I don't think you will find such a list of possible ways to reduce a surface integral into a nicer form. There are just too many potential integrals. Mostly you will have to work off of experience.
There's always nice reductions that occur if the vector field you are integrating over has symmetries with respect to the surface. For example, if the vector field is radially directed, and the surface is a sphere, then the vector field will always be parallel (or anti-parallel) with the dS vector. In this case, one can remove the dot product. Alternatively, if the vector field is always perpendicular with the surface, you know the surface integral is 0. If the surface is a closed surface, then you can probably use the divergence theorem to turn it into a volume integral, or if the surface is not a closed surface, perhaps use the divergence theorem to turn it into a line integral.
All of this basically comes with experience and practicing problems. Even for regular integrals, you just need some experience in how to evaluate them.
Actually the vast majority of integrals will be impossible to do analytically using elementary methods, but the integrals you see from courses or lectures should be able to be evaluated.
As far as your attached thumbnail goes, there's no need for vectors in that problem, so you wouldn't be dealing with a dS vector, you would be dealing with a dA infinitesimal area.
In this case, the volume would be 3*A where A is the area of the circle. Since the density doesn't depend on z, you can basically turn the volume integral into a surface integral over A multiplied by 3.
If you are asking why is dA=rdzdθ, then this is the area element in polar coordinates.
For those, you can certainly look up, for example:
http://en.wikipedia.org/wiki/Cylindrical_coordinates
http://en.wikipedia.org/wiki/Spherical_coordinates