Find other iterated integrals equal to the given triple integral I want to find all the other five iterated integrals that are equal to the integral
$$\int_{1}^{0}\int_{y}^{1}\int_{0}^{y} f(x,y,z)dzdxdy$$
I have so far found these three
$$\int_{0}^{1}\int_{0}^{x}\int_{0}^{y} f(x,y,z)dzdydx$$
$$\int_{0}^{1}\int_{0}^{y}\int_{y}^{1} f(x,y,z)dxdzdy$$
$$\int_{0}^{1}\int_{z}^{1}\int_{y}^{1} f(x,y,z)dxdydz$$
These three were easy to find as I had to look at the XY plane for the first one and ZY plane for the last two. Now to find the rest, I have to look at the XZ plane. I am not able to do it. I would appreciate if someone can help. 
 A: The bounds of integration determine equations that bound a solid $S$ in three-dimensional space. After you integrate with respect to the first variable, you should orthogonally project $S$ along the axis specified by that first variable onto the plane spanned by the other two variables.  That projection then determines a two-dimensional region which is the domain over which you integrate next.
In your original integral, you have
$$\int_{1}^{0}\int_{y}^{1}\int_{0}^{y} f(x,y,z)\,dz\,dx\,dy = -\int_{0}^{1}\int_{y}^{1}\int_{0}^{y} f(x,y,z)\,dz\,dx\,dy.$$
Note that I have switched the outer bounds of integration and changed the sign of the integral to alleviate the minor annoyance that $1$ is not less than $0$.  Now, the bounds of integration imply three inequalities that specify the domain of integration.
$$
\begin{array}{l}
  0<z<y \\
  y<x<1 \\
  0<y<1.
\end{array}
$$
Now, the solid, together with the projections of interest, looks like so:

Now, I guess the reason that the $yz$ and $xy$ projections are relatively easy is that those projections correspond to cross-sections of the solid, i.e. $z=0$ for the $xy$ projection and $x=1$ for the $yz$ projection.  This allows us to just set the variable that we integrate with respect to first to that constant.  It's even easier here, since neither $x$ nor $z$ appear explicitly in the bounds.  We can't do that here but, when we think of it as a projection as we have here, we can see that it's just the triangle 
$$\left\{(x,z): 0<x<1 \text{ and } 0<z<x \right\}.$$
A: Hint: Rewrite your integral as:
$$\int_0^1 \int_y^1 \int_0^y f(x,y,z) \;dz\,dx\,dy
= \int_0^1 \int_0^1 \int_0^1 f(x,y,z) \,\mathbf 1_{x \ge y \ge z} \;dz\,dx\,dy,$$
where the indicator function $\mathbf 1_{x \ge y \ge z}$ equals $1$ if ${x \ge y \ge z}$, and $0$ otherwise.
You can now easily rearrange the order of integration.  Then adjust the limits for the inner integrals so that you only integrate over the volume where the  $\mathbf 1_{x \ge y \ge z} = 1$ (i.e. where $x \ge y \ge z$), and drop the indicator function from the integrand again (since it's now, once more, identically equal to $1$ over the entire volume of integration).
For example, for the $dy\,dz\,dx$ order, by noting that $x \ge y \ge z \iff x \ge y,\ x \ge z,\ y \ge z$, we get:
$$\int_0^1 \int_0^1 \int_0^1 f(x,y,z) \,\mathbf 1_{x \ge y \ge z} \;dy\,dz\,dx
= \int_0^1 \int_0^x \int_z^x f(x,y,z) \;dy\,dz\,dx.$$
You should be able to deal with the last case ($dy\,dx\,dz$) in the same manner.

More generally, the important thing to realize is that your original integral, where the limits of the inner integrals depend on the value of the outer integration variables, is really just a volume integral over a subset $A = \{(x,y,z) \in \mathbb R^3 \mid 1 \le z \le y \le x \le 1\} \subset \mathbb R^3$ of the whole space.  Thus, we can rewrite it as:
$$\iiint\limits_A f(x,y,z) \;dx\,dy\,dz
= \iiint\limits_{\mathbb R^3} f(x,y,z) \,\mathbf 1_A \;dx\,dy\,dz,$$
where the order of integration no longer matters.
