# How to swap the order of these three integration around? The region is $0<x,y<z<\infty$.

I am considering the integration as follows $$\int_0^\infty\int_0^z\int_0^z f(x,y,z)dxdydz.$$ I wanted the integration such that the $$dz$$ is the inner integral, so I wanted it to be in the form of $$\int\int\int f(x,y,z)dzdxdy.$$

What I have done so far is swapping $$y,z$$ first, so I got $$\int_{0}^{\infty}\int_y^{\infty}\int_0^zf(x,y,z)dxdzdy.$$

Now I focussed on $$\int_y^{\infty}\int_0^zf(x,y,z)dxdz.$$ However I had some trouble swapping this. I drew a diagram and I had to split the region up to two regions and I got this $$\int_y^{\infty}\int_0^zf(x,y,z)dxdz=\int_0^y\int_y^\infty f(x,y,z)dzdx+\int_y^{\infty}\int_x^\infty f(x,y,z)dzdx$$

I am not too sure if I am correct up to now, could someone please advise if I have done alright so far? Many thanks

Let $$J$$ be the integral to be reshaped. Then \begin{aligned} J &=\int_{z\in(0,\infty)}\iint_{(x,y)\in(0,z)\times(0,z)}f(x,y,z)\; dy\; dx\; dz \\ &=\iiint_{0 (One can split differently the two $$\iint$$ at the last step, my choice was the one putting the infinity as upper limit for each occuring integral.)