@Alfred_F Okay, that's a good proof.  In fact, the statement of the problem may be generalized a bit.

Let S be any set of points in R³.  Suppose that for any plane P, the points in the intersection S ∩ P lie on a circle.  Then the points in S lie on a sphere.

The attached worksheet illustrates your proof. 

Download mw.mw

@Earl Let r be the bead's position vector relative to a reference point, and let v be velocity vector.  Then the bead's angular momentum is L = m r x v, where m is the mass and "x" is the cross product.

Hello Earl, I looked through your worksheet but couldn't make heads or tails of it.  It seems to me that you wish to analyze the motion of a bead along a wire track.  That's pretty much the same problem that was discussed in an earlier question. How is this one any different?

You may want to state the problem clearly, in words, without an appeal to a worksheet.

@sand15 Also note that his equations involve both lambda and lambda[1].  Then he sets lambda=1 and lambda[1]=0.1. 

@acer Ah, that explains why implicitplot does not complain in nm's answer.

@nm Needn't we worry about the imaginary 7/5*Pi*I term?

@Earl You have asked an interesting question and I am glad to have been able to provide a solution that you have found satisfactory.  Cheers!

@C_R That's a pretty curve and works nicely in the context of this problem.  And you are correct in pointing out that the the curve's parametrization need not be arclength parametrization.

This is extracted from your worksheet:

Earl, in your worksheet you write:

    Both bobs slide down according to the downward tangential
    force of gravity and the opposing tangential force of friction.
    The latter is the product of the component of the gravitational
    force normal to the arc and the coefficient of friction.

That does not look correct.  The normal component of the reaction force should be the sum of two forces:

1. The component of the gravitational  force normal to the arc, as you have correctly stated;
2. The centripetal force due to the bob's curved path.

To see the significance of the #2 above, imagine turning off the force of gravity and letting the bob move solely due to imparted initial velocity.  According to your formulation, the force of friction will be zero.  But as the bob slides along the curve, the centripetal force will press it against the curve, making it feel the force of friction. That's where #2 comes in.

@Kitonum The help page on ListTools:-Collect says: 

    Note that the order of the output is session dependent, and is therefore not defined.

Isn't that a problem with implementing your suggestion?

Interesting subject and good work!

For whatever it's worth, here is a different take to what you have done.

Download EML.mw

@Aliocha Ah, I see the reason for the confusion.  The f(n,x) in my answer stands for the antiderivative of your f(n,x),  so I should have called it by some other name, like F:

F := (n,x) -> n*x + 2*sum((n-k)/k*sin(k*x), k=1..n-1);

You wish to integrate your f(n,x) over the interval (0,Pi).  That amounts to evaluating

F(n,Pi) - F(n,0);

But sin(k*x) is zero at 0 and Pi, therefore the summation evaluates to zero and we are left with n*Pi - n*0, that is n*Pi, as you had expected.
 

@Aliocha I don't have Maple 2026 but the formula works just fine in Maple 2024 and 2025.  Here it is:

Download mw.mw

@C_R My comment applies to your worksheet 2d_BVP_Haberman.mw where BC3 specifies zero temperature on the outer boundary.  That's where you need the compatibility condition that I have noted.  That wouldn't be an issue in the case of an insulated boundary.

@janhardo In the solutions that you have provided, you are taking q = q(t), but the original statement says q = q(r,t).