3
$\begingroup$

What are the limits of the integral: $$\iiint_{D} \frac {\mathrm{d}x\mathrm{d}y\mathrm{d}z}{(x + y + z + 1)^3}$$ where $$ D =\{ x > 0 , y > 0 , z > 0 , x + y + z < 2\}$$

I have previously done double integrals and was wondering if someone could help me derive the limits of this triple one.

Help is appreciated!

$\endgroup$
  • $\begingroup$ Hello, welcome to Math.SE. For some basic information about writing maths at this site see e.g. here, here, here and here. $\endgroup$ – Lord_Farin Oct 5 '13 at 10:02
2
$\begingroup$

Hint: $(x,y,z)\in D$ if and only if $0<x<2$, $0<y<2-x$, and $0<z<2-x-y$. Then, $$\underset{D}{\iiint}\frac{\mathrm{d}x\,\mathrm{d}y\,\mathrm{d}z}{(x+y+z+1)^3}=\int_{x=0}^{x=2}\left[\int_{y=0}^{y=2-x}\left(\int_{z=0}^{z=2-x-y}\frac{1}{(x+y+z+1)^3}\mathrm{d}z\right)\mathrm{d}y\right]\mathrm{d}x.$$ Be careful: since the limits depend on other variables in some of the integrals on the right-hand side, the order of integration must not be changed! That this trick works is a consequence of Fubini's famous theorem.

Detailed calculations: $$\int_{z=0}^{z=2-x-y}\frac{1}{(x+y+z+1)^3}\mathrm{d}z=\left[-\frac{1}{2(x+y+z+1)^2}\right]_{z=0}^{z=2-x-y}=-\frac{1}{18}+\frac{1}{2(x+y+1)^2}.$$

\begin{align*}\int_{y=0}^{y=2-x}\left(-\frac{1}{18}+\frac{1}{2(x+y+1)^2}\right)\mathrm{d}y=&\left[-\frac{y}{18}-\frac{1}{2(1+x+y)}\right]_{y=0}^{y=2-x}\\=&\,\frac{x-2}{18}-\frac{1}{6}+\frac{1}{2(1+x)}\end{align*}

\begin{align*}\int_{x=0}^{x=2}\left(\frac{x-2}{18}-\frac{1}{6}+\frac{1}{2(1+x)}\right)\mathrm{d}x=&\left[\frac{\dfrac{x^2}{2}-2x}{18}-\frac{x}{6}+\frac{\log (1+x)}{2}\right]_{x=0}^{x=2}\\=&\,-\frac{2}{18}-\frac{2}{6}+\frac{\log 3}{2}-0=-\frac{4}{9}+\frac{\log3}{2}.\end{align*}

$\endgroup$
  • $\begingroup$ Thank you for your response! I'm attempting to use Matlab in order to solve this integral (more specifically the 'triplequad' command) and thus I need constant limits which are not allowed to have bounds that depend upon variables. How would one go about in order to obtain that? $\endgroup$ – Soakr Oct 5 '13 at 11:13
  • $\begingroup$ @Soakr No need for MATLAB. This function is not hard to integrate symbolically (i.e., using the usual rules to compute indefinite integrals) and then use the Newton–Leibniz rule. See my computations above in the edited answer. $\endgroup$ – triple_sec Oct 5 '13 at 11:40
  • $\begingroup$ Thank you for your detailed calculations! Indeed you prove that MATLAB is a bit excessive for this function. The reason behind me asking for the constant limits is due to that this integral is part of an exercise which is supposed to be solved both by hand and by the use of the 'triplefunction' of MATLAB. $\endgroup$ – Soakr Oct 5 '13 at 12:07
  • $\begingroup$ I think if you insist on constant limits, then maybe should change coordinates—I'm not sure. Alternatively, you may want to figure out how the triplefunction command can handle variable limits (type “help triplefunction” to explore the options or try googling up alternative MATLAB commands). $\endgroup$ – triple_sec Oct 5 '13 at 12:30
  • $\begingroup$ Ok, I will look into it! Thank you for all your help and time. $\endgroup$ – Soakr Oct 5 '13 at 12:55
3
$\begingroup$

Method 1

A trivial rescaling of the original integral ( substitute $x,y,z$ by $2x,2y,2z$ ) can bring it to the form of a type 1 Dirichlet integral

Let $f$ be a function defined on $[0,1]$ and $\Delta \subset \mathbb{R}^d$ be the $d$-simplex $0 \le x_1, \ldots x_d$; $x_1+\cdots+x_d \le 1$, we have

$$\int_{\Delta} f(\sum_{i=1}^d x_i) \prod_{i=1}^{d} x_i^{\alpha_i-1} d^d x =\frac{\prod_{i=i}^d \Gamma(\alpha_i)}{\Gamma(\sum_{i=1}^d \alpha_i)} \int_0^1 f(x) x^{\sum_{i=1}^d a_i)-1} dx$$

In this case $d = 3, f(x) = \frac{1}{(2x+1)^3}$ and $a_1 = a_2 = a_3 = 1$. It is then clear the integral is equal to

$$\frac{2^3\Gamma(1)^3}{\Gamma(3)}\int_0^1 \frac{x^{(3-1)}}{(2x+1)^3} dt = 4 \left[ \frac{8x + 2(2x+1)^2\log(2x+1)+3}{16(2x+1)^2}\right]_0^1 = \frac{\log 3}{2} - \frac{4}{9} $$

Method 2

If one absolutely want to evaluate this as a multiple integral, there is a useful trick to turn the integral as one over a hypercube $[0,1]^3$. Perform the substitution $x, y, z$ by $2x, 2y, 2z$ as before and introduce variables $\lambda, \mu, \nu$:

$$\begin{cases} \lambda & = x + y + z\\ \lambda \mu & = y + z\\ \lambda \mu \nu & = z \end{cases} \quad\iff\quad \begin{cases} x &= \lambda (1-\mu)\\ y &= \lambda \mu (1-\nu)\\ z &= \lambda \mu \nu \end{cases}$$

We have $$\begin{align} dx \wedge dy \wedge dz &= d(x+y+z) \wedge d(y+z) \wedge dz = d\lambda \wedge \lambda d\mu \wedge \lambda\mu d\nu = \lambda^2\mu\,d\lambda \wedge d\mu \wedge d\nu \end{align}$$

and the integral we want becomes

$$2^3 \int_{[0,1]^3} \frac{\lambda^2 \mu}{(2\lambda + 1)^3} d\lambda d\mu d\nu = 4 \int_0^1 \frac{\lambda^2}{(2\lambda + 1)^3} d\lambda$$ The same integral we got as a Dirichlet integral.

$\endgroup$

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.