Changing order of integration for the triple integral $ \int\limits_{0}^{2} \int\limits_{0}^{2z} \int\limits_{y}^{2y} f_{(x,y,z)}\; dx\, dy\, dz $ I need to change order of integration for the following triple integral:
$$
\int\limits_{0}^{2}
\int\limits_{0}^{2z}
\int\limits_{y}^{2y}
f_{(x,y,z)}\; dx\, dy\, dz
$$
The domain of integration is graphically described in the following images:
Y-Z plane

X-Y plane

By setting $f_{(x,y,z)} = 1$, I verified my results, comparing them to the original integral using WA, which yields $\frac{16}{3} \cong 5.33$.
The requirement
What I'm trying to do is to change the order of integration, so that:


*

*The volume will be projected onto the X-Z plane

*Having x to be a function of z (or vice versa - z a function of x).


But I can't find a relation between them.
My accomplishments
I succeeded in converting the given integral to the following equivalents:
Volume projected onto the Z-X plane ($dz\, dx\, dy$)
This requires the limits of y to be scalars.
$$
\int\limits_{0}^{4}
\int\limits_{y}^{2y}
\int\limits_{\frac{y}{2}}^{2}
f_{(x,y,z)}\; dz\, dx\, dy
$$
WA's result.
Volume projected onto the X-Z plane ($dx\, dz\, dy$)
$$
\int\limits_{0}^{4}
\int\limits_{\frac{y}{2}}^{2}
\int\limits_{y}^{2y}
f_{(x,y,z)}\; dx\, dz\, dy
$$
WA's result.
Volume projected onto the Z-Y plane ($dz\, dy\, dx$)
$$
\int\limits_{0}^{4}
\int\limits_{\frac{x}{2}}^{x}
\int\limits_{\frac{y}{2}}^{2}
f_{(x,y,z)}\; dz\, dy\, dx
+
\int\limits_{4}^{8}
\int\limits_{\frac{x}{2}}^{4}
\int\limits_{\frac{y}{2}}^{2}
f_{(x,y,z)}\; dz\, dy\, dx
$$
WA's result for 1st integral and WA's result for 2nd integral.
 A: Here we have a 3D picture of the entire region ($x$ horizontal, $y$ depth, $z$ height):

In blue we see the $x$-$y$ plane, while the pink part stretches both through the $y$-$z$ plane and the $x$-$y$ plane. This side is determined by the equation $x = y$, so the image can be a bit off-putting. Suppose we cut at $y = 2$, then we obtain:

Here we see what's going on: for fixed $y$, our 3D shape actually is a square! In other words: given $y$, the dependence of $z$ on $x$ or vice versa is utterly trivial: the bounds do not change with varying $z$ or $x$.
This is of course already implied by the equations we obtained, but it's probably more convincing to see this actually happening. I hope that clears the air for you.
The images are courtesy of Mathematica 8's RegionPlot3D command:
RegionPlot3D[
 y <= x <= 2 y && 0 <= y <= 2 z && 0 <= z <= 2, {x, 0, 8}, {y, 0, 
  8}, {z, 0, 2}, Mesh -> None, PlotPoints -> 200]

RegionPlot3D[
 y <= x <= 2 y && 2 <= y <= 2 z && 0 <= z <= 2, {x, 0, 8}, {y, 0, 
  8}, {z, 0, 2}, Mesh -> None, PlotPoints -> 200]

