The Potential Energy Function in Three-Dimensional Motion

Terrycho
Homework Statement:
A particle is attracted toward the z-axis by a force F vector proportional to the square of its distance from the xy-plane and inversely proportional to its distance from the z-axis. Add an additional force perpendicular to F vector in such a way to make the total force conservative, and find the potential energy.
Relevant Equations:
→ → → → →
F = ∇ V, and a conservative force F satisfies ∇ X F = 0
I set the location of the particle (x,y,z); therefore,

the force F_1 is (z^2/root(x^2+y^2) * x/root(x^2+y^2) , z^2/root(x^2+y^2) * y/root(x^2+y^2), 0), since cosΘ is x/root(x^2+y^2).
→ → → →
And also, the force F_1 and the additional force F_2 are perpendicular so, F1 ⋅ F2 =0.

So, I got F_x=-F_y. (I set F_2 as (F_x, F_y, F_z)

The total Force F_tot is ( z^2/root(x^2+y^2) * x/root(x^2+y^2) +F_x , z^2/root(x^2+y^2) * y/root(x^2+y^2) -F_y , F_z )
→ →
Then, I used Del operator ∇ X F = 0, which gave me the following result.

(1) ∂F_z/∂y + ∂F_x/∂z = 2yz/(x^2+y^2)

(2) ∂F_x/∂z - ∂F_z/∂x = 2zx/(x^2+y^2)

(3) ∂F_z/∂y + ∂F_z/∂x = (2yz-2zx)/(x^2+y^2)

I am kind of able to feel by doing something with the equations (1),(2), and (3), you can figure out the F_tot which leads to get to know the potential Energy but... I got stopped here!

I'd really appreciate if you could help me out. I attached some photos below for those of you who got confused with my messed up complicated symbols!

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