Answer
$$0$$
Work Step by Step
$$\eqalign{
& \int_0^\pi {\int_0^{1 - \sin \theta } {{r^2}\cos \theta } } drd\theta \cr
& = \int_0^\pi {\left[ {\int_0^{1 - \sin \theta } {{r^2}\cos \theta } dr} \right]} d\theta \cr
& {\text{solve the inner integral and treat }}\theta {\text{ as a constant}} \cr
& = \cos \theta \int_0^{1 - \sin \theta } {{r^2}} dr \cr
& {\text{then}} \cr
& = \cos \theta \left[ {\frac{{{r^3}}}{3}} \right]_0^{1 - \sin \theta } \cr
& {\text{evaluating the limits in the variable }}r \cr
& = \cos \theta \left[ {\frac{{{{\left( {1 - \sin \theta } \right)}^3}}}{3} - \frac{{{{\left( 0 \right)}^3}}}{3}} \right] \cr
& = \frac{{{{\left( {1 - \sin \theta } \right)}^3}\cos \theta }}{3} \cr
& {\text{then}} \cr
& \int_0^\pi {\left[ {\int_0^{\cos 3\theta } r dr} \right]} d\theta = \int_0^\pi {\frac{{{{\left( {1 - \sin \theta } \right)}^3}\cos \theta }}{3}} d\theta \cr
& = \frac{1}{3}\int_0^\pi {{{\left( {1 - \sin \theta } \right)}^3}\cos \theta } d\theta \cr
& = - \frac{1}{3}\int_0^\pi {{{\left( {1 - \sin \theta } \right)}^3}\left( { - \cos \theta } \right)} d\theta \cr
& {\text{integrating }} \cr
& = - \frac{1}{3}\left( {\frac{{{{\left( {1 - \sin \theta } \right)}^4}}}{4}} \right)_0^\pi \cr
& = \frac{1}{{12}}\left( {{{\left( {1 - \sin \theta } \right)}^4}} \right)_\pi ^0 \cr
& {\text{evaluate}} \cr
& = \frac{1}{{12}}\left( {{{\left( {1 - \sin 0} \right)}^4} - {{\left( {1 - \sin \pi } \right)}^4}} \right) \cr
& = \frac{1}{{12}}\left( {{{\left( 1 \right)}^4} - {{\left( 1 \right)}^4}} \right) \cr
& = 0 \cr} $$
