Components of Curl, Divergence

[Contingency]

Homework Statement

I'm trying to understand where the Cartesian components of the rotor and the divergence of a vector field derived.

I read that the divergence of a vector field is defined by:
\vec { \nabla } \cdot \vec { F } =\lim _{ V\rightarrow \left\{ P \right\} }{ \frac { 1 }{ \left| V \right| } \iint _{ \partial V }^{ }{ \vec { F } \cdot \vec { dA } } }

Also, that components of the rotor are defined by:
\vec { \nabla } \times \vec { F } \cdot \hat { u } =\lim _{ A(\Gamma )\rightarrow 0 }{ \frac { 1 }{ \left| A(\Gamma ) \right| } \int _{ \Gamma }^{ }{ \vec { F } \cdot \vec { dr } } }

I'm trying to understand where the standard "sum of partial derivatives", and the mnemonic determinant with a row of unit vectors come from. I don't see a correlation between the definitions and these simple representations. How are they derived from the definition?

Homework Equations

\vec { \nabla } \times \vec { F } \cdot \hat { u } =\lim _{ A(\Gamma )\rightarrow 0 }{ \frac { 1 }{ \left| A(\Gamma ) \right| } \int _{ \Gamma }^{ }{ \vec { F } \cdot \vec { dr } } }

\vec { \nabla } \cdot \vec { F } =\lim _{ V\rightarrow \left\{ P \right\} }{ \frac { 1 }{ \left| V \right| } \iint _{ \partial V }^{ }{ \vec { F } \cdot \vec { dA } } }

The Attempt at a Solution

No clue

 

[mfb]

You can choose a cube as volume (if the general limit exists, it is the same for a special type of volume), and get the coordinate representation in the limit.
In the same way, you can use a square as path.

 

[Mark44]

This is obviously not a precalculus question, so I am moving it to the appropriate section.

 

[Contingency]

My bad, didn't double check. Sorry

 

[Contingency]

You can choose a cube as volume (if the general limit exists, it is the same for a special type of volume), and get the coordinate representation in the limit.
In the same way, you can use a square as path.

Can you please be a bit more specific?
I divided my field into orthogonal component functions, and I'm trying to look at the circulation of the field over a square with sides parallel to the axis. In order to get my result i'd have to approximate the field as constant over each side.. Why can I do that?