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. 

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 
$$\left\{(x,z): 0<x<1 \text{ and } 0<z<x \right\}.$$
