Because the height and base of each cross section are equal, the area for any given cross section is [tex]\dfrac12bh=\dfrac12b(x)^2[/tex] where the base of each section occurring along the line [tex]x=x_0[/tex] is the vertical distance between the upper and lower halves of the circle [tex]x^2+y^2=64[/tex].
We can write
[tex]y=\pm\sqrt{64-x^2}[/tex]
so that
[tex]b(x)=\sqrt{64-x^2}-(-\sqrt{64-x^2})=2\sqrt{64-x^2}[/tex]
and so the area of each cross section is
[tex]\dfrac12(2\sqrt{64-x^2})^2=2(64-x^2)=128-2x^2[/tex]
Over the circular base of the solid, we have [tex]-8\le x\le8[/tex], so the volume of the solid is given by the integral
[tex]\displaystyle\int_{x=-8}^{x=8}(128-2x^2)\,\mathrm dx=\dfrac{4096}3[/tex]
