What is the other way to write this triple integral? I need to find all $5$ ways to write this integral and so far I have found $3$ of the $5$ but I need some help to find the last $2$ which are $dy$ $dz$ $dx$ and $dy$ $dx$ $dz$. I assume one of them might require you to add $2$ triple integrals  together. Anyways when evaluated the answer needs to come out to about $12.669$ which can be checked via Desmos. Can anyone help?
$$\int_0^2\int_{e^y}^{e^2}\int_0^{e^\left(\frac{y}{2}\right)} f(x,y,z) \ dz \ dx \ dy$$
 A: Please see this in a $3D$ online grapher tool or do rough $2D$ sketches to visualize better.
If you are going $dy$ first, you need to split the integral into two. For the order of integral $dy \ dz \ dx$, we make following observations.
At $y = 0, 0 \leq z \leq 1$. Also $y$ is not a function of $z$ for $0 \leq z \leq 1$.
a) For $0 \leq z \leq 1$, the lower limit of $y$ is $0$ and the upper limit of $y$ comes from the curve $x = e^y \implies y = \ln x$.
We already know that upper bound of $x$ is $e^2$. For lower bound, we note that at $y = 0, x = e^y = 1$.
b) For $z \geq 1$, the lower limit of $y$ comes from the curve $z = e^{y/2} \implies y = 2 \ln z$. Upper limit continues to be $y = \ln x$.
Also the upper bound of $z$ comes from intersection of curves $x = e^y, z = e^{y/2} \implies z = \sqrt x$.
So the integral becomes,
$\displaystyle \int_1^{e^2}\int_{0}^{1}\int_{0}^{\ln(x)}  f(x, y, z) \ dy  \ dz \ dx + 
\int_1^{e^2}\int_{1}^{\sqrt x}\int_{2 \ln(z)}^{\ln(x)} f(x, y, z) \  dy \ dz \ dx$
Now can you try and do it for the order $dy \ dx \ dz$? Let me know if you get stuck.
A: 
$$\begin{matrix}\begin{matrix}2\log z<y<\log x\\z^2<x<e^2\\1<z<e\end{matrix}&\begin{matrix}0<y<\log x\\1<x<e^2\\0<z<1\end{matrix}\end{matrix}$$
So the integral is $\int_1^e\int_{z^2}^{e^2}\int_{2\log z}^{\log x} dydxdz+\int_0^1\int_1^{e^2}\int_0^{\log x} dydxdz$. It consists of a slice of a rectangular prism on the bottom and a twisted part on the top.
