I need to calculate $\int\limits_Kz^2 dx\ dy\ dz$ with $$K=\{(x,y,z) \in \mathbb R^3 | x^2 + y^2 + z^2 \leq 4, \ \ x^2+y^2\geq 1\}.$$
We use cylindrical coordinates and see: $$r^2+z^2 \leq 4 \; \text{and} \; r^2\geq 1\Rightarrow -2\leq z \leq 2 \; \text{and} \; 1\leq r \leq \sqrt{4-z^2}.$$ So we get $$\begin{align}\int\limits_Kz^2\ dx\ dy\ dz&=\int_0^{2\pi}\int_{-2}^2\int\limits_1^{\sqrt{4-z^2}} z^2 r \cdot dr\ dz\ d\varphi\\ &=\int\limits_0^{2\pi}\int_{-2}^2\int_1^{\sqrt{4-z^2}} z^2 r \cdot dr\ dz\ d\varphi=2\pi\int\limits_{-2}^2z^2\frac{1}{2}[(4-z^2)-1 ]dz\\ &=\pi\int\limits_{-2}^2 3z^2-z^4 dz=\pi[z^3-\frac{1}{5}z^5]_{-2}^2\\ &=\pi[(2^3-2^5/5)-(-8-2^5/5)]=\pi(16-64/5)\end{align}$$
Apparently, and this might be wrong (since I noted it in a hurry), the result is $\frac{6\sqrt{3}}{5}\pi$. No idea where I should get a root from.
With $r^2\ge 1$, the $z$ limit is $z^2\le3$. The last step in the integration is $\pi[z^3-\frac{z^5}{5}]_{-\sqrt{3}}^{\sqrt{3}}=\frac{6\sqrt{3}}{5}\pi$.