Answer
$\mathop \smallint \limits_{\theta = 0}^{\pi /2} \mathop \smallint \limits_{r = 0}^{2\cos \theta } \mathop \smallint \limits_{z = 0}^r f\left( {r\cos \theta ,r\sin \theta ,z} \right)r{\rm{d}}z{\rm{d}}r{\rm{d}}\theta $
Work Step by Step
We have $\mathop \smallint \limits_{x = 0}^2 \mathop \smallint \limits_{y = 0}^{\sqrt {2x - {x^2}} } \mathop \smallint \limits_{z = 0}^{\sqrt {{x^2} + {y^2}} } f\left( {x,y,z} \right){\rm{d}}z{\rm{d}}y{\rm{d}}x$.
From the order of the integration, we obtain the region description:
${\cal W} = \left\{ {\left( {x,y,z} \right)|0 \le x \le 2,0 \le y \le \sqrt {2x - {x^2}} ,0 \le z \le \sqrt {{x^2} + {y^2}} } \right\}$
Notice that this is a $z$-simple region such that the projection of ${\cal W}$ onto the $xy$-plane is the upper half of the unit disk ${\left( {x - 1} \right)^2} + {y^2} \le 1$ centered at $\left( {1,0} \right)$. We verify this statement by writing
$y = \sqrt {2x - {x^2}} $
${x^2} + {y^2} - 2x = 0$
${\left( {x - 1} \right)^2} + {y^2} = 1$
The last equation is a unit circle centered at $\left( {1,0} \right)$. Referring to Example 2, this circle has equation in polar coordinates $r = 2\cos \theta $. Thus, $r$ ranges from $r=0$ to $r = 2\cos \theta $. Since $x \ge 0$ and $y \ge 0$, it is located in the first quadrant, so $\theta$ ranges from $\theta = 0$ to $\theta = \frac{\pi }{2}$.
Since in cylindrical coordinates, $r = \sqrt {{x^2} + {y^2}} $, so the description of ${\cal W}$:
${\cal W} = \left\{ {\left( {r,\theta ,z} \right)|0 \le r \le 2\cos \theta ,0 \le \theta \le \frac{\pi }{2},0 \le z \le r} \right\}$
So, the triple integral in cylindrical coordinates:
$\mathop \smallint \limits_{\theta = 0}^{\pi /2} \mathop \smallint \limits_{r = 0}^{2\cos \theta } \mathop \smallint \limits_{z = 0}^r f\left( {r\cos \theta ,r\sin \theta ,z} \right)r{\rm{d}}z{\rm{d}}r{\rm{d}}\theta $
