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. 

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.
