Answer
$\mathop \smallint \limits_0^1 \mathop \smallint \limits_0^{\sqrt {1 - {x^2}} } \mathop \smallint \limits_0^{\sqrt {{x^2} + {y^2}} } z{\rm{d}}z{\rm{d}}y{\rm{d}}x = \mathop \smallint \limits_{\theta = 0}^{\pi /2} \mathop \smallint \limits_{r = 0}^1 \mathop \smallint \limits_{z = 0}^r rz{\rm{d}}z{\rm{d}}r{\rm{d}}\theta = \frac{\pi }{{16}}$
Work Step by Step
From the triple integral $\mathop \smallint \limits_0^1 \mathop \smallint \limits_0^{\sqrt {1 - {x^2}} } \mathop \smallint \limits_0^{\sqrt {{x^2} + {y^2}} } z{\rm{d}}z{\rm{d}}y{\rm{d}}x$, we obtain the domain of integration:
$0 \le x \le 1$, ${\ \ \ }$ $0 \le y \le \sqrt {1 - {x^2}} $, ${\ \ \ }$ $0 \le z \le \sqrt {{x^2} + {y^2}} $
The first two inequalities imply that the projection of the solid region onto the $xy$-plane is a part of the unit disk: ${x^2} + {y^2} \le 1$ in the first quadrant. The third inequality implies that the region is bounded below by the plane $z=0$ and bounded above by the surface $z = \sqrt {{x^2} + {y^2}} $.
Let ${\cal W}$ denote the region. Using $r = \sqrt {{x^2} + {y^2}} $, we describe ${\cal W}$ in cylindrical coordinates:
${\cal W} = \left\{ {\left( {r,\theta ,z} \right)|0 \le \theta \le \frac{\pi }{2},0 \le r \le 1,0 \le z \le r} \right\}$
Evaluate $\mathop \smallint \limits_0^1 \mathop \smallint \limits_0^{\sqrt {1 - {x^2}} } \mathop \smallint \limits_0^{\sqrt {{x^2} + {y^2}} } z{\rm{d}}z{\rm{d}}y{\rm{d}}x$ in cylindrical coordinates:
$\mathop \smallint \limits_{\theta = 0}^{\pi /2} \mathop \smallint \limits_{r = 0}^1 \mathop \smallint \limits_{z = 0}^r rz{\rm{d}}z{\rm{d}}r{\rm{d}}\theta $
$ = \frac{1}{2}\mathop \smallint \limits_{\theta = 0}^{\pi /2} \mathop \smallint \limits_{r = 0}^1 r\left( {{z^2}|_0^r} \right){\rm{d}}r{\rm{d}}\theta $
$ = \frac{1}{2}\mathop \smallint \limits_{\theta = 0}^{\pi /2} \mathop \smallint \limits_{r = 0}^1 {r^3}{\rm{d}}r{\rm{d}}\theta $
$ = \frac{1}{2}\left( {\mathop \smallint \limits_{\theta = 0}^{\pi /2} {\rm{d}}\theta } \right)\left( {\mathop \smallint \limits_{r = 0}^1 {r^3}{\rm{d}}r} \right)$
$ = \frac{1}{2}\left( {\frac{\pi }{2}} \right)\left( {\frac{1}{4}{r^4}|_0^1} \right) = \frac{\pi }{{16}}$
So, $\mathop \smallint \limits_0^1 \mathop \smallint \limits_0^{\sqrt {1 - {x^2}} } \mathop \smallint \limits_0^{\sqrt {{x^2} + {y^2}} } z{\rm{d}}z{\rm{d}}y{\rm{d}}x = \mathop \smallint \limits_{\theta = 0}^{\pi /2} \mathop \smallint \limits_{r = 0}^1 \mathop \smallint \limits_{z = 0}^r rz{\rm{d}}z{\rm{d}}r{\rm{d}}\theta = \frac{\pi }{{16}}$.
