In orthogonal curvilinear coordinates $(\zeta_1, \zeta_2, \zeta_3)$ with scale factors $h_1, h_2, h_3$ the grad, curl and div can be written as: $$\nabla (\psi)=\frac{\hat e_1}{h_1} \frac{\partial \psi}{\partial \zeta_1}+\frac{\hat e_2}{h_2} \frac{\partial \psi}{\partial \zeta_2}+\frac{\hat e_3}{h_3} \frac{\partial \psi}{\partial \zeta_3} $$ $$\nabla \cdot \vec F=\frac{1}{h_1 h_2 h_3}(\frac{\partial (h_2h_3F_1)}{\partial \zeta_1} +\frac{\partial (h_1h_3F_2)}{\partial \zeta_2}+\frac{\partial (h_1h_2F_3)}{\partial \zeta_3})$$ $$\nabla\times \vec F=\frac{\vec e_1}{h_2h_3}(\frac{\partial h_3F_3}{\partial \zeta_2}-\frac{\partial h_2F_2}{\partial \zeta_3} )+\frac{\vec e_2}{h_1h_3}(\frac{\partial h_1F_1}{\partial \zeta_3}-\frac{\partial h_3F_3}{\partial \zeta_1} )+\frac{\vec e_3}{h_2h_1}(\frac{\partial h_2F_2}{\partial \zeta_1}-\frac{\partial h_1F_1}{\partial \zeta_2} )$$ Although not that hard to remember, I was wondering if there was a quick way to derive these equations? (I don't mind how complex the maths is as long as it is quick).
Note: The scale factor for the coordinate $\zeta_i$ is the square root of the coefficient of $d\zeta^2_i$ in the first fundamental such that: $$ds^2=\sum_i h_i^2 d \zeta_i^2$$