The outer boundary is y = 8-x^2.

The inner boundary is y = x^2.


These two curves intersect when  

    8-x^2 = x^2,  8 = x^2+x^2,  8 = 2x^2,  x^2 = 8/2 = 4,  x =  = +/- 2.


In the interval  -2 <= x <= 2, both curve are above the x-axis.


Each vertical section  x= const  of this volume  (of this solid body)  is a RING 
with the inner radius  r= x^2  and the outer radius R = 8-x^2.


The area of this ring is   =  = .


The antiderivative is  of this function    is  F(x) = .  


The integral from  -2  to  2  is the difference  

    F(2) - F(-2) = () - () =  =  =  = .


Thus the desired volume is   =  = 536.1651  (rounded).    ANSWER