First find the region $R_1$ consisting of all points $\langle x,y,z\rangle$ such that $x\ge 0$, $y\ge 0$, and $\sqrt x\le z\le 1$. This has the same cross-section perpendicular to the $y$-axis for each $y\ge 0$; that cross-section projects to the curvilinear triangle in the $xz$-plane bounded by $z=\sqrt x$, $x=0$, and $z=1$, so it lies above the curve $z=\sqrt x$.
Now revolve the parabola $y=x^2$ about the $y$-axis to produce a paraboloid of revolution; its cross-sections perpendicular to the $y$-axis have the form $x^2+z^2=y$, so $z=\sqrt{y-x^2}$ describes the part of this paraboloid lying on and above the $xy$-plane. Let $R_2$ be the region consisting of all points $\langle x,y,z\rangle$ such that $x\ge 0$, $z\ge 0$, and $0\le y\le x^2+y^2$; $R_2$ is the set of points in the first octant between the paraboloid and the $xz$-plane.
Finally, let $R=R_1\cap R_2$; then $R$ is a finite volume bounded by $x=0$, $y=0$, $z=1$, and $z=\sqrt{y-x^2}$ and actually having each of these four surfaces as part of its boundary.