# 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?

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 $$\int_{x=0}^8\int_{z= x^{2/3}}^{4} \int_{y=z^{1/2}}^2$$
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 so that we know what the maximum is. This is what they have done.