The base is a disk of radius $a$. Slices perpendicular to the base are squares.
If we position the center of the disk to the origin, we have the equation $x^2+y^2=a^2$. But we need to integrate using square slices.
Since there is symmetry, we can integrate the part where $y\ge0$ and double: $$2\int_0^a\left(\sqrt{a^2-x^2}\right)^2\;dx=2\left.\left(a^2x-\frac{x^3}3\right)\right|_0^a=\frac{4a^3}3$$
However, the key at the back of the book says: $$\int_{-a}^a\left(2\sqrt{a^2-x^2}\right)^2\;dx=\frac{16a^3}3$$
Why does it put that $2$ inside the square? The equation is really $y=\sqrt{a^2-x^2}$ for $y\ge0$ and $y=-\sqrt{a^2-x^2}$ for $y<0$.