Find the volume: $$\{(x,y,z)\mid x^2+y^2 \leq (z-1)^2 \leq 4-\frac{x^2}{2} - 2y^2, z\geq 1 \}$$
I've got the intersection of the following two basically: \begin{align} 1. & & & (z-1)^2 \leq 4-\frac{x^2}{2} - 2y^2, z\geq 1 \\ 2. & & & (z-1)^2 \geq x^2+y^2 , z\geq 1 \end{align} with 2. you can already tell it implies the "outer" part of a cone.
Simplifying $1.$ I get a certain inside area of a moved ellipsoid because $$1. \Leftrightarrow \frac{x^2}{(\sqrt{8})^2} + \frac{y^2}{(\sqrt{2})^2} + \frac{z^2}{2^2} \leq 1$$
I think it would be best to use some sort of polar or cilindrical or spherical coordinates substitution, but I don't know which or how exactly, mainly because this is an intersection of two areas. I need some unifying parameterization, and would also like to know the bounderies of the integral with these substitutions. All help is much appreciated.