Answer
$$V = \frac{1}{2}\pi $$
Work Step by Step
$$\eqalign{
& {\text{We have }}{y^2} = x,\,\,\,y = 1,\,\,\,x = 0 \cr
& {\text{The volume of the solid can be calculated using cylindrical shells}} \cr
& V = \int_c^d {2\pi y\left[ {f\left( y \right) - g\left( y \right)} \right]dy} \cr
& {\text{Let }}f\left( y \right) = {y^2}{\text{ and }}g\left( y \right) = 0 \cr
& V = \int_0^1 {2\pi y\left( {{y^2} - 0} \right)dy} \cr
& V = 2\pi \int_0^1 {{y^3}dy} \cr
& V = \frac{1}{2}\pi \left[ {{y^3}} \right]_0^1 \cr
& {\text{Integrating}} \cr
& V = \frac{1}{2}\pi \left[ {{{\left( 1 \right)}^3}} \right]_0^1 \cr
& V = \frac{1}{2}\pi \cr} $$
