Each cross section has a side length equal to the distance (parallel to the [tex]y[/tex]-axis) between the curves [tex]y=1[/tex] and [tex]y=x^2[/tex], which is [tex]1-x^2[/tex], and hence area of [tex](1-x^2)^2[/tex]. The two curves intersect at [tex]x=-1[/tex] and [tex]x=1[/tex]. Then the volume is equal to the integral
[tex]\displaystyle\int_{-1}^1(1-x^2)^2\,\mathrm dx[/tex]
The integrand is even, since
[tex](1-(-x)^2)^2=(1-x^2)^2[/tex]
so we can make use of symmetry to simplify this to
[tex]\displaystyle2\int_0^1(1-x^2)^2\,\mathrm dx[/tex]
Computing the integral is trivial:
[tex]\displaystyle2\int_0^1(1-2x^2+x^4)\,\mathrm dx=2\left(x-\frac23x^3+\frac15x^5\right)\bigg|_0^1=\boxed{\frac{16}{15}}[/tex]
