Answer
$\displaystyle{V=\frac{10\sqrt2\pi }{3}}\\ $
Work Step by Step
$\displaystyle{y^2=1-y^2}\\
\displaystyle{2y^2=1}\\
\displaystyle{y=-\frac{1}{\sqrt2}\qquad y=\frac{1}{\sqrt2}}$
$\displaystyle{A\left(y\right)=\pi \left(3-y^2\right)^2-\pi \left(3-\left(1-y^2\right)\right)^2}\\
\displaystyle{A\left(y\right)=\pi \left(5-10y^2\right)}\\$
$\displaystyle{V=\int_{-\frac{1}{\sqrt2}}^{\frac{1}{\sqrt2}}A\left(y\right)\ dy}\\
\displaystyle{V=\int_{-\frac{1}{\sqrt2}}^{\frac{1}{\sqrt2}}\pi \left(5-10y^2\right)\ dy}\\
\displaystyle{V=\pi \int_{-\frac{1}{\sqrt2}}^{\frac{1}{\sqrt2}}5-10y^2\ dy}\\$
By symmetry
$\displaystyle{V=2\pi \int_{0}^{\frac{1}{\sqrt2}}5-10y^2\ dy}\\
\displaystyle{V=\pi\left[5y-\frac{10}{3}y^3\right]_{0}^{\frac{1}{\sqrt2}}}\\
\displaystyle{V=\pi\left(\left(5\times\frac{1}{\sqrt2}-\frac{10}{3}\left(\frac{1}{\sqrt2}\right)^3\right)-\left(0\right)\right)}\\
\displaystyle{V=\frac{10\sqrt2\pi }{3}}\\ $
