The volume of the solid (call it S) in Cartesian coordinates is
[tex]\displaystyle\iiint_S\mathrm dV=\int_{-1}^1\int_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}}\int_{(x^2+y^2)^2-1}^{4-4(x^2+y^2)}\mathrm dz\,\mathrm dy\,\mathrm dx[/tex]
but I suspect converting to cylindrical coordinates would make the integral much more tractable.
Take
[tex]\begin{cases}x=r\cos\theta\\y=r\sin\theta\\z=z\end{cases}\implies\mathrm dV=r\,\mathrm dr\,\mathrm d\theta\,\mathrm dz[/tex]
Then
[tex]4-4(x^2+y^2)=4-4r^2=4(1-r^2)[/tex]
[tex](x^2+y^2)^2-1=(r^2)^2-1=r^4-1[/tex]
and the integral becomes
[tex]\displaystyle\iiint_S\mathrm dV=\int_0^{2\pi}\int_0^1\int_{r^4-1}^{4(1-r^2)}r\,\mathrm dz\,\mathrm dr\,\mathrm d\theta[/tex]
[tex]=\displaystyle2\pi\int_0^1r(4(1-r^2)-(r^4-1))\,\mathrm dr[/tex]
[tex]=\displaystyle2\pi\int_0^1r(5-4r^2-r^4)\,\mathrm dr[/tex]
[tex]=\displaystyle2\pi\int_0^15r-4r^3-r^5\,\mathrm dr[/tex]
[tex]=2\pi\left(\dfrac52-1-\dfrac16\right)=\boxed{\dfrac{8\pi}3}[/tex]
