(to follow up on where this was left in August 2012):

Here is a graph of the region in question, made just incomplete enough to allow the interior to be viewed. The surface $ \ x = y^2 \ $ is a parabolic cylinder extending indefinitely in the $ \ z-$ directions; the surface $ \ z = x \ $ is an oblique plane; and $ \ z = 0 \ $ is, of course, the $ \ xy-$ plane. The volume is then a sort of wedge with a tilted "roof" and a parabolic "wall".
The integrals for the integration orders $ \ dz \ dy \ dx \ $ , which is sort of the "natural" order many people would use, and $ \ dz \ dx \ dy \ $ are shown in the other posts. One should not be too quick to arbitrarily re-arrange the order of integration in multiple integrals for a variety of reasons, sometimes because of the geometric configuration of the integration region, sometimes because of the integrands one would be left to grapple with.
In this problem, the amount of symmetry of the region permits us to choose alternative orders of integration without producing any "crisis". For the order $ \ dx \ dy \ dz \ $ , we are able to make use of the fact that one of the boundary surfaces is $ \ z \ = \ x \ $ . Since we want to "herd" the integration toward working in the variable $ \ z \ $ , we can carry out the integration in $ \ x \ $ from $ \ 0 \ $ to $ \ z \ $ . For the integration in $ \ y \ $ , we can therefore also "replace" $ \ x \ $ by $ \ z \ $ to express the relevant portion of the parabolic surface as $ \ z \ = \ y^2 \ $ . The integration limits on $ \ z \ $ become $ \ 0 \ $ to $ \ 1 \ $ , as they were for $ \ x \ $ .
We can now write the "re-ordered" integration as
$ \int_0^1 \int_{-\sqrt{z}}^{\sqrt{z}} \int_0^z \ \ dx \ dy \ dz \ \ \ \text{or} \ \ \ 2 \ \int_0^1 \int_0^{\sqrt{z}} \int_0^z \ \ dx \ dy \ dz \ \ , $
by exploiting the symmetry of the region about the $ \ xz-$ plane. Since this new integral simply looks like a "relabeled" version of the first integral in the original post, using the order $ \ dz \ dy \ dx \ $ , this will plainly give the same result for the volume of $ \ \frac{4}{5} \ $ .