Hello, KK!
List five other iterated integrals that equals to:
. . $\displaystyle a)\;\int_0^1 \int_y^1 \int_0^y f(x,y,z)\, dz\, dx\, dy$
I write in the limits "completely".
. .$\displaystyle \int^{y = 1}_{y = 0}\int^{x=1}_{x=y}\int^{z=y}_{z=0}f(x,y,z)\,dz\,dx \,dy$
I start with the "outside" limits: .$\displaystyle y = 0,\;y = 1$
That is, $\displaystyle y$ ranges from $\displaystyle 0$ to $\displaystyle 1.$ Code:
y

  1+     
::::::::::::::
::::::::::::::
::::::::::::::
+ x

The next limits are: .$\displaystyle x = y,\;x = 1$
That is, $\displaystyle x$ ranges between the line $\displaystyle x = y$ and $\displaystyle x = 1$ Code:
y

  1+    * 
 /:
 /:::
 /:::::
/+ x
 1
Now lay this diagram "on the floor"; the zaxis goes straight up. Code:
z



    * y
*::*
*:::::*
*::::::::*
* * * * * * *
/
x
I can't draw this next part . . . hope you can follow.
With no restriction on $\displaystyle z$, we have a triangular prism
. . with the above triangle as a crosssection, extending up and down infinitely.
But the limits are: $\displaystyle z = 0$ and $\displaystyle z = y.$
. . That is, $\displaystyle z$ ranges between the "floor" and the slanted plane $\displaystyle z = y.$
So our triangular prism is cut off at floorlevel below
. . and the slanted plane $\displaystyle z = y$ above.
And that is the solid we are dealing with.
Now if we change the order of integration, the limits are changed.
With the first of the answers, we have: $\displaystyle dz\,dy\,dx$
We will try to describe the same solid using this order of limits.
Starting outside: $\displaystyle x$ goes from $\displaystyle x = 0$ to $\displaystyle x = 1$
Then: $\displaystyle y$ goes from $\displaystyle y = 0$ to $\displaystyle y = x$
Then: $\displaystyle z$ goes from $\displaystyle z = 0$ to $\displaystyle z = y$
So the integral is: .$\displaystyle \int_{x=0}^{x=1}\int_{y=0}^{y=x}\int_{x=0}^{z=y}f( x,y,z)\,dz\,dy,\,dx$
. . I hope this helps . . .