S casts a "shadow" on the x-y plane given by the set
[tex]S' = \left\{(x,y) ~:~ -1 \le x \le 1 \text{ and } x^2 \le y \le 1\right\}[/tex]
Each cross section is a square whose side length is determined by the vertical distance between y = 1 and y = x², which is |1 - x²|. But since -1 ≤ x ≤ 1, this distance simplifies to 1 - x².
The volume of an infinitesimally thin section is then (1 - x²)² ∆x (where ∆x represents its thickness), and so the volume of S is
[tex]\displaystyle \int_{-1}^1 (1-x^2)^2 \, dx = \int_{-1}^1 (1 - 2x^2 + x^4) \, dx[/tex]
The integrand is even, so this integral is equal to twice the integral over [0, 1] :
[tex]\displaystyle \int_{-1}^1 (1-x^2)^2 \, dx = 2 \int_0^1 (1 - 2x^2 + x^4) \, dx = 2 \left(1 - \frac23 + \frac15\right) = \boxed{\frac{16}{15}}[/tex]
