Answer
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal W}^{} z{\rm{d}}V = \frac{{25}}{{12}}$
Work Step by Step
We have $f\left( {x,y,z} \right) = z$ over the region ${\cal W}$ as in Figure 11.
Referring to Figure 11 or the figure attached, we see that ${\cal W}$ is a region between the plane $z=0$ and the surface $z = \sqrt {9 - {x^2} - {y^2}} $. So, it is a $z$-simple region which lies over ${\cal D}$ located in the first quadrant of the $xy$-plane. Since ${\cal D}$ is bounded by $x=1$, $y=0$, and $x=y$; we can consider it as a vertically simple region defined by
${\cal D} = \left\{ {\left( {x,y} \right)|0 \le x \le 1,0 \le y \le x} \right\}$
Thus, the triple integral is equal to the iterated integral:
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal W}^{} f\left( {x,y,z} \right){\rm{d}}V = \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal D}^{} \left( {\mathop \smallint \limits_{z = 0}^{\sqrt {9 - {x^2} - {y^2}} } z{\rm{d}}z} \right){\rm{d}}A$
$ = \mathop \smallint \limits_{x = 0}^1 \mathop \smallint \limits_{y = 0}^x \left( {\mathop \smallint \limits_{z = 0}^{\sqrt {9 - {x^2} - {y^2}} } z{\rm{d}}z} \right){\rm{d}}y{\rm{d}}x$
$ = \frac{1}{2}\mathop \smallint \limits_{x = 0}^1 \mathop \smallint \limits_{y = 0}^x \left( {{z^2}|_0^{\sqrt {9 - {x^2} - {y^2}} }} \right){\rm{d}}y{\rm{d}}x$
$ = \frac{1}{2}\mathop \smallint \limits_{x = 0}^1 \mathop \smallint \limits_{y = 0}^x \left( {9 - {x^2} - {y^2}} \right){\rm{d}}y{\rm{d}}x$
$ = \frac{1}{2}\mathop \smallint \limits_{x = 0}^1 \left( {\left( {9y - {x^2}y - \frac{1}{3}{y^3}} \right)|_0^x} \right){\rm{d}}x$
$ = \frac{1}{2}\mathop \smallint \limits_{x = 0}^1 \left( {9x - \frac{4}{3}{x^3}} \right){\rm{d}}x$
$ = \frac{1}{2}\left( {\left( {\frac{9}{2}{x^2} - \frac{1}{3}{x^4}} \right)|_0^1} \right)$
$ = \frac{1}{2}\left( {\frac{9}{2} - \frac{1}{3}} \right)$
$ = \frac{{25}}{{12}}$
Thus, $\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal W}^{} z{\rm{d}}V = \frac{{25}}{{12}}$.
