Answer
$$\frac{3}{{32}}{\pi ^2} + \frac{9}{8}$$
Work Step by Step
$$\eqalign{
& \int_0^{\pi /2} {\int_0^{1 + \sin \theta } {\theta rdr} } d\theta \cr
& {\text{The region }}R{\text{ is:}} \cr
& R = \left\{ {\left( {r,\theta } \right):0 \leqslant r \leqslant 1 + \sin \theta ,{\text{ 0}} \leqslant \theta \leqslant \frac{\pi }{2}} \right\}{\text{ }}\left( {{\text{Graph below}}} \right) \cr
& \cr
& \int_0^{\pi /2} {\int_0^{1 + \sin \theta } {\theta rdr} } d\theta \cr
& {\text{Integrate with respect to }}r \cr
& = \int_0^{\pi /2} {\left[ {\frac{{\theta {r^2}}}{2}} \right]_0^{1 + \sin \theta }} d\theta \cr
& = \int_0^{\pi /2} {\left[ {\frac{{\theta {{\left( {1 + \sin \theta } \right)}^2}}}{2}} \right]} d\theta \cr
& = \int_0^{\pi /2} {\frac{1}{2}\theta {{\left( {1 + \sin \theta } \right)}^2}} d\theta \cr
& = \int_0^{\pi /2} {\frac{1}{2}\theta \left( {1 + 2\sin \theta + {{\sin }^2}\theta } \right)} d\theta \cr
& = \int_0^{\pi /2} {\frac{1}{2}\theta \left( {1 + 2\sin \theta + \frac{{1 - \cos 2\theta }}{2}} \right)} d\theta \cr
& = \int_0^{\pi /2} {\left( {\frac{1}{2}\theta + \theta \sin \theta + \frac{1}{4}\theta - \frac{1}{4}\theta \cos 2\theta } \right)} d\theta \cr
& = \int_0^{\pi /2} {\left( {\frac{3}{4}\theta + \theta \sin \theta - \frac{1}{4}\theta \cos 2\theta } \right)} d\theta \cr
& {\text{Integrate with respect to }}\theta ,{\text{ use integration by parts}} \cr
& = \left[ {\frac{{3{\theta ^2}}}{8} + \sin \theta - \theta \cos \theta - \frac{1}{{16}}\cos 2\theta - \frac{1}{8}\theta \sin 2\theta } \right]_0^{\pi /2} \cr
& {\text{Evaluating the limits}} \cr
& = \left[ {\frac{{3{{\left( {\pi /2} \right)}^2}}}{8} + \sin \left( {\frac{\pi }{2}} \right) - \frac{1}{{16}}\cos \left( \pi \right)} \right] - \left[ { - \frac{1}{{16}}\cos \left( 0 \right)} \right] \cr
& = \frac{3}{{32}}{\pi ^2} + 1 + \frac{1}{{16}} + \frac{1}{{16}} \cr
& = \frac{3}{{32}}{\pi ^2} + \frac{9}{8} \cr} $$
