Perhaps you mean to find the volume under [tex]z=\sqrt{64-x^2-y^2}[/tex]? It seems like you have to rely on calculus otherwise (in the case that you indeed mean [tex]z=64-x^2-y^2[/tex]).
Assuming this, note that [tex]z=\sqrt{64-x^2-y^2}[/tex] describes the top half of a sphere with radius 8, which means the volume below this surface over the disk [tex]x^2+y^2\le64[/tex] is simply half the volume of the sphere. Thus the volume is
[tex]\dfrac43\pi8^3=\dfrac{2048}3\pi[/tex]
