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

\noindent
\large
\centerline{{\bf Overview of Multivariate Integration}}


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

Final Sections: 15.2-15.4, 15.6-15.8, 16.1-16.3, 16.6-16.7 (graphs only), 16.9

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

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\normalsize
\noindent
\underline{\bf DOUBLE INTEGRALS : Cartesian, Polar}

\vspace{0.1in}
\noindent
Iterated integrals in cartesian coordinates using via 
\underline{x-slice and y-slice} are, respectively:
\eq
\int\!\!\int_R f(x,y) \ dA = 
\int_a^b\!\!\int_{c(x)}^{d(x)} f(x,y) \ dydx= \int_c^d\!\!\int_{a(y)}^{b(y)} f(x,y) \ dxdy
\endeq

\vspace{0.1in}
\noindent
Iterated integrals in cartesian coordinates using via \underline{polar coordinates}
\eq
\int\!\!\int_R f(x,y) dA = 
\int_{\theta_1}^{\theta_2}\!\!\int_{R_1(r)}^{R_2(r)}
f(r \cos\theta,r \sin\theta) \ r dr d\theta
\endeq

\vspace{0.1in}
\noindent
Area elements
\[ dA = dy dx = dx dy = r dr d\theta\]

\vspace{0.1in}
\noindent
Coordinate conversions
\begin{eqnarray*}
x=r \cos\theta  \quad & \ & y=r \sin\theta \quad  , \quad \\
r = \sqrt{x^2+y^2} & \ & \theta = arctan(y/x)
\end{eqnarray*}

\vspace{0.2in}
\noindent
\underline{\bf VOLUME INTEGRALS:
{\normalsize Cartesian, Cylindrical, Spherical over solid $E$}}

\vspace{0.1in}
\noindent
{\bf Cartesian:} One ordering is $dV = dzdA=dzdydx$
\eq
\int\!\!\int\!\!\int_E f(x,y,z) \ dV  = 
\int\!\!\int\!\!\int_E f(x,y,z)\  dz dA  =
\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

\noindent
{\bf Cylindrical}{$\quad \theta\in [0,2\pi),r>0$}
\eq
\int\!\!\int\!\!\int_E f(x,y,z)\  dV  =
\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

\noindent
{\bf Spherical}{$\quad \theta\in [0,2\pi),\phi\in [0,\pi],\rho>0$}
\eq
\int\!\!\int\!\!\int_E f(x,y,z)\  dV  =
\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) \
\rho^2 \sin\phi \ d\rho d\phi d\theta
\endeq

\vspace{0.1in}
\begin{center}
\begin{tabular}{|l|l|l|l|}
\hline 
Coordinate System & $dV$ & conversion formulae & notes \\
\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} &  $r^2=x^2+y^2$ \\
\hline
Spherical & $\rho^2 \sin\phi \ d\rho d\phi d\theta$ &
\begin{tabular}{lll}
$x$ & $=$ & $\rho\sin\phi\cos\theta$ \\
$y$ & $=$ & $\rho\sin\phi\sin\theta$ \\
$z$ & $=$ & $\rho\cos\phi$  
\end{tabular} 
& $\rho^2 \sin^2 \phi = x^2+y^2$ \\
\hline
\end{tabular}
\end{center}

\newpage
\noindent
\underline{\bf LINE INTEGRALS}
\vspace{0.1in}

\noindent
Let $\vec{\bf r}(t) = x(t) \hat{\bf i} + y(t) \hat{\bf j}+ z(t) \hat{\bf k}
=<x(t),y(t),z(t)>$  with $t \in(t_1,t_2)$ be a parametrization of the curve $C$.

\vspace{0.1in}
\noindent
{\bf Line integrals of scalars $f(x,y,z)$}
\eq
\int_C f(x,y,z) \ ds = 
\int_{t_1}^{t_2} f(x(t),y(t),z(t))  \ 
| \vec{\bf r} \ '(t) | \ dt
\endeq

\noindent
Noting the arclength element $ds$ is given by
\eq
ds = 
| \vec{\bf r} \ '(t) | \ 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 $C$.

\vspace{0.1in}
\noindent
{\bf Line integrals of vector fields $\vec{\bf F}$} 

\vspace{0.1in}
\noindent
Let $\vec{\bf F} = F_1(x,y,z) \hat{\bf i} + 
F_2(x,y,z) \hat{\bf j}+ 
F_3(x,y,z) \hat{\bf k} = <F_1,F_2,F_3>$
be some vector field.

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

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

\vspace{0.1in}
\noindent
\underline{\bf SURFACE INTEGRALS: Scalars and Vector Fields} 

\vspace{0.1in}
\noindent
Let $S$ be a surface and $\vec{\bf N}$ be normal to $S$. The normal vectors and surface elements $dS$ are:
\vspace{0.1in}

\vspace{0.1in}
\begin{tabular}{|l|l|c|c|l|}
\hline & & & \\
\ & Description & $\vec{\bf N}$ & $dS$ & Note: \\
\hline & & & & \\
Graph & $z=f(x,y)$ &  $<-f_x,-f_y,1>$  &
$| \vec{\bf N} | \ dA = \sqrt{1+f_x^2+f_y^2} \ dA$  & On final\\
\hline & & & & \\
Parametrized surface & $\vec{\bf r}(u,v)$ &  
$\frac{\partial \vec{\bf r}}{\partial u} \times
\frac{\partial \vec{\bf r}}{\partial v}$ &
$| \vec{\bf N} | \ du dv $ = 
$| \frac{\partial \vec{\bf r}}{\partial u} \times
\frac{\partial \vec{\bf r}}{\partial v}
| \  du dv$ & Not on final\\
 \ & \ & \ &  &\\ \hline 
\end{tabular}


\vspace{0.2in}
\noindent
{\bf Scalar surface integrals of $G(x,y,z)$ on a graph $z=f(x,y)$} 

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

\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 $S$.

\vspace{0.1in}
\noindent
{\bf Vector surface integrals of $\vec{\bf F}$ thru the graph $z=f(x,y)$} 
\vspace{0.1in}

\noindent
Let $S$ be an oriented surface, i.e., where the unit
normal $\hat{\bf N}$ has been uniquely specified.
\eq
\Phi = 
\int\!\!\int_S \vec{\bf F}\cdot \vec{\bf dS} = 
\int\!\!\int_S \vec{\bf F}\cdot\hat{\bf N} dS  = 
\int\!\!\int_R \vec{\bf F}(x,y,f(x,y))\cdot\vec{\bf N} \ dA
\endeq
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{0.1in}
\noindent
\underline{\bf DIVERGENCE (GAUSS) THEOREM}

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


\eq
\int\!\!\int_{\partial V} \vec{\bf F}\cdot\hat{\bf N} \ dS = 
\int\!\!\int\!\!\int_V \vec{\nabla}\cdot\vec{\bf F} dV
\quad , \quad
\vec{\nabla}\cdot\vec{\bf F} =
\frac{\partial F_1}{\partial x} +
\frac{\partial F_2}{\partial y} +
\frac{\partial F_3}{\partial z}
\endeq

\newpage
\noindent
\underline{\bf STOKES THEOREM}

\vspace{0.1in}
\noindent
Let $S$ be some oriented surface with 
unit normal $\hat{\bf  N}$ and boundary curve $C=\partial S$.
Then for any (smooth) vector field $\vec{\bf F}$  we have

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

\noindent
says that certain surface integrals can be converted to line
integrals and vice versa. Here, the curl of $\vec{\bf F}$ is:
\eq
\vec{\nabla}\times \vec{\bf 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 $\hat{\bf N}=\hat{\bf k}$.
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