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

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?

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 $S$.

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Absolutely. The operations of div, grad, and curl are independent of the coordinate system, and any identity that one of them satisfies will continue to hold no matter how you express these operations in local coordinates. – Jim Belk Oct 12 '10 at 15:36
If they didn't, there would be no reason to call them the same thing. – Qiaochu Yuan Oct 12 '10 at 15:52
This is no different than the fact that a vector such as a particle velocity is defined without reference to a coordinate system, and the dot product with another vector is independent of coordinates. The coordinate expression will change with system. – Ross Millikan Oct 12 '10 at 16:42
As everyone tolds you, those identities are independent of the coordinate system, but the proof I know is somewhat long: you have to change the language of vector fields and grad, curl, div for the one of differential forms and its differential, using the isomorphism between an Euclidean vector space and its dual. With differential forms, the independence of coordinate system follows in one go from the formula relating the differential and the pull-bakc of forms: $d(f^*w) = f^*(dw)$. – a.r. Oct 12 '10 at 17:24
@muad: I've done it. – Américo Tavares Oct 12 '10 at 21:15

[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.)

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+1: I agree with your key point. – Américo Tavares Oct 12 '10 at 20:11

[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 $S$.

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