Answer :
Use the disk method:
[tex]\displaystyle\pi\int_{-1}^1(4-x^2)^2\,\mathrm dx=2\pi\int_0^1(16-8x^2+x^4)\,\mathrm dx[/tex]
where we take advantage of symmetry to rewrite the limits, and expand the integrand.
[tex]=\displaystyle2\pi\left(16x-\frac83x^3+\frac15x^5\right)\bigg|_0^1=2\pi\left(16-\frac83+\frac15\right)=\frac{406\pi}{15}[/tex]