I want to calculate a volume of the following region.
$2\sqrt{x^2+y^2}+x\leq x^2+y^2\leq 9-z^2$
First inequality can be transformed $2r+r\cos \theta \leq r^2$ and so $2+\cos \theta \leq r$. Thus $Volume=\int_{0}^{2\pi}\int_{2+\cos \theta}^{3}\left ( 2\int_{0}^{\sqrt{9-r^2}}dz \right )rdrd\theta =\int_{0}^{2\pi}\int_{2+\cos \theta}^{3}2r{\sqrt{9-r^2}}drd \theta=\int_{0}^{2\pi}\left [ -\frac23\left (9-r^2 \right )^\frac32 \right ]^3_{2+\cos\theta}d\theta=\int_{0}^{2\pi}\frac23\left ( 9-\left ( 2+cos\theta \right )^2 \right )^\frac32d\theta$.
I'm stuck in here. Please complete this calculation. Helpmehelpme