Do Vector Calculus Cartesian coordinates identities with Div, Grad, Curl hold in cylindrical and spherical coordinates?

Question

These operators are written in different forms in Cartesian, cylindrical and spherical coordinates. For instance, in spherical coordinate system, one has

$$\nabla \cdot \overrightarrow{F}=\frac{1}{r^{2}}\frac{\partial }{\partial r} \left( r^{2}F_{r}\right) +\frac{1}{r\sin \theta }\frac{\partial }{\partial \theta }\left( \sin \theta \cdot F_{\theta }\right) +\frac{1}{r\sin \theta } \frac{\partial F_{\varphi }}{\partial \varphi }.$$

Question: Do identities such as

$$\nabla \cdot \left( \overrightarrow{A}\times \overrightarrow{B}\right) =% \overrightarrow{B}\cdot \nabla \times \overrightarrow{A}-\overrightarrow{A}% \cdot \nabla \times \overrightarrow{B}$$

hold in general when cylindrical and spherical coordinate systems are used or do they have to be adapted?

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Added: After having read the comments it occured to me that the invariance of these identities with regard to the coordinate system is a consequence of the definitions of the mentioned operators in terms of integrals, e.g.:

$$\nabla \cdot \overrightarrow{F}=\underset{V\rightarrow 0}{\lim }\frac{1}{V}% \underset{S}{\int \int }\overrightarrow{F}\cdot \overrightarrow{n}\;dA$$

where $V$ is the volume of a bounded closed region $T$, $S$ is the surface of $T$, and $\overrightarrow{n}$ the unit outer normal vector to $S$.

Answer 1

[This answer doesn't add much that is new, but it became too painful trying to post is as a comment.]

One doesn't need to talk about differential forms, or integral identities, in order to prove the validity of vector calculus formulas in spherical coordinates (although both those points of view are very nice!). If $\vec{F}(x,y,z)$ is a vector field expressed in Cartesian coordinates, say, and $\vec{F}(r,\theta,\varphi)$ is the same vector field but now expressed in spherical coordinates, then the formula for $\nabla$ in spherical coordinates is determined by the requirement that $\nabla\cdot \vec{F}(x,y,z)$ and $\nabla\cdot \vec{F}(r,\theta,\varphi)$ correspond to one another as vector fields when you change from Cartesian to shperical coordinates.

Similarly, if $\vec{A}(x,y,z)$ and $\vec{B}(x,y,z)$ are a pair of vector fields expressed in Cartesian coordinates, with $\vec{A}(r,\theta,\varphi)$ and $\vec{B}(r,\theta,\varphi)$ being the same vector fields but now expressed in spherical coordiantes, then $\vec{A}\times \vec{B}$ (expressed in Cartesian coordinates) will correspond to $\vec{A}\times \vec{B}$ (expressed in spherical coordinates). So the left-hand side of the identity to be checked, when computed in Cartesian coordinates, will match (under the change from spherical to Cartesian coordinates) with the same expression when computed in spherical coordiantes. Similarly for the right-hand side of the identity. Thus the identity is equally valid in Cartesian or spherical coordinates. (Again, the key point is that the formulas for grad, div, and curl under a coordinate change are defined so as to make the above argument valid.)

Answer 2

[I post this answer as suggested in a comment]

The Vector Calculus identities with Div, Grad, Curl are independent of the coordinate system.

A possible explanation is that they may be expressed in terms of integrals whose values do not depend on the particular coordinate system one uses. As an example, one has the following limit for the divergence of the vector field $\overrightarrow{F}$:

$$\nabla \cdot \overrightarrow{F}=\underset{V\rightarrow 0}{\lim }\frac{1}{V}\underset{S}{\int \int }\overrightarrow{F}\cdot \overrightarrow{n}\;dA$$

where $V$ is the volume of a bounded closed region $T$, $S$ is the surface of $T$, and $\overrightarrow{n}$ the unit outer normal vector to $S$.