(Unless i'm wrong) If you want to map it to a plane or to a sphere, I guess Liouville's theorem for comformal mapping says you're out of luck. The only c.m. that are possible for 3D surfaces are with a combination of similitudes and inversions by a sphere. So it does not look possible as we could not obtain a paraboloïd from a sphere or a plan by a combination of permitted inverse transformations.

3D comformal mappings are much more restricted than 2D's.

Else, we could have used an extension of strereographic projection with special case of north pole at infinity, so all points would map by perpendicular projection of each point of the curve by a line parallel to the axis of the paraboloid to a point in the base plan. But I deduce from above that this would not be a comformal map.

Dear Christian..! 

Thank you for your suggestion. If we assume that the paraboloid is made of rotating a parabola ( 2D) about symmetric axis  . Then is it possible to find a conformal map?

My immediate guess would be that it would be possible. But of course it requires some thought. If we are talking about the rotated parabola, we would have a natural system of latitudes and longitudes cut out by planes perpendicular to respectively containing the axis of rotation.  A natural candidate would be that the latitudes are mapped onto concentic circles with appropriate spacing, i.e. that the stretching along the latutudes is infitesimally the same as the stretching along longitudes. If this approach would not work, I am pessimistic.

A wild starting idea, just for rough estimation: a  direct physical experiment. Make a paraboloid-like hard model (for example, out of wood). Smooth a thin transparent plastic sheet on it. (The shape of some drinking glasses is also close to a paraboloid.) Draw any geometric shape on its surface with markers, such as circles or the system of coordinates that Ulf proposes. Take the transparent shape off from the hard model. Use a spot-lamp to project the drawings on a sheet of paper or on the inside surface of a sphere. I use the same method to demonstrate stereographic projection for my students.

Consider the general case of a surface of revolution given by x=f(y) where y is the co-ordinate f the y-axis and corresponding for x, which is thus the radius of the meridan at heighy y.

A map along the lines above is given by specifying $R(y)$ the radius of the image of the  meridan at height y. Now $R$ should satisfy the following differentiak equation

$\frac{R\prime(y)}{\sqrt{1+(f\prime(y)^2}}=\frac{R(y)}{f(y)}$

(equaity of scaling along longitudes and latitudes)

which can easily be rewritten as

$\frac{dlog(R(y)}{dy}=`\frac{\sqrt{1+(f\prime(y)^2}}{f(y)}$

A warm up would be the case of $f(y)=\sqrt{1-y^2}$

István,

Points above the lamp will not map on the plane of the paper so you don't map the whole paraboloïd and this is not bijective.

Looks more like a diffeomorphism on a local chart.

The map given by me is not a geometric projection (i.e. given by a lamp) Of course the solution for R may be limited as far as possible y-values. I have not worked out what it is in the case of the paraboloid, in the case of the sphere it ought to give the usual stereographic projection.

Mr Persson, my last post was made before I read yours. I was not commenting on it.

I will look into it. I was still on the comformal mapping issue of original question.

Have a good day.

To Christian: I realized that later. Sorry about that.

Christian and Ulf, your remarks are correct. I proposed the lamp model because it may be useful, not to find infinite truth but to avoid infinite mistakes (Brecht). I have weak space perception, and need this type of experimentation whenever it is at all possible. Even in teaching about stereographic projection, I use a hemisphere instead of the whole sphere. It gives a better picture on the flat sheet.

If you don't need perfection of a comformal mapping, it should do the job as near the bottom, a paraboloïd resembles a half sphere, so by adjusting the "openness" of the surface, height of the lamp and projecting a small part of the surface, should be good enough. Have you ever watdhed the wonderful "Dimensions" animation film?

This is worth it !

http://www.dimensions-math.org/Dim_regarder.htm

Thank you, Christian!