Answer
The volume of the wedge-shaped region is $18$.
Work Step by Step
We have the wedge-shaped region ${\cal W}$ contained within the cylinder ${x^2} + {y^2} = 9$ and $0 \le z \le x$.
From the figure attached, we see that the region has axial symmetry with respect to the $z$-axis. Thus, to find the volume as an iterated integral we use the cylindrical coordinates.
Referring to the figure attached, the projection of ${\cal W}$ onto the $xy$-plane is ${\cal D}$, a half of a disk of radius $3$, located at the right-hand side of the $y$-axis.
The description of ${\cal D}$ in polar coordinates:
${\cal D} = \left\{ {\left( {r,\theta } \right)|0 \le r \le 3, - \frac{\pi }{2} \le \theta \le \frac{\pi }{2}} \right\}$
Since $z$ varies from $z=0$ to $z = x = r\cos \theta $, the cylindrical description of ${\cal W}$ is
${\cal W} = \left\{ {\left( {r,\theta ,z} \right)|0 \le r \le 3, - \frac{\pi }{2} \le \theta \le \frac{\pi }{2},0 \le z \le r\cos \theta } \right\}$
Using Eq. (5) of Theorem 2, we evaluate the volume of ${\cal W}$:
$V = \mathop \smallint \limits_{\theta = - \pi /2}^{\pi /2} \mathop \smallint \limits_{r = 0}^3 \mathop \smallint \limits_{z = 0}^{r\cos \theta } r{\rm{d}}z{\rm{d}}r{\rm{d}}\theta $
$ = \mathop \smallint \limits_{\theta = - \pi /2}^{\pi /2} \mathop \smallint \limits_{r = 0}^3 r\left( {z|_0^{r\cos \theta }} \right){\rm{d}}r{\rm{d}}\theta $
$ = \mathop \smallint \limits_{\theta = - \pi /2}^{\pi /2} \mathop \smallint \limits_{r = 0}^3 {r^2}\cos \theta {\rm{d}}r{\rm{d}}\theta $
$ = \frac{1}{3}\mathop \smallint \limits_{\theta = - \pi /2}^{\pi /2} \left( {{r^3}|_0^3} \right)\cos \theta {\rm{d}}\theta $
$ = 9\mathop \smallint \limits_{\theta = - \pi /2}^{\pi /2} \cos \theta {\rm{d}}\theta $
$ = 9\left( {\sin \theta |_{ - \pi /2}^{\pi /2}} \right)$
$ = 9\left( {1 + 1} \right) = 18$
Thus, the volume of ${\cal W}$, the wedge-shaped region is $18$.
