Answer
$\mathop \smallint \limits_{\theta = - \pi /2}^{\pi /2} \mathop \smallint \limits_{r = 0}^1 \mathop \smallint \limits_{z = 0}^4 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 $\mathop \smallint \limits_{x = 0}^1 \mathop \smallint \limits_{y = - \sqrt {1 - {x^2}} }^{\sqrt {1 - {x^2}} } \mathop \smallint \limits_{z = 0}^4 f\left( {x,y,z} \right){\rm{d}}z{\rm{d}}y{\rm{d}}x$.
From the order of the integration, we obtain the region description:
${\cal W} = \left\{ {\left( {x,y,z} \right)|0 \le x \le 1, - \sqrt {1 - {x^2}} \le y \le \sqrt {1 - {x^2}} ,0 \le z \le 4} \right\}$
Notice that this is a $z$-simple region. Since $0 \le x \le 1$, the projection of ${\cal W}$ onto the $xy$-plane is a half of the unit disk ${x^2} + {y^2} \le 1$, located at the right-hand side of the $y$-axis (namely, the first and the fourth quadrant). Thus, the description of ${\cal W}$ in cylindrical coordinates:
${\cal W} = \left\{ {\left( {r,\theta ,z} \right)|0 \le r \le 1, - \frac{\pi }{2} \le \theta \le \frac{\pi }{2},0 \le z \le 4} \right\}$
So, the triple integral in cylindrical coordinates:
$\mathop \smallint \limits_{\theta = - \pi /2}^{\pi /2} \mathop \smallint \limits_{r = 0}^1 \mathop \smallint \limits_{z = 0}^4 f\left( {r\cos \theta ,r\sin \theta ,z} \right)r{\rm{d}}z{\rm{d}}r{\rm{d}}\theta $
