I am getting stuck in a detail in a paper. It's about the axi symmetric Navier Stokes equations $$u_t - \nu\,\Delta u + u\cdot \nabla u + \nabla p=0$$ We consider in cylindrical coordinates $u=(u^r, u^\theta,u^z)$. And we have the following vorticity equation in cylindrical form. $$\omega^r = \frac{1}{r}\frac{\partial u^z}{\partial \theta} - \frac{\partial u^\theta}{\partial z} = - \frac{\partial u^\theta}{\partial z}, \\ \omega^\theta = \frac{\partial u^r}{\partial z} - \frac{\partial u^z}{\partial r}, \\ \omega^z = \frac{1}{r}\frac{\partial}{\partial r}(r u^\theta) - \frac{1}{r} \frac{\partial u^r}{\partial \theta} = \frac{1}{r}\frac{\partial}{\partial r}(r u^\theta).$$ My question goes as follows: Suppose $J=\frac{w^r}{r}$, then how to derive the following steps?

\begin{align} & \int J(\omega^r \partial_r+\omega^z \partial_z)\frac{u^r}{r} r \, dr \, dz \\[8pt] = {} & \int [\nabla\times (u^\theta e_\theta)] \left(J \, \nabla \frac{u^r}{r}\right) r \, dr\,dz \\[8pt] = {} & \int (u^\theta e_\theta) \left(\nabla J\times\nabla \frac{u^r}{r}\right) r \, dr \, dz \end{align}

I'm assuming some integration by parts is involved but couldn't derive them; besides, where does the cross product come from? The equality above that bugs me is from the bottom of page 11 of this paper.