Answer
$$\frac{2}{3} r^{3} \tan \theta$$
Work Step by Step
The circle in the xy plane has an equation of $x^{2}+y^{2}=r^{2}$
NOTE: I also refer to the height of the wedge as $y$.
The base of the triangular cross section is $\sqrt{r^{2}-y^{2}}$
And then $\tan \theta=\frac{y}{\sqrt{r^{2}-y^{2}}}$
Consequently we can write $\sqrt{r^{2}-y^{2}} \tan \theta=y$
The cross sectional area is
$$
\begin{array}{l}
\frac{1}{2} \sqrt{r^{2}-y^{2}}\left(\sqrt{r^{2}-y^{2}}\right) \tan \theta=A(y) \\
\frac{1}{2}\left(r^{2}-y^{2}\right) \tan \theta=A(y)
\end{array}
$$
The volume of the wedge is
$2 \int_{0}^{r} \frac{1}{2} A(y) d y$
$2 \int_{0}^{r} \frac{1}{2}\left(r^{2}-y^{2}\right) \tan \theta d y$
$\int_{0}^{\tau}\left(r^{2}-y^{2}\right) \tan \theta d y$
Integrating and evaluating we get
$\left[\left(r^{2} y-\frac{y^{3}}{3}\right) t a n \theta\right]_{0}^{r}$
$\left(r^{3}-\frac{r^{3}}{3}\right) \tan \theta$
$\frac{2}{3} r^{3} \tan \theta=\left(\frac{3 r^{3}}{3}-\frac{r^{3}}{3}\right) \tan \theta$
