 Solution

Using the product rule for directional derivatives,Thus $f=|u{|}^2/2-P$ and $g=-h$. If $\nabla \cdot u=0$ and $w=\nabla \times u$, then $\nabla \cdot w=0$. Integrating the two directional derivatives as divergences gives only boundary fluxes, which vanish because $u\cdot n=w\cdot n=0$. This is helicity conservation by tangent boundary conditions, so $dH/dt=0$.

For the stated field, direct use of the cylindrical curl gives $u\cdot w=-2a{\rho }^4{\sin }^{2}z$. Therefore