Question: Consider a triple integral of the following form \begin{equation} \int_{x=0}^1 \int_{y=0}^{1} \int_{z=0}^{1} f(x,y,z)dzdydx. \end{equation} Because of the specific $f(\cdot,\cdot,\cdot)$ function I am dealing with, I would like to convert the above integral into the following form \begin{equation} \int_{x=0}^1 \int_{y=0}^{x} \int_{z=0}^{y} g(x,y,z)dzdydx, \end{equation} where $g(\cdot,\cdot,\cdot)$ is an appropriately defined function. This form is easier to work with because the integrands are nicely ordered as $x\ge y \ge z.$

Example: I am able to do a similar trick for a double integral. Consider \begin{equation} \int_{x=0}^1 \int_{y=0}^{1} f(x,y)dydx. \end{equation} This integral is equivalent to the sum of two integrals: \begin{equation} \int_{x=0}^1 \int_{y=0}^{1} f(x,y)dydx=\int_{x=0}^1 \int_{y=0}^{x} f(x,y)dydx+\int_{x=0}^1 \int_{y=x}^{1} f(x,y)dydx. \end{equation} By changing the order of the integration in the last integral above, we obtain \begin{equation} \int_{x=0}^1 \int_{y=x}^{1} f(x,y)dydx = \int_{y=0}^1 \int_{x=0}^{y} f(x,y)dxdy. \end{equation} Renaming $x$ as $y$ and vice-versa on the integral in the right hand side, we obtain \begin{equation} \int_{x=0}^1 \int_{y=0}^{x} h(x,y)dydx \end{equation} for an appropriate $h(x,y).$ Thus, the original integral is given as \begin{align} \int_{x=0}^1 \int_{y=0}^{1} f(x,y)dydx &=\int_{x=0}^1 \int_{y=0}^{x} f(x,y)dydx+\int_{x=0}^1 \int_{y=0}^{x} h(x,y)dydx\\ &= \int_{x=0}^1 \int_{y=0}^{x} \left[ f(x,y)+h(x,y) \right] dydx. \end{align}

  • $\begingroup$ This question (particularly the example) bears a lot of similarity to my earlier question: math.stackexchange.com/questions/1403199/… solved by @Brian M. Scott $\endgroup$ – emper Mar 29 '17 at 6:53
  • $\begingroup$ Why don't you use the same 'trick' on $dz,dy$ then $dy,dx$ respectively ? $\endgroup$ – Zaid Alyafeai Mar 29 '17 at 10:38
  • $\begingroup$ @ZaidAlyafeai Thanks for the suggestion but that does not really work. In particular, after the first step of changing the order between $z$ and $y,$ I obtain $\int_{x=0}^{1}\int_{y=0}^{1}\int_{z=0}^{y}.$ Applying the trick again on $y$ and $x$ causes me to rename the upper bound on the final integral to $x.$ That defeats the purpose of the exercise to obtain ordered integrands in the form of $x\ge y \ge z.$ $\endgroup$ – emper Mar 29 '17 at 19:27

We iteratively apply the two-dimensional transformation. In order to do so we recall the slightly more generalized formula \begin{align*} \int_{x=0}^z\int_{y=0}^zf(x,y)\,dx\,dy =\int_{x=0}^z\int_{y=0}^xf(x,y)\,dy\,dx+\int_{y=0}^z\int_{x=0}^yf(x,y)\,dx\,dy\tag{1} \end{align*}

We obtain \begin{align*} \int_{x=0}^1&\left(\int_{y=0}^1\int_{z=0}^1f(x,y,z)\,dz\,dy\right)\,dx\\ &=\int_{x=0}^1\left(\int_{y=0}^1\int_{z=0}^yf(x,y,z)\,dz\,dy\right)\,dx\\ &\qquad+\int_{x=0}^1\left(\int_{z=0}^1\int_{y=0}^zf(x,y,z)\,dy\,dz\right)\,dx\tag{2}\\ &=\int_{x=0}^1\int_{y=0}^1g(x,y)\,dy\,dx +\int_{x=0}^1\int_{z=0}^1h(x,z)\,dz\,dx\tag{3}\\ &=\int_{x=0}^1\int_{y=0}^xg(x,y)\,dy\,dx +\int_{y=0}^1\color{blue}{\int_{x=0}^yg(x,y)\,dx}\,dy\\ &\qquad+\int_{x=0}^1\int_{z=0}^xh(x,z)\,dz\,dx +\int_{z=0}^1\color{red}{\int_{x=0}^zh(x,z)\,dx}\,dz\tag{4}\\ &=\int_{x=0}^1\int_{y=0}^x\int_{z=0}^yf(x,y,z)\,dz\,dy\,dx\\ &\qquad+\int_{y=0}^1\left(\color{blue}{\int_{x=0}^y\int_{z=0}^yf(x,y,z)\,dz\,dx}\right)\,dy\\ &\qquad+\int_{x=0}^1\int_{z=0}^x\int_{y=0}^zf(x,y,z)\,dy\,dz\,dx\\ &\qquad+\int_{z=0}^1\left(\color{red}{\int_{x=0}^z\int_{y=0}^zf(x,y,z)\,dy\,dx}\right)\,dz\tag{5}\\ &=\int_{x=0}^1\int_{y=0}^x\int_{z=0}^yf(x,y,z)\,dz\,dy\,dx\\ &\qquad+\int_{y=0}^1\left(\color{blue}{\int_{x=0}^y\int_{z=0}^xf(x,y,z)\,dz\,dx}\right)\,dy\\ &\qquad+\int_{y=0}^1\left(\color{blue}{\int_{z=0}^y\int_{x=0}^zf(x,y,z)\,dx\,dz}\right)\,dy\\ &\qquad+\int_{x=0}^1\int_{z=0}^x\int_{y=0}^zf(x,y,z)\,dy\,dz\,dx\\ &\qquad+\int_{z=0}^1\left(\color{red}{\int_{x=0}^z\int_{y=0}^xf(x,y,z)\,dy\,dx}\right)\,dz\\ &\qquad+\int_{z=0}^1\left(\color{red}{\int_{y=0}^z\int_{x=0}^yf(x,y,z)\,dx\,dy}\right)\,dz\tag{6} \end{align*} and we are done.


  • In (2) we apply (1) to the bracketed double integral on the left-hand side.

  • In (3) write $g(x,y)=\int_{z=0}^yf(x,y,z)\,dz$ and $h(x,z)=\int_{y=0}^zf(x,y,z)\,dy$

  • In (4) we apply again (1) twice, once for the double integral with integrand $g(x,y)$ and once for the double integral with integrand $h(x,z)$.

  • In (5) we substitute back for $g(x,y)$ and $h(x,z)$ and observe the bracketed double-integrals need one more transformation according to (1)

  • In (6) we finally do this last transformation to the bracketed double integrals.

  • $\begingroup$ @emper: Thanks a lot for accepting my answer and granting the bounty! :-) $\endgroup$ – Markus Scheuer Apr 10 '17 at 6:40
  • $\begingroup$ Sure, very well deserved! $\endgroup$ – emper Apr 11 '17 at 4:06

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.