Coordinate systems/Derivation of formulas/Differential normal areas

Why this formula for surface area element seems either wrong or misleading

The question

The question is whether it is appropriate to write the vector surface element as the sum of three perpendicular elements.

1. $dy\,dz{\hat {\mathbf {x} }}+dx\,dz{\hat {\mathbf {y} }}+dx\,dy{\hat {\mathbf {z} }}$
2. $\rho \,d\phi \,dz{\hat {\boldsymbol {\rho }}}+d\rho \,dz{\hat {\boldsymbol {\phi }}}+\rho \,d\rho \,d\phi {\hat {\mathbf {z} }}$
3. $r^{2}\sin \theta \,d\theta \,d\phi {\hat {\mathbf {r} }}+r\sin \theta \,dr\,d\phi {\hat {\boldsymbol {\theta }}}+r\,dr\,d\theta {\hat {\boldsymbol {\phi }}}$

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 w: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.