The usual way to solve this is by using double integrals.
The volume is given by the summation of zdxdy, which is the volume of a tiny cuboid with length z, width dx and height dy. The limits are those of the rectangular width and height. For the purposes of illustration S will be used as the integration sign and square brackets will be used for the limits: [lower,upper]. S[A,B](...)dv represents the integral between A and B of the expression in the round brackets with respect to variable v. When integration takes place the variable which is not the one implied in the integration is treated as a constant, so ax^2 is treated as a constant when integrating with respect to y.
z=ax^2+by; volume=S[0,3](S[1,2]zdy)dx=S[0,3](S[1,2](ax^2+by)dy)dx=
S[0,3](ax^2y+by^2/2)[1,2])dx=S[0,3](2ax^2-ax^2+2b-b/2)dx=S[0,3](ax^2+3b/2)dx=
(ax^3/3+3bx/2)[0,3]=9a+9b/2=volume.
The same answer is generated if x is integrated first before y.