Calculate the volume of the solid bounded by $y=2-z^{2}$, $z=y^{2}$, $x+y=4$ and $x=0$.
I tried to solve it and I got: $$\int_{0}^{2}\int_{0}^{2}\int_{y^{2}}^{\sqrt{2-y}}\ dzdydx+\int_{2}^{4}\int_{0}^{4-x}\int_{y^{2}}^{\sqrt{2-y}}\ dzdydx=\frac{64\sqrt{2}}{15}-\frac{20}{3}\approx -0.6327$$
Draw the $(y,z)$-plane, and imagine the positive $x$-axis coming vertically out of the paper. The equations $y=2-z^2$ and $z=y^2$ describe two parabolas intersecting at $(1,1)$ and $(\eta,\eta^2)$, where $\eta\approx-1.35321$ is the solution of a third degree equation. The two parabolas generate two parabolic cylinders with generatrices parallel to the $x$-axis. The planes $x=0$ and $x=4-y$ cut these two cylinders. Hopefully a finite piece $B$ of space is created thereby.
As I see it there is one such $B$: In the $(y,z)$-plane there is a lens shaped finite domain bounded by the two parabolas. This domain creates a cylinder volume, whereby this volume is bounded below by the "floor" $x=0$ and above by $x=4-y$. The volume of $B$ then computes as follows: $${\rm vol}(B)=\int_\eta^1\int_{y^2}^{\sqrt{2-y}}\int_0^{4-y}1\>dx\>dz\>dy\ .$$ The numerical end result is $9.46409$.