The volume of the solid whose base is the square in the x y- plane 
[0,1] ×[1,2] and whose height is the function f(x,y) = (x^2)y is represented by the double integral, or nested integral (∫_0^1)(∫_1^2)(x^2)y dy dx =(∫_0^1)((∫_1^2)(x^2)y dy) dx .  Note that this base is "simple".Is it a

RegionPlot3D[z<x^2*y ,{x,0,1},{y,1,2},{z,0,3},Mesh->None,PlotRange->{{-1,2},{0,3},{0,2}},AxesOrigin->{0,0,0},Boxed->False,BoxRatios->{1,1,.5},AxesLabel->{x,y,z}]

This solid has as its base the half-disk of radius 1, where y≥0, in the xy- plane, and height given by the function g(x,y) = 4-y.  It volume is represented by the double integral,  (∫_(-1)^1)(∫_0^(Sqrt[1-(x^2)]))(4-y) dy dx =(∫_(-1)^1)((∫_0^(Sqrt[1-(x^2)]))(4-y) dy) dx.  The base here is y-simple.  Is it also x-simple?

RegionPlot3D[z≤4-y&& y≤Sqrt[1-x^2],{x,-1,1},{y,0,2},{z,0,5},Mesh->None,PlotRange->{{-1.5,1.5},{0,3},{0,5}},AxesOrigin->{0,0,0},Boxed->False,BoxRatios->{1,1,.5},AxesLabel->{x,y,z}]

Here is an example of an x-simple region.  The base is the region between the two curves x=0 and x=2(Sqrt[y]), for 0≤y≤(Sqrt[π]).  Find (∫_0^(Sqrt[π]))(∫_0^(2(Sqrt[y])))x sin((y^2)) dx dy.

RegionPlot3D[z≤x*Sin[y^2]&& x≤2*Sqrt[y],{x,0,4},{y,0,Sqrt[Pi]},{z,0,5},Mesh->None,PlotRange->{{-1.5,1.5},{0,3},{0,5}},AxesOrigin->{0,0,0},Boxed->False,BoxRatios->{1,1,.5},AxesLabel->{x,y,z}]