Particle in a cylinder with a spring, sign convention in potential energy (Lagrangian multipliers)

Question

I'm trying to get the force of constraint. The problem I have is when defying the sign of the potential energy using cylindrical coordinates $(rho,phi,z)$, what I have is:
$$
V=mgy-frac{1}{2}kleft(rho^2+Rrho sinphi+left(frac{R}{2}right)^2+z^2right)=
$$
$$
=mgrhosin{phi}-frac{1}{2}kleft(rho^2+Rrho sinphi+left(frac{R}{2}right)^2+z^2right).
$$
But the solution in theory is:
$$
V=-mgrhosin{phi}-frac{1}{2}kleft(rho^2+Rrho sinphi+left(frac{R}{2}right)^2+z^2right).
$$
I don't get why $mgsin{phi}$ is negative, considering the axes in the frame of reference from the diagram below, shouldn't be positive if the $y$-axis has the same direction as the gravity.

AFG · Accepted Answer

Since the $y-$axis is pointing downwards, the height will increase when the $y$ coordinate decreases, so the correct gravitational potential energy is $V_g=-mgy$, as it is greater at higher height.
You can check it this way:
The azimuthal angle $phi$ is usually defined from the $x-$axis, so $V_g=-mgy=-mgrhosinphi$ will be minimum when $phi=pi/2$ (the situation of the diagram) and maximum when $phi=3pi/2.$

Particle in a cylinder with a spring, sign convention in potential energy (Lagrangian multipliers)

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