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\def\eq{\begin{equation}}
\def\endeq{\end{equation}}

\noindent
\large
\centerline{{\bf Overview of Multivariate Integration}}
\vspace{0.1in}

\noindent
\rule{6.5in}{0.01in}\\

\setcounter{equation}{0}
\large
\baselineskip=22pt

\noindent
{\bf (I) DOUBLE INTEGRALS - 
{\normalsize CARTESIAN, POLAR}}

\eq
\int\!\!\int_R f(x,y) dA = 
\int_a^b\!\!\int_{c(x)}^{d(x)} f(x,y) dydx= 
\int_{\theta_1}^{\theta_2}\!\!\int_{R_1(r)}^{R_2(r)}
f(r \cos\theta,r \sin\theta) r dr d\theta
\endeq

\begin{center}
{\bf coordinate conversion}
\end{center}

\begin{center}
\begin{tabular}{|l|}
\hline 
$dA = dx \ dy = r dr d\theta$ \\
\hline
$x=r \cos\theta \quad , \quad 
y=r \sin\theta \quad  , \quad 
r2 = x2+y2$ \\
\hline
\end{tabular}
\end{center}

\begin{center}
{\bf meaning}
\end{center}


\begin{center}
\begin{tabular}{|l|l|}
\hline 
$f$ & $\int\!\!\int_R f(x,y) dA$ \\
\hline 
1 &   area or region $R$ in $xy$-plane \\
mass density ($kg/m2$) &  mass of region $R$ \\
\hline
\end{tabular}
\end{center}

\vspace{0.2in}
\noindent
{\bf (II) VOLUME INTEGRALS - 
{\normalsize CARTESIAN, CYLINDRICAL, SPHERICAL}}

\eq
\int\!\!\int\!\!\int_V f(x,y,z) dV 
\endeq

{\bf Cartesian}
\eq
\int_a^b\!\!\int_{c(x)}^{d(x)}\!\!\int_{F_B(x,y)}^{F_T(x,y)} 
f(x,y,z)  dz dy dx
\endeq

{\bf Cylindrical}{$\quad \theta\in [0,2\pi),r>0$}
\eq
\int_{\theta_1}^{\theta_2}\!\!
\int_{R_1(r)}^{R_2(r)}\!\!
\int_{F_B(r,\theta)}^{F_T(r,\theta)} 
f(r\cos\theta,r\sin\theta,z) r dz dr d\theta
\endeq

{\bf Spherical}{$\quad \theta\in [0,2\pi),\phi\in [0,\pi],\rho>0$}
\eq
\int_{\theta_1}^{\theta_2}\!\!
\int_{\phi_1(\theta)}^{\phi_2(\theta)}\!\!
\int_{\rho_1(\phi,\theta)}^{\rho_2(\phi,\theta)} 
f(\rho\sin\phi\cos\theta,\rho\sin\phi\sin\theta,\rho\cos\phi)
\rho2 \sin\phi d\rho d\phi d\theta
\endeq

\vspace{0.1in}
\begin{center}
\begin{tabular}{|l|l|l|}
\hline 
Coordinate System & $dV$ & conversion formulae \\
\hline
Cartesian & $dz dy dx$ & * \\
\hline
Cylindrical & $r dz dr d\theta$ &
\begin{tabular}{lll}
$x$ & $=$ & $r\cos\theta$ \\
$y$ & $=$ & $r\sin\theta$ 
\end{tabular} \\
\hline
Spherical & $\rho2\sin\phi d\rho d\phi d\theta$ &
\begin{tabular}{lll}
$x$ & $=$ & $\rho\sin\phi\cos\theta$ \\
$y$ & $=$ & $\rho\sin\phi\sin\theta$ \\
$z$ & $=$ & $\rho\cos\phi$  
\end{tabular} \\
\hline
\end{tabular}
\end{center}

\vspace{0.1in}
\noindent
{\bf (III) LINE INTEGRALS}

\noindent
Let $\vec{r}(t) = x(t) \hat{\bf i} + y(t) \hat{\bf j}+ z(t) \hat{\bf k}
=(x(t),y(t),z(t))$ be a parametrization of the curve $C$.

\noindent
\underline{\bf scalars $f$}
\eq
\int_C f(x,y,z) ds = 
\int_{t_1}^{t_2} f(x(t),y(t),z(t)) 
\parallel \vec{r} \ '(t) \parallel dt
\endeq

\noindent
Noting the arclength element $ds$ is given by

\eq
ds = 
\parallel \vec{r} \ '(t) \parallel dt = \sqrt{
\left(\frac{dx}{dt}\right)2+
\left(\frac{dy}{dt}\right)2+
\left(\frac{dz}{dt}\right)2} dt
\endeq

\noindent
If $f=1$, the integral is the arclength of $C$. If $f$=mass density
per unit length, the line integral is mass of $C$.

\noindent
\underline{\bf vector fields $\vec{F}$} 

\noindent
Let $\vec{F} = F_1(x,y,z) \hat{\bf i} + 
F_2(x,y,z) \hat{\bf j}+ 
F_3(x,y,z) \hat{\bf k} = (F_1,F_2,F_3)$
be some vector field.

\eq
\int_C \vec{F}\cdot\vec{dr} = \int_C F_1 dx + F_2dy +F_3 dz=
\int_{t_1}^{t_2} \vec{F}(x(t),y(t),z(t)) \cdot
\frac{d\vec{r}}{dt} dt 
\endeq

\noindent
If $\vec{F}$ is force, the line integral is the work done
by the force along path $C$.

\vspace{0.1in}
\noindent
{\bf (IV) SURFACE INTEGRALS} 

\vspace{0.2in}
\noindent
\underline{\bf Surface elements and normal vectors}

\noindent
Let $S$ be a surface and $\vec{N}$ be normal to $S$.

\vspace{0.1in}
\begin{tabular}{|l|l|c|c|}
\hline & & & \\
\ & Description & $\vec{N}$ & $dS$ \\
\hline & & & \\
Graph & $z=f(x,y)$ &  $(-f_x,-f_y,1)$  &
$\parallel \vec{N} \parallel dA = \sqrt{1+f_x2+f_y2} dA$ \\
\hline & & & \\
Parametrized & $\vec{\phi}(u,v)$ &  
$\frac{\partial \vec{\phi}}{\partial u} \times
\frac{\partial \vec{\phi}}{\partial v}$ &
$\parallel \vec{N} \parallel du dv $ = 
$\parallel \frac{\partial \vec{\phi}}{\partial u} \times
\frac{\partial \vec{\phi}}{\partial v}
\parallel \  du dv$ \\
 \ & \ & \ &  \\ \hline 
\end{tabular}


\vspace{0.1in}
\noindent
\underline{\bf scalars $F$} 

\eq
\int\!\!\int_S F(x,y,z) dS = 
\int\!\!\int_R F(x,y,f(x,y)) \parallel \vec{N} \parallel dA =
\int\!\!\int_{R_{uv}} F(\vec{\phi}) \parallel \vec{N} \parallel dudv
\endeq

\noindent
If $F=1$, the surface integral is the surface area of $S$.
If $F=$mass density per unit area, the surface integral
is the mass of $S$.

\newpage
\noindent
\underline{\bf Flux $\Phi$ of vector fields $\vec{F}$} 

\noindent
Let $S$ be an oriented surface, i.e., where the unit
normal $\hat{N}$ has been uniquely specified.

\eq
\Phi = 
\int\!\!\int_S \vec{F}\cdot \vec{dS} = 
\int\!\!\int_S \vec{F}\cdot\hat{N} dS
\endeq

\noindent
For graphs and parametrized surfaces we have 

\eq
\Phi = 
\int\!\!\int_R \vec{F}(x,y,f(x,y))\cdot\vec{N} dA =
\int\!\!\int_{R_{uv}} \vec{F}(\vec{\phi})\cdot\vec{N} dudv
\endeq

\noindent
If $\vec{F}=\rho \vec{v}$ where $\rho$ is fluid density,
and $\vec{v}$ is the velocity field of the fluid, then
the flux is the net rate of mass flow through $S$ per unit
time.

\vspace{0.2in}
\noindent
{\bf (V) DIVERGENCE (GAUSS) THEOREM}

\noindent
Let $V$ be some solid region in space and $S=\partial V$ be its bounding
surface. If $\hat{N}$ is the \underline{outward} unit normal to
the surface then


\eq
\int\!\!\int_{\partial V} \vec{F}\cdot\hat{N} dS = 
\int\!\!\int\!\!\int_V \vec{\nabla}\cdot\vec{F} dV
\endeq

\noindent
says that the flux of $\vec{F}$
through a \underline{closed} surface $S$
can be converted into a volume integral over the solid $V$
which it bounds.

\eq
\vec{\nabla}\cdot\vec{F} =
\frac{\partial F_1}{\partial x} +
\frac{\partial F_2}{\partial y} +
\frac{\partial F_3}{\partial z}
\endeq

\vspace{0.2in}
\noindent
{\bf (VI) STOKES THEOREM}

Let $S$ be some oriented surface with 
unit normal $\hat{N}$ and boundary curve $C=\partial S$.
Then for any (smooth) vector field $\vec{F}$  we have

\eq
\int\!\!\int_{S} (\vec{\nabla}\times \vec{F})\cdot \hat{N} dS =
\int_C \vec{F}\cdot \vec{dr}
\endeq

\noindent
says that certain surface integrals can be converted to line
integrals and vice versa.
\eq
\vec{\nabla}\times \vec{F} = 
\left|
\begin{array}{lll}
\hat{i} & \hat{j} & \hat{k} \\
\partial_x & \partial_y & \partial_z \\
F_1 & F_2 & F_3
\end{array}
\right|
\endeq

\vspace{0.1in}
\noindent
{\bf Green's Theorem}
\noindent
Is the special case of Stokes Theorem where the surface $S$ is a region in the
$xy$-plane, i.e. $S=R$ and therefore $\hat{N}=\hat{k}$.
\end{document}