@AndreaAlp It seems that in your my_quadric, you have written z where you should have y.  If we make that change, then graph that we obtain agrees with the photo of the bone that you have shown.

But this raises the question of what you mean by "the center of the ellipse".  If you rotate the graph in the worksheet (given in the link below) you will see that I have calculated and marked the true center of the ellipse.  (That mark is not visible in the picture shown above).   I am afraid, however, that's not what you are really looking for.  I suspect that you are looking for some sort of a "center" on the red surface.  If so, the meaning of such a "center" is unclear.  You may want to rethink it.

mw3.mw

@AndreaAlp Youwrite: "Still, those surfaces should intersect and create an ellipse":

Which surfaces?  We seem to have agreed that g(x,z)=0 and g(x,z)=y are different.

@AndreaAlp You seem to be confusing f(x,y)=0 with f(x,y)=z.  They are certainly not the same things!  For instance, x^2 + y^2 - 1 = 0 is a cylinder, while x^2 + y^2 - 1 = z is a paraboloid.

Similarly, g(x,z)=0 and g(x,z)=y are quite different things.   For instance, x^2 + z^2 - 1 = 0 is a cylinder, while x^2 + z^2 - 1 = y is a paraboloid.

@AndreaAlp In the x,y,z Cartesian coordinates, any surface of the form f(x,z)=0 is a cylinder aligned with the y axs.

If you expect something other than a cylinder, then there must be something wrong with your LinearFit calculations.

@EoM007 Here is how.  Change terminating colons to semicolons if you wish to see the intermediate steps.

Download mw2.mw

All we can say from what you have given is that this is a bad approximation.

Perhaps there is a typographical error somewhere?

I have two books here on my shelf that contain the derivation of the equations of motion of a homogeneous ball that rolls on an arbitrary surface without slipping:

1. Neimark and Fufaev, Dynamics of Nonholonomic Systems, pages 76-86.
2. L. A. Pars, A Treatise on Analytical Dynamics, pages 211-212.

The equations describe the motion in terms of a coordinate system that move with the ball.  It will take some effort to relate them to a stationary coordinate system which would be necessary for producing a Maple animation.  I do not have the time to do it now.  Perhaps at some other time.

Aside: The surface cos(abs(x)+abs(y)) is a rather odd choice for this problem because it has sharp ridges due to the presence of the absolute values.  Is there a reason you are interested in that particular surface?

@EoM007 Details in the worksheet mw.mw

@EoM007 In your worksheet you have reversed the left and right hand sides of  the equations that gave.  The order matters.  Keep omega(t) on the right-hand side.

As to differentiating a vector, use the tilde operator, as in:
V := < a(t), b(t) >;
diff~(V,t);

@EoM007 Showing a picture of your worksheet isn't very helpful because I cannot work with it.  See if you can post your actual worksheet.  In the window where you edit your message, click on the big fat green arrow to upload it.

@Christianwm I am unable to read the file that you have uploaded.  It's possilbe that it's due to non-ASCII characters in its name.   See if you can rename it to something with English characters only.

Additionally, have a look at this recent exhange.  It may have some useful informaiton:

https://www.mapleprimes.com/questions/222566-How-To-Make-Danish-Letters-Work-In-MapleCloud

It's hard to tell without further information.  It will help if you upload your worksheet here so that people may have a look.

In the screen where you edit your message, click on the big fat green arrow to upload your worksheet.

Linux 64 bit, Maple 2017.3 64 bit — crashes.

Mahmood, In the title you refer to "two equations" and later you ask about "these equations" but I can't tell what equations you are talking about.

To receive helpful answers, try to clearly state the problem that you are attempting to solve, and how you intend to solve it.  Then the rest of your calculations may make better sense.

@John Fredsted The M that enters the right-hand side of the differential equation is not the mass of Mercury, it is the mass of the sun, and therefore M = 1,99*10^30 is correct.

As to h, it is equal to r^2*diff(phi(t),t) which is a constant during the motion.  But r*diff(phi(t),t)  is the orbital velocity, let's call it v.  Therefore h = r*v.  We know r at perihelion, all we need is v at perihelion. A good approximation to v may be obtained by assuming that the orbit is circular, therefore the orbital velocity is the length of the orbit, that is 2*Pi*r divided by the time it takes to go around which is about 83 days if I remember it correctly.