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\Lab{Change of Variables}
\SecMark
\label{cov}
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%*==== CHANGE OF VARIABLES ====
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\subsection{Essentials}
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%*=== Essentials ===
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\subsubsection{Main ideas}
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%*== Main ideas ==
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\Goal{
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\begin{itemize}
\item
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There are many ways to solve this problem!
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\item
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Using Jacobians (and inverse Jacobians)
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\end{itemize}
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}
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\subsubsection{Prerequisites}
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%*== Prerequisites ==
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\Req{
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\begin{itemize}
\item
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Surface integrals
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\item
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Jacobians
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\item
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Green's/Stokes' Theorem
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\end{itemize}
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}
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\subsubsection{Warmup}
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%*== Warmup ==
Perhaps a discussion of single and double integral techniques for solving this
problem.
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\subsubsection{Props}
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%*== Props ==
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\begin{itemize}
\item
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\end{itemize}
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\subsubsection{Wrapup}
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%*== Wrapup ==
This is a good conclusion to the course, as it reviews many integration
techniques. We emphasize that (2-dimensional) change-of-variable problems are
a special case of surface integrals.
Here are some of the methods one could use to do these integrals:
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\begin{itemize}
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change of variables (at least 2 ways)
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\item
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Area Corollary to Green's Theorem (at least 2 ways)
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\item
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ordinary single integral (at least 2 ways)
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\item
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ordinary double integral (at least 2 ways)
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\item
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surface integral
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\end{itemize}
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\newpage
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\subsection{Details}
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%*=== Details ===
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\subsubsection{In the Classroom}
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%*== In the Classroom ==
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\item
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Some students will want to simply use Jacobian formulas; encourage such
students to try to solve this problem both by computing
$\Partial{(x,y)}{(u,v)}$ and by computing $\Partial{(u,v)}{(x,y)}$.
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\item
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Other students will want to work directly with $d\rr_1$ and $d\rr_2$. This
works fine if one first solves for $x$ and $y$ in terms of $u$ and $v$.
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\item
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Students who compute $d\rr_1$ and $d\rr_2$ directly can easily get confused,
since they may try to eliminate $x$ or $y$, rather than $u$ or $v$.
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\footnote{
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%*((
Along the curve $v=\hbox{constant}$, one has $dy=v\,dx$, so that
$d\rr_1 = dx\,\ii + dy\,\jj = (\ii + v\,\jj)\,dx$,
which some students will want to write in terms of $x$ alone. But one needs
to express this in terms of $du$! This can be done using
$du = x\,dy + y\,dx = x (v\,dx) + y\,dx = 2y\,dx$,
so that
$d\rr_1 = (\ii + v\,\jj) \,\frac{du}{2y}$.
A similar argument leads to
$d\rr_2 = (-\frac{1}{v}\,\ii+\jj)\,\frac{x\,dv}{2}$ for $u=\hbox{constant}$,
so that
$d\SS
= d\rr_1\times d\rr_2
= \kk \,\frac{x}{2y}\,du\,dv
= \kk \,{du\,dv\over2v}$.
This calculation can be done without solving for $x$ and $y$, provided one
recognizes $v$ in the penultimate expression.
%*))
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}
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Emphasize that one must choose parameters, both on the region, and on each
curve, and that $u$ and $v$ are chosen to make the limits easy.
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\end{itemize}
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\subsubsection{Subsidiary ideas}
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%*== Subsidiary ideas ==
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\Sub{
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Review of Green's Theorem
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Review of single integral techniques
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Review of double integral techniques
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\end{itemize}
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}
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\subsubsection{Homework \None}
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%*== Homework ==
%* (none yet)
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\HW{
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}
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\subsubsection{Essay questions \None}
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%*== Essay questions ==
%* (none yet)
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\Essay{
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}
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\subsubsection{Enrichment}
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%*== Enrichment ==
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\Rich{
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\begin{itemize}
\item
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Discuss the 3-dimensional case, perhaps relating it to volume integrals.
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\end{itemize}
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}
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