Need a lot of help with vector fields/vector operators

[Blkmage]

Homework Statement

http://img818.imageshack.us/f/screenshot20110423at733.png/
http://img856.imageshack.us/f/screenshot20110423at733.png/

If it'll help you guys help me understand this, here are the solutions:
http://img828.imageshack.us/f/screenshot20110423at752.png/

Homework Equations

\text{curl}\bold{F} = \nabla \times \bold{F}
\text{div}(\bold{F}) = \nabla \cdot \bold{F}

The Attempt at a Solution

My problem is that I don't understand what is meant by a "fixed, but arbitrary function" or a "fixed, but arbitrary vector field." Is a fixed function one that is constant? ie g(x,y,z) = 2 Is a fixed vector fixed a constant one, like \bold{F}(x,y,z) = 2\hat{i} + 3\hat{j} - 5\hat{k}?

My problem is that I'm not really understanding the nature of these vector operators. I have the solution and it says:

\text{curl} \nabla g is a constant vector whereas
\text{curl} \bold{F} is a vector field

How is it possible that they are not both vector fields. Same with this:

\text{div} (\text{curl} \bold{F}) is a constant scalar whereas
\text{div} (\bold{v} \times \bold{F}) is a scalar function...

How are these not both scalar functions?

 

[HallsofIvy]

=No, a "fixed" function is just a "given" one- for the purposes of this problem you are talking about a specific ("fixed") function.

For g a scalar function, you can say more than just that "curl \nabla g vis a constant function"- it is a very specific constant!

If g(x,y,z) is a real valued function, then
\nabla g= \frac{\partial g}{\partial x}\vec{i}+ \frac{\partial g}{\partial y}\vec{j}+ \frac{\partial g}{\partial z}\vec{k}.

And so
curl \nabla g= \nabla\times\nabla g= \left|\begin{array}{ccc}\vec{i} & \vec{j} & \vec{k} \\ \frac{\partial }{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ \frac{\partial g}{\partial x} & \frac{\partial g}{\partial y} & \frac{\partial g}{\partial z}\end{array}\right|

Now what is that?
(Assuming that g has continuous second derivatives.)

 

[Blkmage]

OH. I completely forgot my rules. The curl of a gradient is a the zero vector and the divergence of the curl is the scalar 0. Thanks.

Also, for the very last integral, does that not not exist because d\bold{s} denotes a line integral, but since there are two integral symbols, it doesn't mean anything? Or is it because there is not dot product between \nabla g and the differential d\bold{s}?

 

[HallsofIvy]

It should be dS, the differential of surface area, not ds, the differential or arc length.