Easier/Faster Way to Calculate $\iint \operatorname{curl}\mathbf{G \cdot} \; d\mathbf{S} \quad$?

Question

Is it always correct to rewrite $\iint \operatorname{curl}\mathbf{G \cdot} \; d\mathbf{S}$ as

$\iint \color{green}{\operatorname{curl}\mathbf{G} \cdot} \, d\mathbf{S} = \iint \color{green}{\operatorname{curl}\mathbf{G}}{\; \cdot \; (\partial_u \mathbf{r} \times \partial_u \mathbf{r}) } \; dA \qquad ?$

I substituted $\color{green}{\mathbf{F} = \operatorname{curl}\mathbf{G}}$ by virtue of Equation 9 below (from James Stewart P1087).

Answer 1

Yes; its correct.

By the way: The double (and finally cancelling) appearance of the square root $|{\bf r}_u\times{\bf r}_v|$ in some explanations of the flux integral is superfluous. By its nature the flux is a volume, namely the amount of fluid traversing the surface per second, and being a volume it should appear as a determinant (or triple vector product) under the integral sign.