We see that the inferior limit of Z is 0. But if I plug X=0 in your integration bounds we will end up with the volume of the cylinder which has the same circle (base) as the cone and the same height.
Because what you did was taking theta from 0 to 2pi with r from 0 to 1 (so you calculated the area of the circle) and then if you do z from 0 to 2 it is the same as if you multiplied the area of the circle by the height (2) (since all 3 parts of your triple integral are independent) so it is really the volume of the cylindre we find (the one which has the same base and height as the cone).
So we must use something else for the bounds.
Theta and r are fine since they help us find the area of the base which is the same in cylinder and cone. For z :
The line representing the edge of the cone does not depend on theta. So for any theta we will have the line z=2-2r. So z goes from 0 to the line 2-2r. We can understand that z does not go from 0 to 2 for any r since z is stopped by the line 2-2r. For example : when r=0, z goes until 2-2*0 = 2 (it is the top of the cone) and when r=1 (i.e. at the edge of the circle) z=2-2*1=0 (we can see that because the edge of the circle corresponds with the edge of the upper part of the cone).
If that is ok, then your intergal is as shown in the figure. You can verify the result since you know V(cone) = Area(base)*Height /3 = pi r^2 * Height /3 = pi *2/3.
Ok. I hope it helps you. (the integral drawing is done by OpenOffice.org 2.0 Math)