Doubt about triple integrals

Question

I'm doing a triple integral where the function to be integrated is $z$ and the domain is: $ 1 < x^2 + y^2 + z^2 < 4 $ and $ z < 0 $. So, I thought about cylindrical coordinates (although I know spherical ones would do too). I got this:
$ -\sqrt(1-r^2) < z < -\sqrt(4-r^2)$
$0< \theta <2\pi$
$ 1<r<2 $

The final result (calculated by Wolfram Alpha) is $-9\pi/2$. Solutions say that the final result is $-15\pi/4$

What am I doing wrong?

Answer 1

Spherical coordinates is the way to go here. You are wrong when you claim that $1\leqslant r\leqslant2$. In cylindrical coordinatex, $r$ is the distance of the point to the $z$-axis. Therefore, $0\leqslant r\leqslant2$ and then the limits for $z$ are not as simple as you wrote them.

Answer 2

Indeed David K, -$\sqrt(1-r^2)$, with $ r > 1$ does not make when when we are talking about real calculus. Thank you José for your advice, solving this problem using cylindrical coordinates is way harder than using spherical ones.

I managed to solve it properly using spherical coordinates.