This volume integration looks rather daunting, but proves to be one of those that works out rather nicely if one perseveres. We should locate the volume we will be working with; here, a couple of three-dimensional graphs show that this volume is in the first octant, well away from the origin (which will be important to know):
These views down the $ \ y-$ (left) and $ \ z-$ axes (right) show that the three pairs of parabolic cylinders do not enclose a volume near the coordinate axes. Instead, we find the enclosed region surrounded by all six surfaces lies in the general vicinity of $ \ ( \ \frac{3}{2} , \frac{3}{2} , \frac{3}{2} \ ) \ $ :
A graph with an incomplete "plot" of the surface $ \ z^2 \ = \ 2y \ $ permits us to see the interior of the region. By solving for intersection points of the surfaces by forming various "chains" of three surface equations among the six involved, we find the coordinates of the four vertices of the region visible here (those on the surface $ \ x^2 \ = \ z \ $ ) as
$$ \mathbf{A} : \ (1,1,1) \ \ , \ \ \mathbf{B} : \ (2^{4/7},2^{2/7},2^{1/7}) \ \ , \ \ \mathbf{C} : \ (2^{5/7},2^{6/7},2^{3/7}) \ \ , \ \ \mathbf{D} : \ (2^{1/7},2^{4/7},2^{2/7}) \ \ . $$
If we adjust the graph "window" to see the intersections with the surface $ \ z^2 \ = \ 2y \ $ , we find the "upper" vertices to be
$$ \mathbf{E} : \ (2^{6/7},2^{3/7},2^{5/7}) \ \ , \ \ \mathbf{F} : \ (2^{2/7},2^{1/7},2^{4/7}) \ \ , \ \ \mathbf{G} : \ (2^{3/7},2^{5/7},2^{6/7}) \ \ , \ \ \mathbf{H} : \ (2,2,2) \ \ . $$
The specific coordinates will be of some use in our making the volume calculation; what they indicate at present is that there is no common projection of opposite "faces" of this volume onto any of the coordinate planes. The volume is not a region of the sorts some call Types I, II, or III. This means that the integration can only be made convenient by a coordinate transformation.
We need to choose a set of transformed variables that will take on constant values at the edges of the "faces" of the enclosed volume, so that we can have numerical limits of integration in the calculation. Since there is a coefficient in each of the surface equations, we can plan our transformations around those numbers. If we choose the variables
$$ u \ = \ \frac{x^2}{z} \ \ , \ \ v \ = \ \frac{y^2}{x} \ \ , \ \ w \ = \ \frac{z^2}{y} \ \ , $$
we find (after some computation) that all of the "edges" of the region lie on the lines $ \ u \ = \ 1 \ , $ $ u \ = \ 2 \ , \ v \ = \ 1 \ , \ v \ = \ 2 \ , \ w \ = \ 1 \ , \ \ \text{and} \ \ \ w \ = \ 2 \ $ . These will serve as the integration limits for the three transformed variables; these choices present no difficulties as none of the coordinate variables can be zero in this region. [This is along the lines of what ELEC was pursuing.]
To complete the set-up for the volume integral, we will need to determine the Jacobian determinant $ \ \mathfrak{J} \ $ for this transformation. As is often the case, it is easier to find the determinant of the inverse transformation first:
$$ \mathfrak{J}^{-1} \ = \ \left| \ \begin{array}{ccc}
\frac{\partial u}{\partial x} & \frac{\partial u}{\partial y} & \frac{\partial u}{\partial z} \\ \frac{\partial v}{\partial x} & \frac{\partial v}{\partial y} & \frac{\partial v}{\partial z} \\ \frac{\partial w}{\partial x} & \frac{\partial w}{\partial y} & \frac{\partial w}{\partial z} \end{array} \ \right| \ \ = \ \ \left| \ \begin{array}{ccc}
\frac{2x}{z} & 0 & -\frac{x^2}{z^2} \\ -\frac{y^2}{x^2} & \frac{2y}{x} & 0 \\ 0 & -\frac{z^2}{y^2} & \frac{2z}{y} \end{array} \ \right| $$
$$ = \ \left( \frac{2x}{z} \right) \ \left( \frac{2y}{x} \right) \ \left( \frac{2z}{y} \right) \ \ + \ \ \left( -\frac{x^2}{z^2} \right) \ \left( -\frac{y^2}{x^2} \right) \ \left( -\frac{z^2}{y^2} \right) $$
$$ = \ \frac{8xyz}{xyz} \ \ - \ \ \frac{x^2 \ y^2 \ z^2}{x^2 \ y^2 \ z^2} \ = \ 7 \ \ . $$
Almost incredibly, the determinant is a constant value! The Jacobian for the transformation itself is $ \ \mathfrak{J} \ = \ \frac{1}{\mathfrak{J}^{-1}} \ = \ \frac{1}{7} \ $ , giving us the volume integral
$$ \iiint_V \ | \ \mathfrak{J} \ | \ \ du \ dv \ dw \ \ = \ \ \frac{1}{7} \ \int_1^2 \int_1^2 \int_1^2 \ \ du \ dv \ dw \ = \ \frac{1}{7} \ \ . $$
We can check to see if this result is plausible. The relevant fractional-powers of 2 are approximately
$$ 2^{1/7} \ \approx \ 1.10 \ \ , \ \ 2^{2/7} \ \approx \ 1.22 \ \ , \ \ 2^{3/7} \ \approx \ 1.35 \ , \ $$
$$ 2^{4/7} \ \approx \ 1.49 \ \ , \ \ 2^{5/7} \ \approx \ 1.64 \ \ , \ \ 2^{6/7} \ \approx \ 1.81 \ \ . $$
The distances from vertex $ \ \mathbf{H} \ $ to its neighbors $ \ \mathbf{E} \ , \ \mathbf{G} \ , \ \ \text{and} \ \ \mathbf{C} \ $ are all $ \ \sqrt{ \ 0.65^2 \ + \ 0.36^2 \ + \ 0.19^2 } \ \approx \ 0.77 $ . If we crudely approximate the volume of the region as a cube, then it is certainly smaller than $ \ 0.77^3 \ \approx \ 0.45 \ . $
But we can manage a better estimation than that. If we work out the vectors from vertex $ \ \mathbf{A} \ $ to each of its neighbors and do the same for $ \ \mathbf{H} \ $ , in order to approximate the somewhat slab-like volume as a parallelopiped, we can use vector triple products to bracket the size of our enclosed volume:
$$ \left| \ \ \left| \ \begin{array}{ccc}
0.49 & 0.22 & 0.10 \\ 0.22 & 0.10 & 0.49 \\ 0.10 & 0.49 & 0.22 \end{array} \ \right| \ \ \right| \ < \ V \ < \ \left| \ \ \left| \ \begin{array}{ccc}
0.65 & 0.36 & 0.19 \\ 0.19 & 0.65 & 0.36 \\ 0.36 & 0.19 & 0.65 \end{array} \ \right| \ \ \right| $$
$$ \Rightarrow \ \ 0.097 \ < \ V \ < 0.195 \ \ . $$
Our result of $ \ V \ = \ \frac{1}{7} \ $ is close to the midpoint of this interval.
