What I wanted to know is if its the same this equation than the other:
This one:

And this one:

I think I'm getting it now... The second one is to derive the function under the curve (or between the curve and the X axis) around the Y axis, and the first derives between the curve and the Y axis around the Y axis. Is that right?

Yes, you have it. If you draw the corresponding dy (in the first one) and dx (in the second) rectangle and revolve it about the given axis, you will see the first one generates a disk of thickness dy and the second generates a cylinder or shell of thickness dx. The integrands are just the standard formulas.

Disk: [itex]dV = \pi r^2 dy[/itex] where r = R(y)

Shell: [itex]dV = 2\pi r h dx [/itex] where r = x and h = f(x).

You seem to be getting the basic Idea of it. I find for any volume question, it is always useful to draw a diagram including the bounded area being revolved to produce the solid, a typical approximating rectangle and either a disk/washer or cylindrical shell. The diagram will help you determine which method is best for the particular question. Additionally, when you are trying to understand the difference between cylindrical shells and revolution disk method, accompanying diagrams really help.

Here is a generic example of a solid by revolution disk method:
[PLAIN]http://img85.imageshack.us/img85/944/diski.png [Broken]

Notice how the rectangle is perpendicular to the axis of rotation.

Here is a generic example of a solid by cylindrical shells method:
[PLAIN]http://img341.imageshack.us/img341/1791/cylindrical.png [Broken]

Notice how the approximating rectangle is now parallel to the axis of rotation.

Thanks to both of you. Sorry I haven't posted anything till now. I've already had the exam (and I've approved), and I went on a travel to Buenos Aires, so I've been away for a few days.

Well, about this topic, on the exam I had one error on the solids of revolution. When it asked to revolve an integral round the line y=6, I've revolved it on the line x=6, it was an misinterpretation due to the reading, and if I would used a diagram as you said I wouldn't commit that mistake, cause I thought of the line x=6 due to a fast read, and as I went direct to the equations I didn't realize of it, cause first it asked to make the solid with the integral of the equation around the x axis, and then around the line y=6, and when I road "y" I thought of a line parallel to the y axis and there was the mistake. A diagram would cleared what the exercise was asking.

I had two more mistakes, one was with the definite integral of an equation involving absolute value, I didn't consider the two curves on [tex]f(x)=x|x|[/tex]. And the other was setting the equations (didn't asked to solve, just setting it) for the perimeter given by two curves and a line. In the last case, when I set the equations, I didn't realize that in one case I had a denominator involving an x on the derivative on the equation that gives the perimeter. One of the curves, which was [tex]y=x^{1/2}[/tex], didn't realize that I couldn't consider the derivative on the point 0, needed to set the perimeter of the definite integral, which goes from the point zero to an other number. I think that I should redefined the equation of the derivative on that point.