Derive the curl in general coordinates. Many mathematical physics texts inform the orthogonal curvilinar coordinates system and differential operators.
But, they don't use the precise mathematical method for the derivation of the curl and divergence in orthogonal curvilinear coordinats. They only give the intuative pictures.
Example: Arfken-texts.
I want to find the precise derivation of these forms.
 A: Let a position be $x = (x^1, x^2, x^3)$.  Any vector field $F$ can be written as
$$F(x) = F^1 \frac{\partial x}{\partial x^1} + F^2 \frac{\partial x}{\partial x^2} + F^3 \frac{\partial x}{\partial x^3}$$
Each $\partial x/\partial x^i$ is a basis vector.
The vector derivative operator $\nabla$ can always be written as
$$\nabla = (\nabla x^1) \frac{\partial}{\partial x^1} + (\nabla x^2) \frac{\partial}{\partial x^2} + (\nabla x^3) \frac{\partial}{\partial x^3}$$
This may seem circular; in practice, the basis covectors $\nabla x^i$ can be found by computing the reciprocal (or dual) basis to the basis vectors.
The divergence and curl then follow using the usual vector algebra rules for dot and cross products.  The thing that may be a "gotcha" here is that $F^i$ and $\partial x/\partial x^i$ both may depend on position in a general coordinate system, and as such, you must use the product rule when differentiating.
Computing this relations for a given coordinate system can always be done by writing the $\partial x/\partial x^i$ and other vectors in terms of some fiducial basis (usually the standard basis, which makes the math simplest).  However, it's also possible to use some geometric intuition.  For instance, if we're in a cylindrical coordinate system, then $\partial x/\partial x^2 = r \hat \phi$, and $\partial(r \hat \phi)/\partial x^2$ can be seen as $-x$ rather quickly, without having to break down into another basis.
Edit: also, when the coordinates are orthogonal, you should be able to conclude that $(\nabla x^i) \parallel \partial x/\partial x^i$ and similarly that $(\nabla x^i) \perp \partial x/\partial x^j$ for $i \neq j$.  This simplifies the dot and cross products considerably, of course.  In this case, the basis vectors and covectors differ from each other only by some set of scale factors, which is how it all tends to be presented.  One choice is a symmetric set of scale factors, so that $\partial x/\partial x^i = h_i \hat e_i$ and $\nabla x^i = \frac{1}{h_i} \hat e_i$.
