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**lllll** I'm having some trouble finding the area under $\displaystyle z= 18-2x^2-2y^2$ and above the xy-plane

starting off $\displaystyle z= 18-2(x+3)(y-3)$ and $\displaystyle z= 18-2(y+3)(x-3)$

so the paraboloid intersects the xy-plane $\displaystyle -3 \leq x \leq 3$ and $\displaystyle -3 \leq y \leq 3$ Mr F says: No. This implies that the region of integration in the xy-plane is a square and, as we both know, it's not. It's a disk (see below).

taking the integral would wield $\displaystyle \iint_D 18-2x^2-2y^2 \ dA$

(this is the part I'm not sure off) coveting to polar form yields $\displaystyle \int_0^{2\pi} \int_0^{{\color{red}3}}(18-2r^2)r \ dr \ d \theta$