Answer
$$\frac{{5\sqrt 5 \pi }}{6}$$
Work Step by Step
$$\eqalign{
& \int_0^{\pi /2} {\int_2^3 {\sqrt {9 - {r^2}} rdr} } d\theta \cr
& {\text{The region }}R{\text{ is:}} \cr
& R = \left\{ {\left( {r,\theta } \right):2 \leqslant r \leqslant 3,{\text{ 0}} \leqslant \theta \leqslant \frac{\pi }{2}} \right\}{\text{ }}\left( {{\text{Graph below}}} \right) \cr
& \cr
& \int_0^{\pi /2} {\int_2^3 {\sqrt {9 - {r^2}} rdr} } d\theta \cr
& = - \frac{1}{2}\int_0^{\pi /2} {\int_2^3 {{{\left( {9 - {r^2}} \right)}^{1/2}}\left( { - 2r} \right)dr} } d\theta \cr
& {\text{Integrate with respect to }}r \cr
& = - \frac{1}{2}\int_0^{\pi /2} {\left[ {\frac{2}{3}{{\left( {9 - {r^2}} \right)}^{3/2}}} \right]_2^3} d\theta \cr
& = - \frac{1}{3}\int_0^{\pi /2} {\left[ {{{\left( {9 - {3^2}} \right)}^{3/2}} - {{\left( {9 - {2^2}} \right)}^{3/2}}} \right]} d\theta \cr
& = - \frac{1}{3}\int_0^{\pi /2} {\left[ {{{\left( 0 \right)}^{3/2}} - {{\left( 5 \right)}^{3/2}}} \right]} d\theta \cr
& = \frac{{5\sqrt 5 }}{3}\int_0^{\pi /2} {d\theta } \cr
& {\text{Integrate with respect to }}\theta \cr
& = \frac{{5\sqrt 5 }}{3}\left[ \theta \right]_0^{\pi /2} \cr
& = \frac{{5\sqrt 5 \pi }}{6} \cr} $$
