2
$\begingroup$

In Mathematica how can I compute this integral:$$ \iiint_{D}\sqrt{(1-9z^2)(1-4y^2-9z^2)}\,dx\,dy\,dz$$ where D is the domain:

$$D: x^2 +4y^2+9z^2\le1$$

Please I need help!!!

$\endgroup$
  • $\begingroup$ I asked how to in fact... $\endgroup$ – user14738 Jun 3 '14 at 15:26
  • $\begingroup$ Where.., there are Integrate Sqrt and all.. :) $\endgroup$ – Öskå Jun 3 '14 at 15:27
  • $\begingroup$ I know...but which is the syntax I've to use? $\endgroup$ – user14738 Jun 3 '14 at 15:28
  • $\begingroup$ Yeah...but no one talks about triple integrals $\endgroup$ – user14738 Jun 3 '14 at 15:29
  • 3
    $\begingroup$ Please, check the documentation of Integrate. Specifically, look into the Scope section, subsection Integrals over Regions $\endgroup$ – Sjoerd C. de Vries Jun 3 '14 at 15:31
4
$\begingroup$

I just did this:

Integrate[Boole[x^2 + 4*y^2 + 9*z^2 <= 1]*
 Sqrt[(1 - 9*z^2)*(1 - 4*y^2 - 9*z^2)], {z, -Infinity, Plus[Infinity]},{y, -Infinity, Plus[Infinity]}, {x, -Infinity, Plus[Infinity]}]

MMA quickly returned

64/135
| improve this answer | |
$\endgroup$
  • 1
    $\begingroup$ The order in which you did the integrations is important — doing x then y then z gives the quickest route through the nested integral. Other orderings seem not to evaluate. $\endgroup$ – Stephen Luttrell Jun 3 '14 at 19:42
  • 1
    $\begingroup$ Heh. Sometimes it's better to be lucky than to be smart! $\endgroup$ – Cassini Jun 3 '14 at 19:56
4
$\begingroup$

This is perhaps too "creative". Some health checks needed for the series behavior:

Graphics`Region`RegionInit[];
region = (x x + 4 y y + 9 z z <= 1);
paregion = Region`ParametricRegion[{{x, y, z}, region}];
k = FullSimplify@
     Normal@Series[Sqrt[(1-9 z^2) (1-9 z^2-4 y^2)], {z,0, #},{y,0, #}] &/@ Range[1, 10, 2];
res = N@Integrate[#, {x, y, z} ∈ paregion] & /@ k

(* {0.698132, 0.488692, 0.480154, 0.477423, 0.476201} *)

So the result is near to 0.476

ListLinePlot@res

Mathematica graphics

This is where I've read first about this way for using Integrate[]

| improve this answer | |
$\endgroup$
  • $\begingroup$ I don't think this is the most practical way to integrate this. I'm posting it more as a curiosity. $\endgroup$ – Dr. belisarius Jun 3 '14 at 19:10
  • $\begingroup$ +1 for the undocumented integrate use :) $\endgroup$ – RunnyKine Jun 3 '14 at 19:10
  • $\begingroup$ @RunnyKine Thanks for remembering me that! Added a link. $\endgroup$ – Dr. belisarius Jun 3 '14 at 19:51

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