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!!!
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2 Answers
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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
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1$\begingroup$ The order in which you did the integrations is important — doing
xthenythenzgives the quickest route through the nested integral. Other orderings seem not to evaluate. $\endgroup$ Commented Jun 3, 2014 at 19:42 -
1$\begingroup$ Heh. Sometimes it's better to be lucky than to be smart! $\endgroup$– CassiniCommented Jun 3, 2014 at 19:56
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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
This is where I've read first about this way for using Integrate[]
answered Jun 3, 2014 at 16:57
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$\begingroup$ I don't think this is the most practical way to integrate this. I'm posting it more as a curiosity. $\endgroup$ Commented Jun 3, 2014 at 19:10
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$\begingroup$ +1 for the undocumented integrate use :) $\endgroup$ Commented Jun 3, 2014 at 19:10
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$\begingroup$ @RunnyKine Thanks for remembering me that! Added a link. $\endgroup$ Commented Jun 3, 2014 at 19:51
IntegrateSqrtand all.. :) $\endgroup$Integrate. Specifically, look into the Scope section, subsection Integrals over Regions $\endgroup$