Answer
$$\frac{4}{9}$$
Work Step by Step
$$\eqalign{
& \int_0^\pi {\int_0^{\sin \theta } {{r^2}} dr} d\theta \cr
& {\text{The region }}R{\text{ is:}} \cr
& R = \left\{ {\left( {r,\theta } \right):0 \leqslant r \leqslant \pi ,{\text{ 0}} \leqslant \theta \leqslant \sin \theta } \right\}{\text{ }}\left( {{\text{Graph below}}} \right) \cr
& \cr
& \int_0^\pi {\int_0^{\sin \theta } {{r^2}} dr} d\theta \cr
& {\text{Integrate with respect to }}r \cr
& = \int_0^\pi {\left[ {\frac{{{r^3}}}{3}} \right]_0^{\sin \theta }} d\theta \cr
& = \int_0^\pi {\left[ {\frac{{{{\left( {\sin \theta } \right)}^2}}}{3} - \frac{{{{\left( 0 \right)}^2}}}{2}} \right]} d\theta \cr
& = \frac{1}{3}\int_0^\pi {{{\sin }^3}\theta } d\theta \cr
& = \frac{1}{3}\int_0^\pi {{{\sin }^2}\theta \sin \theta } d\theta \cr
& = \frac{1}{3}\int_0^\pi {\left( {1 - {{\cos }^2}\theta } \right)\sin \theta } d\theta \cr
& {\text{Integrate with respect to }}\theta \cr
& = \frac{1}{3}\left[ { - \cos \theta + \frac{1}{3}{{\cos }^3}\theta } \right]_0^\pi \cr
& = \frac{1}{3}\left[ { - \cos \pi + \frac{1}{3}{{\cos }^3}\pi } \right] - \frac{1}{3}\left[ { - \cos 0 + \frac{1}{3}{{\cos }^3}0} \right] \cr
& = \frac{1}{3}\left( {1 - \frac{1}{3}} \right) - \frac{1}{3}\left( { - 1 + \frac{1}{3}} \right) \cr
& = \frac{2}{9} + \frac{2}{9} \cr
& = \frac{4}{9} \cr} $$
