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What are the constant limits (obtained through change of variables) of the integral:

$$\iiint\limits_D \frac{dx \ dy \ dz}{(x+y+z+1)^3}, \quad \text{where} \; \; D=\left\{x>0,y>0,z>0,x+y+z<2 \right\}$$

I am supposed to use the 'triplequad' command in MATLAB to solve the above integral. To be correctly executed the triplequad command requires the bonds to be constant, so they cannot depend on variables. I believe the key lies in changing the variables but this is new to me so I was wondering if someone could lend me a helping hand!

Thanks in advance!

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  • $\begingroup$ Are you allowed to do the first integration analytically and the other two numerically, or should you do the three integrations numerically in one go? $\endgroup$ – Rogelio Molina Oct 12 '13 at 8:57
  • $\begingroup$ All three in one go :). $\endgroup$ – Soakr Oct 12 '13 at 9:00
  • $\begingroup$ I have thought about it, but all I can come up with is that you need to find a transformation from a tetrahedron (your original integration region) into a rectangular prism (where the integration limits will be constant for each variable). However I do not know how to find this transformation, I am curious to know what the answer will look like. $\endgroup$ – Rogelio Molina Oct 12 '13 at 9:05
  • $\begingroup$ I can also say that the change of variable you are looking for will not be a linear transformation, because bijective linear transformations map straight lines to straight lines, which is not what you need. It seems you need to "tailor" a transformation that maps your region into a rectangular 3D prism. $\endgroup$ – Rogelio Molina Oct 12 '13 at 10:56
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$\displaystyle{% {\cal I}\left(\mu\right) \equiv \int\int\int_{D_{\mu}} {{\rm d}x\,{\rm d}y\,{\rm d}z \over \left(x + y + z + 1\right)^{3}}\,, \qquad D_{\mu} \equiv \left\{\left(x,y,z\right)\ \ni\ x, y, z>0\,;\ x + y + z < \mu\right\}}$

We'll calculate ${\cal I}\left(2\right)$.

$$ {\cal I}\left(\mu\right) = \int_{0}^{\infty}\int_{0}^{\infty}{\rm d}x\,{\rm d}y\int_{0}^{\infty} {\Theta\left(\mu - x - y - z\right) \over \left(x + y + z + 1\right)^{3}} \,{\rm d}z $$
\begin{align} {\cal I}'\left(\mu\right) &= \int_{0}^{\infty}\int_{0}^{\infty}{\rm d}x\,{\rm d}y\int_{0}^{\infty} {\delta\left(\mu - x - y - z\right) \over \left(x + y + z + 1\right)^{3}} \,{\rm d}z \\[3mm]&= {1 \over \left(\mu + 1\right)^{3}} \int_{0}^{\infty}\int_{0}^{\infty}{\rm d}x\,{\rm d}y\int_{0}^{\infty} \delta\left(\mu - x - y - z\right) \,{\rm d}z \\[3mm]&= {1 \over \left(\mu + 1\right)^{3}} \int_{0}^{\infty}{\rm d}x\int_{0}^{\infty}{\rm d}y\, \Theta\left(\mu - x - y\right) = {1 \over \left(\mu + 1\right)^{3}} \int_{0}^{\infty}{\rm d}x\,\Theta\left(\mu - x\right)\int_{0}^{\mu - x}{\rm d}y\, \\[3mm]&= {1 \over \left(\mu + 1\right)^{3}} \int_{0}^{\mu}{\rm d}x\,\left(\mu - x\right) = {1 \over 2}\,{\mu^{2} \over \left(\mu + 1\right)^{3}} \end{align}
\begin{align} {\cal I}\left(\mu\right) &= {1 \over 2}\int_{0}^{\mu}{t^{2} \over \left(t + 1\right)^{3}}\,{\rm d}t = {1 \over 2}\int_{1}^{\mu + 1}{\left(t - 1\right)^{2} \over t^{3}}\,{\rm d}t = {1 \over 2}\int_{1}^{\mu + 1}\left(% {1 \over t} - {2 \over t^{2}} + {1 \over t^{3}} \right)\,{\rm d}t \\[3mm]&= {1 \over 2}\left[% \ln\left(\mu + 1\right) + {2 \over \mu + 1} - 2 - {1 \over 2\left(\mu + 1\right)^{2}} + {1 \over 2} \right] \end{align}

Set $\mu = 2$. The $\large\tt\mbox{final result}$ is $$\color{#ff0000}{\large% {1 \over 2}\left[\ln\left(3\right) - {8 \over 9}\right] \color{#000000}{\ \approx\ } 0.1049} $$

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Perform the following two changes of variables, they transform the tetrahedron region into a cube, first let $\xi = x$, $\eta=y,$ $\zeta= 2z/(2-x-y)$ (this transforms the region into a triangular prism of height 2 along the $z$ axis), then the transformation $x'=\xi$, $y'= 2\eta/(2-\xi)$, $z'=\zeta$ (which transforms the triangular prism into a cube of side 2), the total effect of both transformations is then

\begin{eqnarray} x'&=&x \\ y'&=&\frac{2y}{2-x} \\ z'&=& \frac{2z}{2-x-y} \end{eqnarray}

which can be inverted in all interior points of the cube (not in the boundary) \begin{eqnarray} x&=&x' \\ y&=&y' - \frac{x'y'}{2} \\ z&=&z'-\frac{(x'z'+y'z')}{2} + \frac{x'y'z'}{4} \end{eqnarray} The Jacobian of the transformation is

$$ J= \frac{1}{4}(2-x')(2-x'-y' + \frac{x'y'}{2}) $$ so the integral you want is:

$$ 2 \int_{0}^2 \int_{0}^2 \int_{0}^2 \frac{(2-x')(2-x'-y' + \frac{x'y'}{2})}{(2(x'+y'+z') - x'y'-x'z'-y'z' + \frac{x'y'z'}{2} +2)^3}dx'dy'dz' \approx .104862 $$

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