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.