Constant limits of integral: ∭D dxdydz/(x+y+z+1)^3, where D={x>0,y>0,z>0,x+y+z<2}

Question

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!

Answer 1

$\displaystyle{% {\cal I}\left(\mu\right) \equiv \int\int\int_{D_{\mu}} {{\rm d}x\,{\rm d}y\,{\rm d}z \over \left(x + y + z + 1\right)^{3}}\,, \qquad D_{\mu} \equiv \left\{\left(x,y,z\right)\ \ni\ x, y, z>0\,;\ x + y + z < \mu\right\}}$

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 $$\color{#ff0000}{\large% {1 \over 2}\left[\ln\left(3\right) - {8 \over 9}\right] \color{#000000}{\ \approx\ } 0.1049} $$

Answer 2

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:

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