A curvilinear system of coordinates (u, v, w) is related to Cartesian coordinates (x, y, z) by $$x=\frac12(u^2-v^2), y=uv, z=w$$ $$0\le u \le1, 0\le v\le1, 0\le w\le1$$
So I find the unit vectors and the scalar factors to find the volume of the region. $$ h_u=\sqrt{u^2+v^2}, h_v=\sqrt{u^2+v^2}, h_w=1 $$ $$ dV=h_uh_vh_wd_ud_vd_w = (u^2+v^2)d_ud_vd_w $$
Here is my triple integral: $$ \iiint{dV} = \int_0^1\int_0^1\int_0^1{u^2+v^2}d_ud_vd_w =\frac23$$
But when I tried to sketch the region on the xyz plane, it's a cube (well, a rectangle in 3D, what you guys call it?) with a volume of $\frac12$. Is my sketch wrong or it's my calculation or both? Thanks!
The shape would only appear to be a prism if you plotted it treating $u$, $v$, and $w$ as the orthogonal axes. If you plot it where $x$, $y$, $z$ are on the orthogonal axes, it is something else. Fix $u=1$, and let $v$ run from $0$ to $1$. As this happens, $x=\frac12\left(1-v^2\right)$ and $y=v$. This traces out one of four boundary curves for the shape in the $xy$-plane, and it is not a straight line. Repeat this fixing $u=0$, then $v=0$, then $v=1$.