# Thread: Volume under solid and polar integration

1. ## Volume under solid and polar integration

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

starting off $z= 18-2(x+3)(y-3)$ and $z= 18-2(y+3)(x-3)$ so the paraboloid intersects the xy-plane $-3 \leq x \leq 3$ and $-3 \leq y \leq 3$

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

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

2. Originally Posted by lllll
I'm having some trouble finding the area under $z= 18-2x^2-2y^2$ and above the xy-plane

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

so the paraboloid intersects the xy-plane $-3 \leq x \leq 3$ and $-3 \leq y \leq 3$ 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 $\iint_D 18-2x^2-2y^2 \ dA$

(this is the part I'm not sure off) coveting to polar form yields $\int_0^{2\pi} \int_0^{{\color{red}3}}(18-2r^2)r \ dr \ d \theta$
See the correction (in red). The area of the xy-plane you're integrating over is the disk $x^2 + y^2 \leq 9$. The radius of this disk is 3, not $\sqrt{3}$.

To get the region, solve $z = 0 \Rightarrow 0= 18-2x^2-2y^2$.