Answer
$$0$$
Work Step by Step
$$\eqalign{
& \int_0^{2\pi } {\int_0^6 {3{r^2}} \sin \theta dr} d\theta \cr
& {\text{The region }}R{\text{ is:}} \cr
& R = \left\{ {\left( {r,\theta } \right):0 \leqslant r \leqslant 6,{\text{ 0}} \leqslant \theta \leqslant 2\pi } \right\}{\text{ }}\left( {{\text{Graph below}}} \right) \cr
& \cr
& \int_0^{2\pi } {\int_0^6 {3{r^2}} \sin \theta dr} d\theta \cr
& {\text{Integrate with respect to }}r \cr
& = \int_0^{2\pi } {\left[ {{r^3}\sin \theta } \right]_0^6} d\theta \cr
& = \int_0^{2\pi } {\left[ {{{\left( 6 \right)}^3}\sin \theta - {{\left( 0 \right)}^3}\sin \theta } \right]} d\theta \cr
& = \int_0^{2\pi } {216\sin \theta } d\theta \cr
& {\text{Integrate with respect to }}\theta \cr
& = - 216\left[ {\cos \theta } \right]_0^{2\pi } \cr
& = - 216\left( {\cos 2\pi - \cos 0} \right) \cr
& = - 216\left( {1 - 1} \right) \cr
& = 0 \cr} $$
