give 5 other equivalent iterated triple integrals I am given the following integral:
$$\int_0^2\int_0^{y^3}\int_0^{y^2}f(x,y,z) dzdxdy $$ 
I was successfully able to rewrite this in its dzdydx, dxdzdy, and dxdydz forms, but I'm having a hard time understanding the iterated triple integrals where dy is the inner integral.
The solution for dydzdx is as follows:
$$\int_0^8\int_0^{x^{2/3}}\int_{x^{1/3}}^2f(x,y,z)dydzdx\,+\,\int_0^8\int_{x^{2/3}}^4\int_{z^{1/2}}^2f(x,y,z)dydzdx $$
I don't understand why the integral was split. How do I figure out what this shape looks like in the zx plane in order to even know that it needed to be split? 
 A: The conditions on $y$ are $y^{2} >z$ and $y^{3} >x$. We have to integrate w.r.t. $y$ from the maximum of $\sqrt z$ and $x^{1/3}$ to $2$. In order to carry this out it is convenient to split the possible values of $x$ and $z$ into two parts: $x>z^{2/3}$ and $x<z^{2/3}$ so that we know what the maximum is. This is what they have done. 
A: The region of the integral is characterized by 
$$
0 \leq y \leq 2
\\ 0 \leq x \leq y^3 \\ 0 \leq z \leq y^2
$$
If we have in mind that $x$ should be the outermost variable, the first thing is to deduce that  $$0 \leq x \leq y^3 \leq 2^3 = 8 \\ 0 \leq z \leq y^2 \leq 4$$
The $0 \leq x \leq 2^3 = 8$ part gives the outer integral limits.
And we also have two lower-boundary conditions on $y$, namely
$$ y \geq x^{1/3} \\ y \geq z^{1/2}$$
And here is why the integral has to be split up -- we have no clue whether $x^{1/3}$ is more or less than $z^{1/2}$.  If it is greater or equal, then $z\leq x^{2/3}$ and the limits will be
$$
\int_{x=0}^8\int_{z= 0}^{x^{2/3}} \int_{y=x^{1/3}}^2
$$
and if  $x^{1/3} < z^{1/2}$ the upper limit on $z$ is given by $z \leq y^2 \leq 4$ and the  limits will be
$$
\int_{x=0}^8\int_{z= x^{2/3}}^{4} \int_{y=z^{1/2}}^2
$$
