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double integral on non-rectangular area

Posted 10 years ago

How can I do it on Mathematica 9?

Ex: integral of sqrt(4-x^2-y^2) dy dx where x^2+y^2-2*x<=0.

POSTED BY: André Carvalho
9 Replies
Posted 10 years ago

Looking at my messy result, it is actually a sum of several integrals which Mathematica returned unevaluated.

POSTED BY: David Keith

Doing the same thing numerically (using NIntegrate) gives the result 19.9868 in 0.34 s. (Mathematica 10.0, Windows 8.1). Actually the result has a tiny imaginary part 5.5 *10^-24 as an artifact of the numerical method ('numerical noise').

POSTED BY: Ulrich Mutze

Where do you get the z from?

POSTED BY: Frank Kampas
Posted 10 years ago

1722 seconds on my laptop, for a very messy result. I wonder if we're solving the same integral.

Timing[
 Integrate[(x^2 + y^2)*Boole[(x - z)^2 + (y - z)^2 <= 1], {x, -1, 
   3}, {y, -1, 3}, {z, 0, 2}]
 ]
POSTED BY: David Keith

took 11.5 seconds on my computer.

POSTED BY: Frank Kampas
Posted 10 years ago

Me too, but it's an interesting 3D problem.

POSTED BY: David Keith

Integrate[(x^2 + y^2)*Boole[(x - z)^2 + (y - z)^2 <= 1], {x, -1, 3}, {y, -1, 3}, {z, 0, 2}]. I left my notebook computing about a half an hour and nothing happened.

POSTED BY: André Carvalho

Wow! Very good. Thanks!

POSTED BY: André Carvalho
In[1]:= Integrate[
 Sqrt[4 - x^2 - y^2]*
  Boole[x^2 + y^2 - 2 x <= 
    0], {x, -\[Infinity], \[Infinity]}, {y, -\[Infinity], \[Infinity]}]

Out[1]= 8/9 (-4 + 3 \[Pi])
POSTED BY: Frank Kampas
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