<s>[INST] How to calculate the volume of the solid described $\frac{x^2}{4}+ \frac{y^2}{4}+z^2 \le 1$ and $z \ge \sqrt{x^2+y^2}-2$ How to calculate the volume of the solid described $\frac{x^2}{4}+ \frac{y^2}{4}+z^2 \le 1$ and $z \ge  \sqrt{x^2+y^2}-2$?
I try
$x=2r \cos \phi$,
$y=2r \sin \phi$,
$z=z$, but but probably not the way to go
 [/INST] Letting $r=\sqrt{x^2+y^2}$ gives $r^2+4z^2=4$ in the 1st equation and $z=r-2$ in the 2nd equation, so
substituting gives $r^2+4(r-2)^2=4$ and so $5r^2-16r+12=0$, 
which gives $(5r-6)(r-2)=0$ so $r=\frac{6}{5}$ or $r=2$.
$\;\;$Using $z=r-2$, we have that 
if $0\le r\le\frac{6}{5}, \;\;-\frac{1}{2}\sqrt{4-r^2}\le z\le \frac{1}{2}\sqrt{4-r^2}$  $\;\;\;$ and $\;\;\;$ if $\frac{6}{5}\le r\le2, \;\;r-2\le z\le \frac{1}{2}\sqrt{4-r^2}$.
Therefore $\displaystyle V=\int_0^{2\pi}\int_0^{\frac{6}{5}}\sqrt{4-r^2}r\;drd\theta+\int_0^{2\pi}\int_{\frac{6}{5}}^2\big(\frac{1}{2}\sqrt{4-r^2}-(r-2)\big)r\;drd\theta$.
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