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Here is the construction of the solid. Take an ellipse, make a copy of it, and put it on top of the original ellipse. Now turn the top ellipse by $90^\circ$ (quarter turn). Glue the two boundaries. I would like the height, volume, and the equation describing the ridge/boundary of this solid in terms of the major and minor axes.

Edit: Instead of two ellipses if you took a rectangle with sides $a$ and $b$ with $ b < a$ and followed the same procedure, you end up with box base a $b \times b$ square and of height $a -b.$ I wish I could take pictures of both solids I made and post it here. Hopefully this helps.

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Your description is far too vague. Put the copy where? Turn about what axis? Glue what where? –  Robert Israel Feb 20 '12 at 19:48
Do you mean just two ellipses like this: $\{(x,y,z): x^2 + (y+1)^2 \leq 1, x^2 + (z-1)^2 \leq 1\}$? Or is it a 3D solid with a 2D ellipse (the first one) added? For the second case, you can easily find the volume with the Pappus-Guldin theorem (planetmath.org/encyclopedia/GuldinsTheorem.html). –  savick01 Feb 20 '12 at 19:51
@savick01: i don't think the solid i described is a solid of revolution. –  abel Feb 20 '12 at 21:01
If you say that you turn sth, there are two cases: you change its position (then it is similar to the set that I described) or you produce a solid of revolution. Here it is quarter of the full revolution, but we can divide by 4 (the theorem is true for any angle). Well, there is the third case if you turn around an axis intersecting the ellipse. –  savick01 Feb 20 '12 at 21:14
Could you elaborate a bit in your question to remove the ambiguities? –  savick01 Feb 20 '12 at 21:16
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If I understand you correctly, in two dimensions the figure would be as shown below Do you want the area of intersection or union of the two ellipses? Note that in your rectangle (which should be rectangular prism) example you assume that two of the dimensions are the same. An ellipsoid has three axes. If you want to specify that two are the same you have a spheroid, prolate if the odd axis is longer than the other two, oblate if the odd axis is short.

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Do you mean like this?enter image description here

If by height you mean the distance AB, then the major axes make a right triangle, so height$=\frac{a}{\sqrt 2}$, where $a$ is the major axis and $b$ is the minor axis.

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