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In specific, given an 5-dimensional ellipsoid with intercepts,(a1, a2, ..., a5), with the Cartesian coordinates, what are the set of linear equations that describe 5 - dimensional rectangle inscribed in ellipsoid with vertices at the intersection of 5-dim ellipsoid with the Cartesian coordinates. Also what is the fraction of volume covered by inscribed rectangle. Further I am looking to generalise this for n -dim and non-euclidean geometry. So any references that builds up for specific cases and generalizes it.

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First do the case when all $a_i = 1$, so the ellipsoid is just the unit Euclidean ball. Then get the general case by applying a linear transformation whose matrix has $a_1,\dotsc,a_n$ on the diagonal and zeroes elsewhere (which changes all volumes by a factor of $a_1\cdot a_2\dotsb a_n$).

The volume of the $n$-dimensional Euclidean ball of radius $1$ is $\pi^{n/2}/\Gamma(1+\frac{n}{2})$. There are many ways to compute this; Wikipedia has a few. Another nice method involves computing $\int_{\mathbb{R}^n} e^{-|x|^2} \,dx$ in two ways; see the first few pages of Keith Ball's An Elementary Introduction to Modern Convex Geometry, available here.

I'm not totally sure what you mean by "5-dimensional rectangle", but since you say it has vertices where the Cartesian coordinate axes meet the ball, I think you mean a cross-polytope. The volume of the standard cross-polytope (the one inscribed in the unit ball) is $2^n/n!$. I think this is computed in Coxeter's Regular Polytopes, but I'm not sure, so here's the proof: cut the cross-polytope into $2^n$ pieces, one in each orthant; each piece is a "right-angled simplex", by which I mean (something congruent to) the convex hull of the origin and the standard basis vectors $e_1,\dotsc,e_n$. This simplex is a cone with vertex at $e_n$, of height $1$, and whose base is a right-angled simplex of the next lower dimension; by induction, it has volume $1/n!$.

So the proportion of the volume of the ball which is contained in the inscribed cross-polytope is $$ \frac{\Gamma(1+\frac{n}{2}) 2^n}{\pi^{n/2} n!} $$ (By Stirling's formula, as $n\to\infty$ this is asymptotically $\frac1{\sqrt2} \left(\frac{2e}{\pi n}\right)^{n/2}$, which goes to zero very rapidly as the dimension grows. The inscribed cube captures much more of the volume.)

The cross-polytope is given by the inequality $$ \sum_{i=1}^n |x_i| \le 1 $$ (where $x_1,\dotsc,x_n$ are the Cartesian coordinates). You can resolve this into $2^n$ linear inequalities if you like, one for each way of choosing signs for the $x_i$. For example, in the $3$-dimensional case, you'd have $$ x_1 + x_2 + x_3 \le 1 $$ $$ x_1 + x_2 - x_3 \le 1 $$ $$ x_1 - x_2 + x_3 \le 1 $$ $$ x_1 - x_2 - x_3 \le 1 $$ $$ - x_1 + x_2 + x_3 \le 1 $$ $$ - x_1 + x_2 - x_3 \le 1 $$ $$ - x_1 - x_2 + x_3 \le 1 $$ $$ - x_1 - x_2 - x_3 \le 1 $$ I don't know of a reference that explains this carefully. [Edit: I meant I didn't know a reference that explains the cross-polytope carefully. But the general theory of expressing convex bodies via linear inequalities is standard, and probably any introductory text on convex geometry will discuss it. Try, say, Lay's Convex Sets and Their Applications, the section on separating hyperplanes and the section on polarity. The $2^n$ inequalities above arise because the polar of the cross-polytope is the $n$-dimensional cube, which has $2^n$ vertices.]

I don't know anything about non-Euclidean geometry.

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