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