The top surface is $z=18-16(x^2+y^2)$ and lower surface is $z=16(x^2+y^2)$. When these curves intersect we have $18 - 16(x^2+y^2) = 16(x^2+y^2)$ thus $x^2+y^2 = \tfrac{9}{16}$. This means the area over we are integrating is disk $x^2+y^2\leq \left( \tfrac{3}{4} \right)^2$.
$\iint_A 18 - 16(x^2+y^2) - 16(x^2+y^2) dA = \int_0^{2\pi} \int_0^{3/4} (18 - 32r^2)r ~ dr ~ d\theta$.