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### CHANGE OF VARIABLES

#### Essentials

##### Main ideas

- There are many ways to solve this problem!
- Using Jacobians (and inverse Jacobians)

##### Prerequisites

- Surface integrals
- Jacobians
- Green's/Stokes' Theorem

##### Warmup

Perhaps a discussion of single and double integral techniques for solving this problem.

##### Props

- whiteboards and pens

##### 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:

- change of variables (at least 2 ways)
- Area Corollary to Green's Theorem (at least 2 ways)
- ordinary single integral (at least 2 ways)
- ordinary double integral (at least 2 ways)
- surface integral

#### Details

##### In the Classroom

- 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 $\Partial{(u,v)}{(x,y)}$.
- 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 $v$.
- 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 $v$. 1) 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.

##### Subsidiary ideas

- Review of Green's Theorem
- Review of single integral techniques
- Review of double integral techniques

##### Homework

(none yet)

##### Essay questions

(none yet)

##### Enrichment

- Discuss the 3-dimensional case, perhaps relating it to volume integrals.

1)
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.