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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}

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  • $\begingroup$ 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
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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.

Comment:

  • 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.

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  • $\begingroup$ @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

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