Answer
$$\frac{{16}}{3}$$
Work Step by Step
$$\eqalign{
& \int_0^{\pi /4} {\int_0^4 {{r^2}\sin \theta } \cos \theta dr} d\theta \cr
& {\text{The region }}R{\text{ is:}} \cr
& R = \left\{ {\left( {r,\theta } \right):0 \leqslant r \leqslant 4,{\text{ 0}} \leqslant \theta \leqslant \frac{\pi }{4}} \right\}{\text{ }}\left( {{\text{Graph below}}} \right) \cr
& \cr
& \int_0^{\pi /4} {\int_0^4 {{r^2}\sin \theta } \cos \theta dr} d\theta \cr
& {\text{Integrate with respect to }}r \cr
& = \int_0^{\pi /4} {\left[ {\frac{{{r^3}}}{3}\sin \theta \cos \theta } \right]_0^4} d\theta \cr
& = \frac{1}{3}\int_0^{\pi /4} {\left[ {{r^3}\sin \theta \cos \theta } \right]_0^4} d\theta \cr
& = \frac{1}{3}\int_0^{\pi /4} {\left[ {{{\left( 4 \right)}^3}\sin \theta \cos \theta - {{\left( 0 \right)}^3}\sin \theta \cos \theta } \right]} d\theta \cr
& = \frac{{64}}{3}\int_0^{\pi /4} {\sin \theta \cos \theta } d\theta \cr
& {\text{Integrate with respect to }}\theta \cr
& = \frac{{64}}{3}\left[ {\frac{{{{\sin }^2}\theta }}{2}} \right]_0^{\pi /4} \cr
& = \frac{{64}}{3}\left[ {\frac{{{{\sin }^2}\left( {\pi /4} \right)}}{2} - \frac{{{{\sin }^2}\left( 0 \right)}}{2}} \right] \cr
& = \frac{{64}}{3}\left[ {\frac{1}{4}} \right] \cr
& = \frac{{16}}{3} \cr} $$
