Answer
$\frac{27}{16} \pi$
Work Step by Step
the volume is
$\int_{0}^{\pi / 2} \int_{0}^{3 \sin \theta} r(r \sin \theta) d r d \theta=V$
simplify
$\int_{0}^{\pi / 2} \int_{0}^{3 \sin \theta} r^{2} \sin \theta d r d \theta=V$
integrate with respect to $r$
$\int_{0}^{\pi / 2}\left[\frac{r^{3}}{3} \sin \theta\right]_{0}^{3 \sin \theta} d \theta=V$
evaluate
$\int_{0}^{\pi / 2}\left[\frac{27 \sin ^{3} \theta}{3} \sin \theta\right] d \theta=V$
$V=9 \int_{0}^{\pi / 2} \sin ^{4} d \theta$
integrate
$V=9\left[\left(12 x-8 \sin ^{2} x+4 \sin (4 x)\right)\frac{1}{3^{2}}\right]_{0}^{\pi / 2}$
evaluate and simplify
$(6 \pi-0)V \frac{9}{32}=\frac{27}{16} \pi$
