Question: vector integral in non-orthog coordinates

[7thSon]

reposting, there was some major latex fail

 

[7thSon]

Hi, I think I have a relatively simple question. I'm not that strong in differential geometry at all so a shorter and somewhat heuristic answer might actually be more useful to me than a rigorous one.

Suppose I want to use the change of variables theorem on the following integral, where I will be changing variables into a non-orthogonal curvilinear coordinate system {\xi_1, \xi_2, \xi_3} to describe my domain. Assume Euclidian space, smooth boundary, etc.

\int_{\Gamma} \nabla u \cdot \mathbf{n} \ d\Gamma

I am used to doing this integral only in orthonormal coordinate systems, so my question is that when I change coordinate systems to {\xi_1, \xi_2, \xi_3}, do I need to convert the gradient operator and/or use the induced inner product in this new coordinate system?

It seems to me the answer would be no, because I can expand the vectors in their cartesian components and rid myself of the dot product that way.

\int_{\Gamma} \frac{\partial u}{\partial \xi_a} \frac{\partial \xi_a}{\partial x_i} \mathbf{\tilde{e_i}} \cdot n_j \mathbf{\tilde{e_j}} \ d\Gamma

Obviously in a cartesian frame, \mathbf{\tilde{e_i}}\cdot\mathbf{\tilde{e_j}}=\delta_{ij}

Giving

\int_{\Gamma} \frac{\partial u}{\partial \xi_a} \frac{\partial \xi_a}{\partial x_i} n_i \ d\Gamma

after which point I can change the differential to be w.r.t. \xi, have consistent area size, etc.

Is this valid? I feel like I got something for nothing.

Would it be equally valid to recast the gradient in the skew coordinate system along with the normal, giving
\int_{\Gamma} \nabla_{\xi} u \cdot n_j \mathbf{\tilde{g_j}} \ d\Gamma


And in this case, would the inner product still be diagonal and not some other bilinear form, even though the (skew) basis functions would have off-diagonal metric coefficients?

thanks for any help!