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

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 $\pi(\int_{-1}^{-\frac45}4(1-z^2)dz + \int_{-\frac45}^0(z+2)^2dz + \int_0^14(1-z^2)dz)$.
