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I want to know if the equation I provide actually does indeed accurately model the transformation that I desire to model

Model the ellipsoid by $$f\left(u,v\right)=(A\cos\left(u\right)\sin\left(v\right)\,B\sin\left(u\right)\sin\left(v\right),C\cos\left(v\right)) $$

Model the resultant sphere by $$g\left(u,v\right)=(\sqrt[3]{ABC}\cos\left(u\right)\sin\left(v\right)\,\sqrt[3]{ABC}\sin\left(u\right)\sin\left(v\right),\sqrt[3]{ABC}\cos\left(v\right)) $$

Define:

$$A_{1}=A+T\left(\sqrt[3]{ABC}-A\right)$$

$$B_{1}=B+T\left(\sqrt[3]{ABC}-B\right)$$

$$C_{1}=C+T\left(\sqrt[3]{ABC}-C\right)$$

Then in order to ensure that the volume is constant throughout the transformation, model the transformation by

$$x\left(u,v\right)=A_{1}\sqrt[3]{\left(ABC\right)^{-1}}\sqrt[3]{\left(\left(A+T\left(\sqrt[3]{ABC}-A\right)\right)\left(B+T\left(\sqrt[3]{ABC}-B\right)\right)\left(C+T\left(\sqrt[3]{ABC}-C\right)\right)\right)}\cos\left(u\right)\sin\left(v\right)$$

$$y\left(u,v\right)=B_{1}\sqrt[3]{\left(ABC\right)^{-1}}\sqrt[3]{\left(\left(A+T\left(\sqrt[3]{ABC}-A\right)\right)\left(B+T\left(\sqrt[3]{ABC}-B\right)\right)\left(C+T\left(\sqrt[3]{ABC}-C\right)\right)\right)}\sin\left(u\right)\sin\left(v\right)$$

$$z\left(u,v\right)=C_{1}\sqrt[3]{\left(ABC\right)^{-1}}\sqrt[3]{\left(\left(A+T\left(\sqrt[3]{ABC}-A\right)\right)\left(B+T\left(\sqrt[3]{ABC}-B\right)\right)\left(C+T\left(\sqrt[3]{ABC}-C\right)\right)\right)}\cos\left(v\right)$$

where $T$ goes from zero to 1

Here I have a graph modeling the transformation where we can see the surface continuously deform as $T$ goes from zero to one

This problem is the 3D counterpart to its 2D sister question Modeling the continuous deformation of an arbitrary ellipse centered at origin into a circle of the same area centered at origin (graph included)

Simon M
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    This will be a lot easier to do if you can frame this in terms of linear algebra. An ellipsoid (or hyper-ellipsoid in higher dimensions) can always be written as the set ${x: x^T A x <= 1}$ for a matrix $A$ that characterizes the ellipsoid shape (more on this in the first part of the answer here: https://math.stackexchange.com/a/3678498/3060). To deform the ellipsoid, you can slowly modify the eienvalues of $A$ until they become equal, while keeping the product of the eigenvalues the same – Nick Alger Sep 28 '23 at 20:53
  • @NickAlger that is rather illuminating. It seems to me that the linear algebra approach is closely related to the typical way in which normalized ricci flow is studied. In my experience I've only ever seen ricci flow mathematically formalized in terms of matrices, so it would make a lot of sense that the usual approach taken by mathematicians when doing this type of procedure is to use matrices. I'm curious to learn now about this approach, assuming I can understand it with only an introductory differential geometry background. Is the linear algebra method indeed related to ricci flow? – Simon M Sep 28 '23 at 21:00
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    I don't know much about Ricci flow, but I wouldn't be surprised if there is a connection. – Nick Alger Sep 28 '23 at 21:05

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