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Find a parametrization $\sigma : I \subseteq \mathbb{R}^3 \rightarrow \mathbb{R}^3$, with $I$ a parallelepiped, of $\lbrace (x,y,z) \in \mathbb{R}^3 : |z| \leq 4x^2 + 9y^2 \leq 1 \rbrace $.

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@Alex does my answer make sense to you? –  James S. Cook Sep 7 '12 at 5:45
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up vote 2 down vote accepted

I would start by observing $|z|$ is non-negative hence given your constraint we have $0 \leq |z| \leq 1$. Therefore, $z \in [-1,1]$. This is a good candidate for one of the parameters.

Next, for fixed $z$, the condition $|z| \leq 4x^2+9y^2 \leq 1$ places $(x,y,z)$ in an elliptical annulus. The outer ellipse $4x^2+9y^2=1$ and the inner ellipse is $4x^2+9y^2=|z|$. We can use $x = \frac{1}{2}f(z,r)\cos(\theta)$ and $y = \frac{1}{3}f(z,r)\sin(\theta)$ to obtain $4x^2+9y^2=f(z,r)^2$ and we would like to choose $f(z,r)$ such that $f(z,0)^2=|z|$ and $f(z,1)^2=1$. As $r$ varies from $r=0$ to $r=1$ for fixed $z$ we sweep out the elliptical annulus at $z$. I propose:

$$ f(z,r) = \sqrt{r(1-|z|)+|z|} $$

You can easily see $f(z,0)^2=|z|$ whereas $f(z,1)^2=1$. To summarize,

$$ x(r,\theta,z) = \frac{1}{2}\sqrt{r(1-|z|)+|z|}\cos(\theta) $$ $$ y(r,\theta,z) = \frac{1}{3}\sqrt{r(1-|z|)+|z|}\sin(\theta) $$ $$ z(r,\theta,z) = z $$

where $(r,\theta,z) \in [0,1]\times [0, 2\pi] \times [-1,1]$. Certainly a rectangular solid is a parallel-piped.

Notice the method:

1.pick a parameter

2.freeze the parameter and focus on that cross-section

3.use knowledge about trig., hyperbolic functions, whatever... to disassemble the cross- section.

4.start again when the formulas are too ugly for your purposes...

I hope 4. does not apply to this answer.

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