I assume the base of S is the disk centered at the origin, or x² + y² ≤ r². Solving for y gives two solutions corresponding to the upper and lower halves of the circular boundary,
[tex]x^2+y^2 = r^2 \implies y = \pm \sqrt{r^2-x^2}[/tex]
and the vertical distance between them - i.e. the side length of each cross section - is
[tex]\sqrt{r^2-x^2} - \left(-\sqrt{r^2-x^2}\right) = 2\sqrt{r^2-x^2}[/tex]
Then the volume of each square cross section 4 (r² - x²) ∆x, and the volume of S is
[tex]\displaystyle \int_{-r}^r 4(r^2-x^2) \, dx[/tex]
By symmetry, this is the same as
[tex]\displaystyle 8 \int_0^r (r^2-x^2) \, dx[/tex]
which gives a volume of
[tex]\displaystyle 8 \int_0^r (r^2-x^2) \, dx = 8 \left(r^3 - \frac{r^3}3\right) = \boxed{\frac{16r^3}3}[/tex]
