= Solution
An infinitesimal rotation about $x$ gives $\delta\psi_y=-\epsilon\psi_z$ and $\delta\psi_z=\epsilon\psi_y$. Both the squared transverse velocity and squared transverse slope are rotational invariants. <Noether's theorem> therefore gives
$$
\boxed{\ell_x=\rho(\psi_y\dot\psi_z-\psi_z\dot\psi_y),\qquad
\mathcal J_x=-F(\psi_y\psi_z'-\psi_z\psi_y').}
$$
Differentiating and using the two <wave equations> gives
$$
\partial_t\ell_x=F(\psi_y\psi_z''-\psi_z\psi_y'')
=-\partial_x\mathcal J_x.
$$
Consequently \b[$L_x=\int\ell_x\,dx$ is the <angular momentum> about the equilibrium string axis], conserved when the net endpoint <torque> flux vanishes. For a finite segment, $dL_x/dt=-[\mathcal J_x]_{\rm left}^{\rm right}$. It is not momentum associated with translating the coordinate $x$; its symmetry rotates the two physical transverse displacement components into one another.
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