I'm trying to solve this problem
A particle of mass m moves under the action of gravity on the inner surface of a paraboloid of revolution $x^2+y^2=az$ which assumed frictionless. Obtain the equations of motion.
The Lagrangian in polar coordinates, assuming gravity towards the negative z axis, is $$L=\frac 1 2m\left(\dot\rho^2+\rho^2\dot\varphi^2+\dot z^2\right)-mgz;\qquad \ddot q\equiv\frac {d\dot q} {dt}\equiv\frac{d^2 q}{dt^2}$$constraints imposed condition implies $$f=\rho^2-az=0$$ Lagrange equations for this system are$$\frac d{dt}\left(\frac{\partial L}{\partial \dot q_i} \right)-\frac{\partial L}{\partial q_i}=\lambda\frac{\partial f}{\partial q_i}$$ $\lambda$ is a Lagrange multiplier. Then to $q_1=\rho$, $q_2=z$, $q_3=\varphi$ $$\fbox{$m\ddot\rho-m\rho\dot \varphi^2=2\rho\lambda$}$$ $$\fbox{$m\ddot z+mg=-a\lambda $}$$ $$\fbox{$ \frac{d}{dt}\left(\rho^2\dot\varphi\right)=0$}$$ $$\fbox{$\rho^2-az=0$}$$ $$\rho\in[0,+\infty),\quad\varphi\in[0,2\pi),\quad z\in(-\infty,+\infty)$$ I do not know how to solve this system of equations, and most important is to determine $\lambda$.
From system of equations I could deduce that $$\rho^2\dot\varphi=c_0$$ $$m\ddot\rho-m\frac{c_0^2}{\rho^3}=2\rho\lambda$$ $$az\dot\varphi=c_0$$ $$\dot z\dot\varphi+z\ddot\varphi=0$$
in Cartesian coordinates $$\fbox{$m\ddot y=2y\lambda$}$$ $$\fbox{$m\ddot x=2x\lambda$}$$ $$\fbox{$m\ddot z+mg=-a\lambda $}$$ $$\fbox{$x^2+y^2-az=0$}$$