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So I need to find an volume-preserving mapping from an ellipsoid to a ball (solid sphere). (Specifically: $\dfrac{x^2}9 + y^2 + z^2 \le 3$, but I'd rather understand the general case than just get how to transform this one ellipsoid.)

I think I've got linear transformations (parallelotopes to parallelotopes) down, but apparently I missed something in class because I can't figure out how to approach these (seemingly) non-linear transformations.

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EDITED: SORRY, I MADE A SILLY DETERMINANT MISTAKE, SEE CORRECTIONS.

Also, if you are interested in learning a little more linear algebra you should research singular values and SVD's. It turns out that all linear transformations take unit circles/spheres to ellipses/ellipsoids etc..

The volume of that ellipsoid is $$4/3\pi abc=4/3\pi (3)=4\pi$$ so the sphere with the same volume would have radius $$r=3^{1\over 3}$$ If you want it linear, you could use the matrix $$3^{1\over 3}\begin{bmatrix}{1\over 3}&0&0\\0&1&0\\0&0&1\end{bmatrix}$$ but note that the transformation is not in general volume preserving.

Or perhaps you meant volume-preserving, but not for the ball? I just thought of that...

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so you could use $${1\over 9^{1\over 3}}\begin{bmatrix}1&0&0\\0&3&0\\0&0&3\end{bmatrix}$$ which has determinant 1 and sends the ellipse to a sphere of radius 1/3 (and is the same as the transformation above!!).

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In general, the ellipsoid $${x^2\over a^2}+{y^2\over b^2}+{z^2\over c^2}=1$$ can be transformed into a sphere with the volume-preserving matrix $${1\over (abc)^{2\over 3}}\begin{bmatrix}bc&0&0\\0&ac&0\\0&0&ab\end{bmatrix}$$ So $$(u,v,w)={1\over (abc)^{2\over 3}}(bcx,acy,abz)$$

I found the matrix as follows: the current ellipsoid has sides of length a,b, and c. You want them to be equal, so you scale in the x-direction by bc, in y by ac and in z by ab. Then the lengths all become abc. That gets the scaling matrix I wrote above $$\begin{bmatrix}bc&0&0\\0&ac&0\\0&0&ab\end{bmatrix}$$ I added the constant out front to make the determinant of the matrix one, so that it would be volume-preserving.

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  • $\begingroup$ Okay. Thank you. That's exactly what I needed! $\endgroup$
    – user137731
    Commented Apr 8, 2014 at 1:18

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