Question: I have this integral:

$$\iiint \frac{z}{\sqrt{x^2 + y^2}} \,dx \,dy \,dz$$

And I cant really get the boundaries from this shape:
V: $1 \leqslant x^2 + y^2 \leqslant 4$ and $0 \leqslant z \leqslant \sqrt{x^2+y^2}$

Answer: 
Set
\begin{align}
 & x=r \cos \theta \\ 
 & y=r \sin \theta \\ 
\end{align}
$$dxdy=\left| \frac{\partial (x,y)}{\partial (r ,\theta )} \right|dr\,d\theta={{r}\, dr\,d\theta }$$
we have
$$\iiint\_{V} \frac{z}{\sqrt{x^2 + y^2}} \,dx \,dy \,dz=\int\_{0}^{2\pi}\int\_{1}^{2}\int\_{0}^{r}r\frac{z}{r}dzdrd\theta$$
Thanks for your help Behrouz ! I was thinking about the polar method, but wasn't sure.

Still, after I do the drawing for $\sqrt{1-x^2} <= y <= \sqrt{4-x^2}$ I get two half circles, with radius 1 and 2, but isn't the angle supposed to be from 0 to PI? Because 2pi would be the whole circle.

Also why is the dz integral like $\frac{rz}r$ ? I figured out the boundaries.

Thanks !