The triple integral is bounded by
$$z=4-y^2$$ $$y=2x$$ $$z=0$$ $$x=0$$
I have to rewrite it so that the plane is on the $xz$ plane, and therefore I fix $y$.
First I notice that $z>=0$, and so $-2\leq y\leq 2x$
Now on the $xz$ plane, I have the following:
The red line is $x=0$, the yellow line is $z=0$, and the blue line is $z=4-y^2$, but since $y=0$, we just have $z=4$
I see no "bounded area". What is my mistake here? I know that this integral is supposed to converge. I also took into account all the functions.
Your bounded region doesn’t intersect the $xz$-plane except along the line $x=0$. You want the region with $2x<y<\sqrt{4-z}$. You can see nice cross-sections of this region in the $xy$-plane and in the $yz$-plane. You can also get non-trivial pictures by setting $y=k$ with $0<k<2$.
Edit
To project this region onto the $xz$-plane, it helps a lot to visualize the region in $3$ dimensions first. At any rate, the trick is to look at the intersection curves of the defining surfaces, and then we project those curves by eliminating $y$ algebraically. Thus, the surface $y=2x$ and the surface $z=4-y^2$ project onto the curve $z=4-(2x)^2$, or $z=4-4x^2$. That curve, together with $z=0$ and $x=0$ define a region in the $xz$-plane, which you can use to set up an integral with order of integration $dy\,dz\,dx$. Namely:
$$\int_0^1\int_0^{4-4x^2}\int_{2x}^{\sqrt{4-z}}f(x,y,z)\,dy\,dz\,dx$$
Does that help?