Answer
$$m = \pi + 2$$
Work Step by Step
$$\eqalign{
& \rho \left( x \right) = 1 + \sin x,{\text{ }}0 \leqslant x \leqslant \pi \cr
& {\text{The mass of the object is }}m = \int_a^b {\rho \left( x \right)} dx,{\text{ }}\left( {{\text{See page 460}}} \right) \cr
& m = \int_0^\pi {\left( {1 + \sin x} \right)} dx \cr
& {\text{Integrating}} \cr
& m = \left[ {x - \cos x} \right]_0^\pi \cr
& m = \left[ {\pi - \cos \pi } \right] - \left[ {0 - \cos 0} \right] \cr
& m = \pi - \left( { - 1} \right) + 1 \cr
& m = \pi + 2 \cr} $$
