Answer
$\text{Thus, the volume is}$
\begin{align}
V = \frac{16}{15}
\end{align}
Work Step by Step
$\text{It is given that}$
\begin{align}
x = 1 -y^2
\end{align}
$\text{The intersections with y axis:}$
\begin{align}
1-y^2 = 0 \Rrightarrow y = \pm 1
\end{align}
$\text{To find the volume, we have to find the area of the cross section.}$
$\text{It is given that the cross section is squre, therefore:}$
\begin{align}
A(y) = (1-y^2)^2 = 1 - 2y^2+y^4
\end{align}
$\text{Thus, the volume is}$
\begin{align}
V = \int_{-1}^{1} A(y) \ dy =& \int_{-1}^{1} (1 - 2y^2+y^4) \ dy = \left[ y-2\frac{y^3}{3} +\frac{y^5}{5}\right]_{-1}^1 = \\ & = 2 - \frac{4}{3} +\frac{2}{5} = \frac{16}{15}
\end{align}
