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
 A: Substitute $r^2 = x^2 + y^2$ and make $r^2$ the subject.
The inequalities are then
$4(1 - z^2) \geq r^2$ and
$(z+2)^2 \geq r^2$.
Now we compare the LHS of the two inequalities, to determine which gives a stricter upper bound on $r^2$.
$$(z+2)^2 - 4(1-z^2) = 5z^2 +4z$$
which has roots $z=-\frac45$ and $z=0$.
Thus the second inequality's LHS is smaller than the first only between these two values; therefore the second is stricter between these values and the first is stricter everywhere else.
You also want to ensure that your integral includes only values of $z$ for which the first inequality is possible; that is, for which its LHS is nonnegative. So $|z| \leq 1$.
Now the appropriate integral is $\pi(\int_{-1}^{-\frac45}4(1-z^2)dz + \int_{-\frac45}^0(z+2)^2dz + \int_0^14(1-z^2)dz)$.
A: 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 $\displaystyle V=\int_0^{2\pi}\int_0^{\frac{6}{5}}\sqrt{4-r^2}r\;drd\theta+\int_0^{2\pi}\int_{\frac{6}{5}}^2\big(\frac{1}{2}\sqrt{4-r^2}-(r-2)\big)r\;drd\theta$.
