Answer
$\left( {\text{a}} \right)\int_0^\pi {\pi \sin x} dx,{\text{ }}\left( {\text{b}} \right)2\pi $
Work Step by Step
$$\eqalign{
& {\text{From the graph we have:}} \cr
& y = \sqrt {\sin x} {\text{ on the interval }}0 \leqslant x \leqslant \pi \cr
& \cr
& \left( {\text{a}} \right){\text{Using the disk method about the }}x{\text{ - axis}} \cr
& V = \int_a^b {\pi {{\left[ {f\left( x \right)} \right]}^2}} dx \cr
& {\text{Therefore}}{\text{,}} \cr
& V = \int_0^\pi {\pi {{\left[ {\sqrt {\sin x} } \right]}^2}} dx \cr
& V = \int_0^\pi {\pi \sin x} dx \cr
& \cr
& \left( {\text{b}} \right) \cr
& {\text{Integrating}} \cr
& V = \pi \left[ { - \cos x} \right]_0^\pi \cr
& {\text{Evaluating}} \cr
& V = \pi \left[ { - \cos \pi + \cos 0} \right] \cr
& V = \pi \left( {1 + 1} \right) \cr
& V = 2\pi \cr} $$
