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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

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1 Answer 1

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Let $J$ be the integral to be reshaped. Then $$ \begin{aligned} J &=\int_{z\in(0,\infty)}\iint_{(x,y)\in(0,z)\times(0,z)}f(x,y,z)\; dy\; dx\; dz \\ &=\iiint_{0<x,y<z}f(x,y,z)\;dx\; dy\; dz \\ &=\iiint_{0<x,y\le \max(x,y)<z}f(x,y,z)\;dx\; dy\; dz \\ &=\iint_{(x,y)\in(0,\infty)^2}dx\; dy\int_{z=\max(x,y)}^\infty f(x,y,z)\; dz \\[2mm] &\qquad\text{ (and we can stop here or further split...)} \\[2mm] &= \iint_{\substack{(x,y)\in(0,\infty)^2\\x<y}}dx\; dy\int_{z=\max(x,y)}^\infty f(x,y,z)\; dz % \\ % &\qquad + \iint_{\substack{(x,y)\in(0,\infty)^2\\y<x}}dx\; dy\int_{z=\max(x,y)}^\infty f(x,y,z)\; dz \\ &= \iint_{\substack{(x,y)\in(0,\infty)^2\\x<y}}dx\; dy\int_y^\infty f(x,y,z)\; dz % \\ % &\qquad + \iint_{\substack{(x,y)\in(0,\infty)^2\\y<x}}dx\; dy\int_x^\infty f(x,y,z)\; dz \\[2mm] &\qquad\text{ (and we can stop here or further split...)} \\[2mm] &= \int_0^\infty dx\int_x^\infty dy\int_y^\infty f(x,y,z)\; dz % \\ % &\qquad + \int_0^\infty dy\int_y^\infty dx\int_x^\infty f(x,y,z)\; dz \ . \end{aligned} $$ (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.)

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