Answer
$V = \frac{{{\pi ^2}}}{4}$
Work Step by Step
$$\eqalign{
& {\text{From the graph shown below}}{\text{, we have:}} \cr
& f\left( x \right) = \arcsin x,{\text{ for }}0 \leqslant x \leqslant 1 \cr
& {\text{Using the the Shell method about the }}y{\text{ - axis}} \cr
& V = \int_a^b {2\pi x\left[ {f\left( x \right) - g\left( x \right)} \right]dx} \cr
& {\text{Where }}f\left( x \right) = \arcsin x,{\text{ }}g\left( x \right) = 0{\text{ and }}a = 0,{\text{ }}b = 1 \cr
& V = \int_0^1 {2\pi x\left( {\arcsin x - 0} \right)dx} \cr
& V = 2\pi \int_0^1 {x\arcsin xdx} \cr
& {\text{Integrate by tables using the entry 90:}} \cr
& \int {u{{\sin }^{ - 1}}u} du = \frac{{2{u^2} - 1}}{4}{\sin ^{ - 1}}u + \frac{{u\sqrt {1 - {u^2}} }}{4} + C \cr
& {\text{Therefore}} \cr
& V = 2\pi \left[ {\frac{{2{x^2} - 1}}{4}\arcsin x + \frac{{x\sqrt {1 - {x^2}} }}{4}} \right]_0^1 \cr
& V = 2\pi \left[ {\frac{{2{{\left( 1 \right)}^2} - 1}}{4}\arcsin \left( 1 \right) + \frac{{\left( 1 \right)\sqrt {1 - {{\left( 1 \right)}^2}} }}{4}} \right] - 2\pi \left[ 0 \right] \cr
& V = 2\pi \left[ {\frac{1}{4}\arcsin \left( 1 \right) + \frac{0}{4}} \right] - 0 \cr
& V = 2\pi \left( {\frac{\pi }{8}} \right) \cr
& V = \frac{{{\pi ^2}}}{4} \cr} $$
