Find the rhumb lines of the surface of revolution obtained rotating the curve $xz=1$ around the $z$-axis.
I consider the parametrization $\sigma(u,v)=(u\cos(v), u\sin(v), 1/u)$ with $(u,v)\in (\mathbb{R}-\{0\}) \times \mathbb{R}$.
Since $F=\sigma_u \cdot \sigma_v=0$, a regular curve $\gamma(t)=\sigma(u(t),v(t))$ is a rhumb line iff for some $\theta \in [0,2\pi]$ we have $\cos(\theta)=(\dot{\gamma} \cdot \sigma_u)/(|\dot{\gamma}| \cdot|\sigma_u|)$ and $\sin(\theta)=(\dot{\gamma} \cdot \sigma_v)/(|\dot{\gamma}| \cdot|\sigma_v|)$.
Since the first fundamental form is $E = 1+u^{-4}, F=0, G=u^2$, we obtain $\tan(\theta)= \frac{\dot{v}}{\dot{u}}\cdot \sqrt\frac{G}{E}$, thus $\frac{dv}{du}=\tan(\theta)\cdot\sqrt{\frac{1+u^4}{u^6}}$.
Integrating we get $v$ in function of $u$ and substituting $\gamma(t)=\sigma(t, v(t))$.
Is this correct? Is there another way for finding $\gamma$ without having to solve that integral?
Thanks in advance