# Shape/volume of this solid (don't know the name)

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

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.