Answer
$\displaystyle{V=\frac{248\pi }{45}}\\ $
Work Step by Step
$y^4=2-y^2\\
y^4+y^2-2=0\\
y=-1 \qquad y=1$
$\displaystyle{A\left(y\right)=\pi \left(2-y^2\right)^2-\pi \left(y^4\right)^2}\\
\displaystyle{A\left(y\right)=\pi \left(4+y^4-4y^2-y^8\right)}\\$
$\displaystyle{V=\int_{-1}^1A\left(y\right)\ dy}\\
\displaystyle{V=\int_{-1}^1\pi \left(4+y^4-4y^2-y^8\right)\ dy}\\
\displaystyle{V=\pi \int_{-1}^14+y^4-4y^2-y^8\ dy}\\
\displaystyle{V=\pi\left[4y+\frac{1}{5}y^5-\frac{4}{3}y^3-\frac{1}{9}y^9\right]_{-1}^1}\\
\displaystyle{V=\pi\left(\left(4\times1+\frac{1}{5}1^5-\frac{4}{3}1^3-\frac{1}{9}1^9\right)-\left(4\times-1+\frac{1}{5}{-1}^5-\frac{4}{3}{-1}^3-\frac{1}{9}{-1}^9\right)\right)}\\
\displaystyle{V=\frac{248\pi }{45}}\\ $
