
	\end{itemize}

\medskip

\textsc{\S14.4 Differentiability and Tangent Planes}

	\begin{itemize}
		\item[-] Tangent plane to $z = f(x,y)$ at the point $(a,b)$
		
			$$ z = f(a,b) + f_x(a,b)(x-a) +f_y(a,b)(y-b) $$
			
			\medskip
			
		\item[-] Locally to $z = f(a,b)$ the surface can be approximated by the tangent plane, that is
		
			$$ f(x,y) \approx f(a,b) + f_x(a,b)(x-a) +f_y(a,b)(y-b) $$
			
			\medskip
		
	\end{itemize}

\medskip

\textsc{\S14.5 The Gradient and Directional Derivatives}

	\begin{itemize}
		\item[-] $\ds \nabla = \V{ \frac{\del}{\del x}, \frac{\del}{\del y}, \frac{\del}{\del z} }$
		\item[-] The gradient of a function $f$
		
			$$ \nabla f = \V{ f_x, f_y, f_z } $$
			
			\medskip
			
		\item[-] Chain rule for paths: given a function $f(x,y,z)$ and a path parameterized by $\vec{r}(t)$ then the rate of change of $f$ along the parameterized path is
		
			$$ \frac{d}{dt} f(\vec{r}(t)) = \nabla f(\vec{r}(t))\cdot\vec{r}'(t) $$
			
			\medskip
			
		\item[-] The rate of change of $f$ in the direction of a unit vector $\vec{u}$ is given by
		
			$$ D_\vec{u} f = \nabla f \cdot \vec{u} $$
			
			\medskip
			
		\item[-] The gradient always points in the direction of the maximum rate of increase
		\item[-] The negative gradient always points in the direction of the maximum rate of decrease
		\item[-] The gradient is always orthogonal to the level curves
		\item[-] Given an equation of the form $F(x,y,z) = 0$, which is an implicit surface, then the gradient $\nabla F$ serves as a normal vector to the surface
		\item[-] Given the implicit surface $F(x,y,z) = 0$ then the equation of the tangent plane to this surface at the point $(a,b,c)$ is
		
			$$ F_x(a,b,c)(x-a) + F_y(a,b,c)(y-b) + F_z(a,b,c)(z-c) = 0 $$
			
			\medskip
	\end{itemize}

\medskip

\textsc{\S14.6 The Chain Rule}

	\begin{itemize}
		\item[-] Given a function $f(x,y,z)$ where $x = x(q,r,s)$, $y = y(q,r,s)$, and $z = z(q,r,s)$ then a dependency tree can be drawn
		
			\begin{center}
				\begin{tikzpicture}
%					[every node/.style={circle,fill=blue!20}]
					\node[circle,fill=blue!20] (n1) at (0,10) {$f$};
					\node[circle,fill=blue!20] (n2) at (-5,5) {$x$};
					\node[circle,fill=blue!20] (n3) at (0,5) {$y$};
					\node[circle,fill=blue!20] (n4) at (5,5) {$z$};
					\node[circle,fill=blue!20] (n5) at (-5,0) {$q$};
					\node[circle,fill=blue!20] (n6) at (0,0) {$r$};
					\node[circle,fill=blue!20] (n7) at (5,0) {$s$};
					
					\foreach \from/\to/\top/\bottom/\pos in {
						n1/n2/f/x/0.5, n1/n3/f/y/0.5, n1/n4/f/z/0.5,
						n2/n5/x/q/0.5, n2/n6/x/r/0.8, n2/n7/x/s/0.2,
						n3/n5/y/q/0.15, n3/n6/y/r/0.7, n3/n7/y/s/0.15,
						n4/n5/z/q/0.2, n4/n6/z/r/0.8, n4/n7/z/s/0.5
					}
						\draw (\from) -- (\to) node[pos=\pos,circle,fill=red!20] {$\ds \frac{\del \top}{\del \bottom}$};
				\end{tikzpicture}
			\end{center}
		
			\medskip
			
		Then to write down how $f$ changes with respect to $q$ look at each path from $f$ to $q$ and multiply the partials along each connection and add paths together, yielding
		
			$$ \frac{\del f}{\del q} = \frac{\del f}{\del x}\frac{\del x}{\del q} + \frac{\del f}{\del y}\frac{\del y}{\del q} + \frac{\del f}{\del z}\frac{\del z}{\del q}$$
			
			\medskip
			
		\item[-] Implicit Differentiation: given an equation of the form $\ds F(x,y,z) = 0$ then
		
			$$ \frac{\del z}{\del x} = -\frac{F_x}{F_z} \hspace{1in} \frac{\del z}{\del y} = -\frac{F_y}{F_z} $$
	\end{itemize}

	\medskip
	
\textsc{\S14.7 Optimization in Several Variables}

	\begin{itemize}
		\item[-] Critical Point: A point $(a,b)$ in the domain of $f(x,y)$ is a critical point if $f_x(a,b) = 0$ or DNE AND $f_y(a,b) = 0$ or DNE
		\item[-] Discriminant: $\ds D(a,b) = f_{xx}(a,b) f_{yy}(a,b) - [f_{xy} (a,b)]^2$
		\item[-] Second Derivative Test: if $(a,b)$ is a critical point of $f$ and all second order partials are continuous at $(a,b)$ then
			\begin{itemize}
				\item If $D(a,b) > 0$ and $f_{xx}(a,b) > 0$, then $f(a,b)$ is a local minimum
				\item If $D(a,b) > 0$ and $f_{xx}(a,b) < 0$, then $f(a,b)$ is a local maximum
				\item If $D(a,b) < 0$, then $f(a,b)$ is a saddle
				\item If $D(a,b) = 0$, then the test is inconclusive
			\end{itemize}
		\item[-] Finding absolute extrema of $f$ over a domain $\mathcal{D}$:
			\begin{itemize}
				\item Find local extrema of $f$ that live in $\mathcal{D}$
				\item Restrict $f$ to the boundary of $\mathcal{D}$ by performing some form of a substitution
				\item Find local extrema of this restricted $f$ using Calc I methods
				\item Compare all candidate values: the biggest is the absolute max and the smallest is the absolute minimum
			\end{itemize}
	\end{itemize}

	\medskip

\textsc{\S14.8 Lagrange Multipliers: Optimizing with a Constraint}

	\begin{itemize}
		\item[-] Let $f$ be the function to be optimized (minimized or maximized), this is called the \textit{objective function} subject to the constraint $g(x,y,z) = 0$
		\item[-] Lagrange Condition:
		
			$$ \nabla f = \lambda \nabla g $$
			
			\medskip
			
		\item[-] Find all solutions to the system of equations: $\nabla f = \lambda \nabla g$ and $g(x,y,z) = 0$
		\item[-] Evaluate $f$ at all solutions to the above system; the highest is the maximum and the lowest is the minimum
	\end{itemize}

	\medskip
	
\newpage

\textsc{15.1 Double Integration Over Rectangles}

	\begin{itemize}
		\item[-] Fubini's Theorem: integrating over a rectangle $\mathcal{R} = [a,b] \times [c,d]$ can be determined by evaluating an iterated integral in either order,
	
			$$ \iint\limits_\mathcal{R} f(x,y)\,dA = \int_a^b \int_c^d f(x,y) \,dy\,dx = \int_c^d \int_a^b f(x,y) \,dx\,dy $$
			
			\medskip
			
	\end{itemize}

	\medskip

\textsc{15.2 Double Integration Over More General Regions}
	
	\begin{itemize}
		\item[-] Vertically simple region $\mathcal{D}:  a \leq x \leq b$ and $g_1(x) \leq y \leq g_2(x)$
	
			$$ \iint\limits_\mathcal{D} f(x,y)\,dA = \int_a^b \int_{g_1(x)}^{g_2(x)} f(x,y) \,dy\,dx $$
			
		\medskip
		
		\item[-] Horizontally simple region $\mathcal{D}:  c \leq y \leq d$ and $g_1(y) \leq x \leq g_2(y)$
	
			$$ \iint\limits_\mathcal{D} f(x,y)\,dA = \int_c^d \int_{g_1(y)}^{g_2(y)} f(x,y)\,dx\,dy $$
		
		\medskip
			
		\item[-] Volume between two surfaces. Let $z = z_1(x,y)$ and $z = z_2(x,y)$ be two surfaces such that $ z_1(x,y) \leq z \leq z_2(x,y)$ for all $z \in \mathcal{D}$ where $\mathcal{D}$ is the projection of the bounded region onto the $xy$-plane, then the volume bounded between these two surfaces is given by
		
			$$ \iint\limits_\mathcal{D} (z_2(x,y) - z_1(x,y))\,dA $$
		
		\medskip
			
	\end{itemize}

	\medskip
	
\textsc{15.4 Part I Double Integration in Polar Coordinates}

	\begin{itemize}
		\item[-] Polar conversions: $x = r\cos\theta$ and $y = r\sin\theta$
		\item[-] Polar differential area element: $dA = r\,dr\,d\theta$
		\item[-] Radially simple region $\mathcal{D}: \theta_1 \leq \theta \leq \theta_2$ and $r_1(\theta) \leq r \leq r_2(\theta)$
			
			$$ \iint\limits_\mathcal{D} f(x,y)\,dA = \int_{\theta_1}^{\theta_2} \int_{r_1(\theta)}^{r_2(\theta)} f(r\cos\theta, r\sin\theta) r\,dr\,d\theta $$
			
		\medskip
			
	\end{itemize}
	
\textsc{15.3 Triple Integrals in Cartesian Coordinates}

	\begin{itemize}
		\item[-] Fubini's theorem: integrating over a box $\mathcal{B} = [a,b] \times [c,d] \times [p,q]$ can be determined by evaluating an iterated integral in any order (there are $3! = 6$ possible orders)
		\item[-] Integrating over a $z$-simple region $\mathcal{W}: \, (x,y) \in \mathcal{D}$ where $z_1(x,y) \leq z \leq z_2(x,y)$ where $\mathcal{D}$ is the projection of $\mathcal{W}$ onto the $xy$-plane:
		
			$$ \iiint\limits_\mathcal{W} f(x,y,z)\,dV = \iint\limits_\mathcal{D} \Sp{ \int_{z_1(x,y)}^{z_2(x,y)} f(x,y,z)\,dz }\,dA $$
			
		\medskip
			
		\item[-] Integrating over a $y$-simple region $\mathcal{W}: \, (x,z) \in \mathcal{D}$ where $y_1(x,y) \leq y \leq y_2(x,z)$ where $\mathcal{D}$ is the projection of $\mathcal{W}$ onto the $xz$-plane:
		
			$$ \iiint\limits_\mathcal{W} f(x,y,z)\,dV = \iint\limits_\mathcal{D} \Sp{ \int_{y_1(x,z)}^{y_2(x,z)} f(x,y,z)\,dy }\,dA $$
		
		\medskip
		
		\item[-] Integrating over a $x$-simple region $\mathcal{W}: \, (y,z) \in \mathcal{D}$ where $x_1(y,z) \leq x \leq x_2(y,z)$ where $\mathcal{D}$ is the projection of $\mathcal{W}$ onto the $yz$-plane:
		
		$$ \iiint\limits_\mathcal{W} f(x,y,z)\,dV = \iint\limits_\mathcal{D} \Sp{ \int_{x_1(y,z)}^{x_2(y,z)} f(x,y,z)\,dx }\,dA $$
		
		\medskip
		
	\end{itemize}

	\medskip
	
\textsc{15.4 Part II Triple Integration in Cylindrical and Spherical Coordinates}

	\begin{itemize}
		\item[-] Cylindrical conversions: $x = r\cos\theta$, $y = r\sin\theta$, and $z$ remains the same
		\item[-] Cylindrical differential volume element: $dV = r\,dz\,dr\,d\theta$
		\item[-] Cylindrically simple region $\mathcal{W}: \theta_1 \leq \theta \leq \theta_2$, $r_1(\theta) \leq r \leq r_2(\theta)$, and $z_1(r,\theta) \leq z \leq z_2(r,\theta)$
		
			$$ \iiint\limits_\mathcal{W} f(x,y,z)\,dV = \int_{\theta_1}^{\theta_2}\int_{r_1(\theta)}^{r_2(\theta)}\int_{z_1(r,\theta)}^{z_2(r,\theta)} f(r\cos\theta,r\sin\theta,z)r\,dz\,dr\,d\theta $$
		
		\medskip
		
		\item[-] Spherical conversions: $x = \rho\cos\theta\sin\phi$, $y = \rho\sin\theta\sin\phi$, and $z = \rho\cos\phi$
		\item[-] Spherical differential volume element: $dV = \rho^2\sin\phi\,d\rho\,d\phi\,d\theta$
		\item[-] Spherically simple region $\mathcal{W}: \theta_1 \leq \theta \leq \theta_2$, $\phi_1 \leq \phi \leq \phi_2$, and $\rho_1(\theta,\phi) \leq \rho \leq \rho_2(\theta,\phi)$
		
			$$ \iiint\limits_\mathcal{W} f(x,y,z)\,dV = \int_{\theta_1}^{\theta_2} \int_{\phi_1}^{\phi_2} \int_{\rho_1(\theta,\phi)}^{\rho_2(\theta,\phi)} f(\rho\cos\theta\sin\phi,\rho\sin\theta\sin\phi,\rho\cos\phi)\rho^2 \sin\phi \, d\rho\,d\phi\,d\theta $$
		
		\medskip
		
	\end{itemize}

\textsc{Useful Trigonometric Identities}

	\begin{itemize}
		\item[-] $\ds \sin^2\theta + \cos^2\theta = 1$
		\item[-] $\ds \sin^2\theta = \frac{1-\cos2\theta}{2}$
		\item[-] $\ds \cos^2\theta = \frac{1+\cos2\theta}{2}$
	\end{itemize}

\medskip

\textsc{16.1 Vector Fields}

	\begin{itemize}
		\item[-] "Nabla" $\ds \nabla = \V{ \frac{\del}{\del x}, \frac{\del}{\del y}, \frac{\del}{\del z} } $
		\item[-] Divergence operator: $\ds \text{div}(\vec{F}) = \nabla\cdot\vec{F} = \V{ \frac{\del}{\del x}, \frac{\del}{\del y}, \frac{\del}{\del z} } \cdot \V{ F_1, F_2, F_3 } = \frac{\del F_1}{\del x} + \frac{\del F_2}{\del y} + \frac{\del F_3}{\del z} $
		\item[-] Curl operator: $\ds \text{curl}(\vec{F}) = \nabla \times \vec{F} =  \begin{vmatrix} \ih & \jh & \kh \\ \frac{\del}{\del x} & \frac{\del}{\del y} & \frac{\del}{\del z} \\ F_1 & F_2 & F_3 \end{vmatrix} = \V{ \frac{\del F_3}{\del y} - \frac{\del F_2}{\del z}, \frac{\del F_1}{\del z} - \frac{\del F_3}{\del x}, \frac{\del F_2}{\del x} - \frac{\del F_1}{\del y} } $
		\item[-] A vector field $\vec{F}$ is conservative if there exists a scalar potential function $f$ such that $\nabla f = \vec{F}$
	\end{itemize}

\medskip

\textsc{16.2 Line Integrals}

	\begin{itemize}
		\item[-] A path $\mathcal{C}$ in $\mathbb{R}^3$ can be parameterized by $\vec{r}(t)$ for $t\in[a,b]$
		\item[-] Scalar line integral of a scalar function $f$ over a path $\mathcal{C}$
		
			$$ \int\limits_\mathcal{C} f(x,y,z)\, ds = \int_a^b f(\vec{r}(t))\norm{\vec{r}'(t)}\,dt $$
			
		\medskip
		
		\item[-] If $f(x,y,z) = 1$ then the scalar line integral of $f$ over a path $\mathcal{C}$ is the length of $\mathcal{C}$, i.e. the arc length,
		
			$$ \int\limits_\mathcal{C} ds = \int_a^b \norm{\vec{r}'(t)}\,dt = \text{length}(\mathcal{C}) $$
			
		\medskip
		
		\item[-] Vector line integral of a vector field $\vec{F}$ over a path $\mathcal{C}$
		
			$$ \int\limits_\mathcal{C} \vec{F}\cdot\,d\vec{r} = \int\limits_\mathcal{C} (\vec{F}\cdot \vec{T})\,ds = \int_a^b \vec{F}(\vec{r}(t))\cdot \vec{r}'(t)\,dt $$
			
		\medskip
		
		\item[-] Alternate notation for a vector line integral of $\vec{F} = \V{ F_1, F_2, F_3 }$ over a path $\mathcal{C}$
		
			$$ \int\limits_\mathcal{C} \vec{F} \cdot d\vec{r} = \int\limits_\mathcal{C} F_1\,dx + F_2\,dy + F_3\,dz $$
			
	\end{itemize}

\medskip

\textsc{16.3 Conservative Vector Fields}

	\begin{itemize}
		\item[-] The vector line integral over a closed path (endpoints are equal) $\mathcal{C}$ is called the circulation and is denoted
		
			$$ \oint\limits_\mathcal{C} \vec{F}\cdot\,d\vec{r} $$
			
		\medskip
		
		\item[-] If $\vec{F}$ is conservative, that is $\vec{F} = \nabla f$ for some scalar function $f$, and $\mathcal{C}$ is a path with endpoints $P$ and $Q$ then
		
			$$ \int\limits_\mathcal{C} \vec{F}\cdot\,d\vec{r} = \int\limits_\mathcal{C} \nabla f\cdot\,d\vec{r} = f(Q) - f(P) $$
			
		\newpage
		
		\item[-] If $\vec{F}$ is conservative, that is $\vec{F} = \nabla f$ for some scalar function $f$, and $\mathcal{C}$ is a closed path, then
		
			$$ \oint\limits_\mathcal{C} \vec{F}\cdot\,d\vec{r} = 0 $$
		
		\medskip
		
		\item[-] If $\curl{F} = \vec{0}$ (or if $\ds \frac{\del F_2}{\del x} - \frac{\del F_1}{\del y} = 0$ for 2D fields) and $\vec{F}$ is defined on a simply connected domain, then $\vec{F}$ is conservative and therefore, there exists a scalar potential function $f$ such that $\nabla f = \vec{F}$
		\item[-] If $\vec{F} = \V{ F_1, F_2, F_3 }$ is known to be conservative then $f$ can be found by evaluating the following antiderivatives:
		
		\medskip
		
			\hfill $\ds f = \int F_1\,dx $ \hfill $ \ds f = \int F_2\,dy $ \hfill $\ds f = \int F_3\,dz$ \hfill\Null
			
	\end{itemize}

\medskip

\textsc{17.1 Green's Theorem}

	\begin{itemize}
		\item[-] Let $\mathcal{D}$ be domain whose boundary $\del\mathcal{D}$ is a simple closed curve oriented counterclockwise, then
		
			$$ \oint\limits_{\del\mathcal{D}} \vec{F}\cdot\,d\vec{r} = \iint\limits_\mathcal{D} \Sp{ \frac{\del F_2}{\del x} - \frac{\del F_1}{\del y} }\,dA $$
			
		\medskip
		
		\item[-] When orienting a boundary, if $\mathcal{D}$ lies to the left as the boundary is traversed then this is considered to be oriented positively
		\item[-] The area of $\mathcal{D}$ can be determined using a line integral around the boundary $\del\mathcal{D}$ provided
		
			 $$\ds \frac{\del F_2}{\del x} - \frac{\del F_1}{\del y} = 1$$
		
		\medskip
			 
		A few common fields that have this property are
		
		\medskip
		
		\hfill $\ds \vec{F} = \V{0,x}$ \hfill $\ds \vec{F} = \V{-y,0}$ \hfill $\ds \vec{F} = \V{-\frac{y}{2},\frac{x}{2}}$ \hfill \Null
	
		\medskip
		
		\item[-] If the boundary of $\mathcal{D}$ is composed of multiple curves, then the total boundary $\del \mathcal{D}$ can be written as a sum or difference of the constituent curves. For example, if $\del \mathcal{D}$ is composed of two boundaries $\mathcal{C}_1$ and $\mathcal{C}_2$ then $\del\mathcal{D} = \pm \mathcal{C}_1 \pm \mathcal{C}_2$ where the choice of plus or minus depends on whether that curve has been oriented positively or negatively. Positively oriented curves get a plus sign and negatively oriented curves receive the minus sign.
	\end{itemize}

\medskip

\textsc{\S16.4 Parameterized Surfaces and Surface Integrals}

	\begin{itemize}
		\item[-] Sphere: $\ds \vec{r}(u,v) = \V{ \rho \cos u\sin v, \rho \sin u\sin v, \rho \cos v }$ for $u \in [0,2\pi]$ and $v \in [0,\pi]$ where $\rho$ is the radius of the sphere
		\item[-] Hemispheres: use the same parameterization as a sphere but limit the range of $u$ and/or $v$, as appropriate
		\item[-] Cylinder: $\ds \vec{r}(u,v) = \V{ r\cos u, r\sin u, v}$ for $u \in [0,2\pi]$ and $v$ will depend on the cylinder's height and $r$ is the radius
		\item[-] Cone: assuming it's the cone $z = \sqrt{x^2 + y^2}$ then $\ds \vec{r}(u,v) = \V{ v\cos u, v\sin u, v}$ for $u \in [0,2\pi]$ and $v$ will depend on the cone's height
		\item[-] Function: $z = f(x,y)$ then $\ds \vec{r}(u,v) = \V{ u, v, f(u,v) }$ where the range of $u$ and $v$ is the same as the range for $x$ and $y$, respectively
		\item[-] Scalar surface integral of a function $f(x,y,z)$ over a surface $\mathcal{S}$ parameterized by $\vec{r}(u,v)$ for $(u,v) \in \mathcal{D}$ then
		
			$$ \iint\limits_\mathcal{S} f(x,y,z)\,d\mathcal{S} = \iint\limits_\mathcal{D} f(\vec{r}(u,v))\norm{\vec{r}_u \times \vec{r}_v}\,du\,dv $$
			
			\medskip
			
		\item[-] Surface area of a surface $\mathcal{S}$ parameterized by $\vec{r}(u,v)$ for $(u,v) \in \mathcal{D}$
		
			$$ \text{Area}(\mathcal{S}) = \iint\limits_\mathcal{S} \,dS = \iint\limits_\mathcal{D} \norm{\vec{r}_u \times \vec{r}_v}\,du\,dv $$
			
			\medskip
	\end{itemize}

\medskip

\textsc{\S16.5 Surface Integrals of Vector Fields}

	\begin{itemize}
		\item[-] Vector surface integral of a vector field $\vec{F}$ over a surface $\mathcal{S}$ parameterized by $\vec{r}(u,v)$ for $(u,v) \in \mathcal{D}$
		
			$$ \iint\limits_\mathcal{S} \vec{F} \cdot d\vec{S} = \iint\limits_\mathcal{D} \vec{F}(\vec{r}(u,v))\cdot \vec{n}\,du\,dv $$
			
			\medskip
			
			where $\vec{n}$ is either $\vec{r}_u \times \vec{r}_v$ or $\vec{r}_v \times \vec{r}_u$ and will depend on how $\mathcal{S}$ is chosen to be oriented (the problem will say how, somehow)
	\end{itemize}

\medskip

\textsc{\S17.2 Stokes' Theorem}

	\begin{itemize}
		\item[-] Let $\mathcal{S}$ be a piecewise smooth surface consisting of finitely many positively oriented boundaries, then
		
			$$ \iint\limits_\mathcal{S} \Sp{\nabla \times \vec{F}} \cdot d\vec{S} = \oint\limits_{\del\mathcal{S}} \vec{F} \cdot d\vec{r} $$
			
			\medskip
			
		\item[-] If $\vec{F} = \nabla \times \vec{A}$ then the flux of $\vec{F}$ through a surface $\mathcal{S}$ depends only on the oriented boundary $\del \mathcal{S}$ and not on the surface itself
		
			$$ \iint\limits_\mathcal{S} \vec{F} \cdot d\vec{S} = \iint\limits_\mathcal{S} \Sp{\nabla \times \vec{A} }\cdot d\vec{S} = \oint\limits_{\del \mathcal{S}} \vec{A} \cdot d\vec{r} $$
			
			\medskip
			
			If $\mathcal{S}$ is a closed surface then the boundary of $\mathcal{S}$ is empty, that is $\del\mathcal{S} = \emptyset$ and so the flux is 0.
	\end{itemize}

\medskip

\textsc{\S17.3 Divergence Theorem}

	\begin{itemize}
		\item[-] Let $\del \mathcal{W}$ be a closed surface that encloses a region $\mathcal{W}$ in $\mathbb{R}^3$. Further assume that $\del \mathcal{W}$ is piecewise smooth and oriented with outward pointing normals (i.e. to the outside of $\mathcal{W}$) then,
		
			$$ \iint\limits_{\del\mathcal{W}} \vec{F} \cdot d\vec{S} = \iiint\limits_\mathcal{W} = \nabla \cdot \vec{F} \,dV $$
	\end{itemize}

\end{document}