Answer
$$\frac{1}{6}$$
Work Step by Step
$$\eqalign{
& \int_0^{\pi /2} {\int_0^{1 - \cos \theta } {\left( {\sin \theta } \right)rdr} } d\theta \cr
& {\text{The region }}R{\text{ is:}} \cr
& R = \left\{ {\left( {r,\theta } \right):0 \leqslant r \leqslant 1 - \cos \theta ,{\text{ 0}} \leqslant \theta \leqslant \frac{\pi }{2}} \right\}{\text{ }}\left( {{\text{Graph below}}} \right) \cr
& \cr
& \int_0^{\pi /2} {\int_0^{1 - \cos \theta } {\left( {\sin \theta } \right)rdr} } d\theta \cr
& {\text{Integrate with respect to }}r \cr
& = \int_0^{\pi /2} {\left[ {\frac{{{r^2}}}{2}} \right]_0^{1 - \cos \theta }} d\theta \cr
& = \int_0^{\pi /2} {\left[ {\frac{{\left( {\sin \theta } \right){{\left( {1 - \cos \theta } \right)}^2}}}{2}} \right]} d\theta \cr
& = \frac{1}{2}\int_0^{\pi /2} {{{\left( {1 - \cos \theta } \right)}^2}\left( {\sin \theta } \right)} d\theta \cr
& {\text{Integrate with respect to }}\theta ,{\text{ use integration by parts}} \cr
& = \frac{1}{2}\left[ {\frac{{{{\left( {1 - \cos \theta } \right)}^3}}}{3}} \right]_0^{\pi /2} \cr
& {\text{Evaluating the limits}} \cr
& = \frac{1}{2}\left[ {\frac{{{{\left( {1 - \cos \left( {\pi /2} \right)} \right)}^3}}}{3} - \frac{{{{\left( {1 - \cos \left( 0 \right)} \right)}^3}}}{3}} \right] \cr
& = \frac{1}{6} \cr} $$
