I have the equation ...
Determine the volume for $\displaystyle f(x,y)=x^2+y^2+1$; 0<x<3 (greater than or equal to on the <, I didn't know how to do that); 0<y<3 (also greater than or equal to)
Be careful of that z limit. It wouldn't be 2Pi, unless you're using polar.
$\displaystyle z=x^{2}+y^{2}+1$
Perhaps try this:
$\displaystyle \int_{0}^{3}\int_{0}^{3}\int_{0}^{x^{2}+y^{2}+1}dz dydx$
A double integral may be better:
$\displaystyle \int_{0}^{3}\int_{0}^{3}[x^{2}+y^{2}+1]dydx$