Coordinate systems/Derivation of formulas/Differential normal areas

The question
The question is whether it is appropriate to write the vector surface element as the sum of three perpendicular elements. dy \, dz \hat{\mathbf x} + dx \, dz \hat{\mathbf y} + dx \, dy \hat{\mathbf z} $$ \rho \, d\phi \, dz   \hat{\boldsymbol\rho} +        d\rho \, dz    \hat{\boldsymbol\phi} + \rho \, d\rho \, d\phi \hat{\mathbf z} $$ r^2 \sin\theta \, d\theta \, d\phi  \hat{\mathbf r} + r   \sin\theta \, dr      \, d\phi   \hat{\boldsymbol\theta} + r             \, dr      \, d\theta \hat{\boldsymbol\phi} $$

Answer
Changing notation a bit, we work in Cartesian coordinaes:

(1)  $$d\vec A = dxdy\hat z + dydz\hat x + dzdx\hat y$$

is not how the surface integral is usually set up. Instead one uses a cross product and a surface defined in parametric form:

(2)  $$d\vec A = d\vec\ell_\alpha\, \times d\vec\ell_\beta\, $$ $$= (dy_\alpha\,  dz_\beta\,  -    dz_\alpha\,  dy_\beta\, )\hat x + (dz_\alpha\,  dx_\beta\,  -    dx_\alpha\,  dz_\beta\, )\hat y + (dx_\alpha\,  dy_\beta\,  -    dy_\alpha\,  dx_\beta\, )\hat z $$

Nevertheless, it is possible to force $$d\vec A$$ as shown in (1) to a vector area element with the following constraints:

(3)    $$dz_\alpha\,  =0\qquad dz_\beta\, <0\qquad dx_\beta\, =0$$ (4)  $$d\vec A = d\vec\ell_\alpha\, \times d\vec\ell_\beta\, $$ $$= dy_\alpha\,  dz_\beta\,  \hat x  -    dx_\alpha\,  dz_\beta\,  \hat y + dx_\alpha\,  dy_\beta\, \hat z $$

In this context, writing (1) is putting the proverbial round peg into a square hole. This can be seen by if we examine both differentials:

(5)  $$d\vec\ell_\alpha\,  = dx_\alpha\,  \hat x + dy_\alpha\, \hat y$$  ,   and   $$d\vec\ell_\beta\,  = dy_\beta\,  \hat y + dz_\beta\, \hat z$$

The paths $$d\vec\ell_\alpha\, $$ and $$d\vec\ell_\beta\, $$ must follow the contour of the surface like a net. But one path cannot move in the z direction, while the other is similarly confined to a plane at constant x. Hence we can form a area differential of the form given by (1), but it cannot be integrated over any surface not confined to a flat plane. Hence, we conclude that (1) describes a differential surface element that cannot be integrated over any curved surface. Ultimately, there is no strong reason to replace (1) by

(6)  $$d\vec A = dxdy\hat z \ \dots or \dots \ dydz\hat x  \ \dots or \dots \ dzdx\hat y.$$

Equation (6) has the advantage of not misleading the reader that the differential can be integrated over a surface, while (1) has the advantage of allowing the surface to be a flat plane at any orientation.