
Re: Integrals involving square roots
Posted:
Mar 5, 2007 5:06 AM


In article <es90ol$2lt$1@smc.vnet.net>, "Andre Giroux" <giroux@dms.umontreal.ca> wrote:
> Hi All. > Mathematica is very unreliable when it computes integrals involving square > roots. The following were obtained running Mathematica 5.2 and 5.1 with > Windows XP pro: > > In[1]:= > \!\(\[Integral]\_\(3\)\%3\(\[Integral]\_\(\@\(9  x\^2\)\)\%\(\@\(9  > \x\^2\)\)\(\[Integral]\_0\%\(y  3\)2 y \[DifferentialD]z \[DifferentialD]y > \[DifferentialD]x\)\)\) > Out[1]= > \!\(\(81\ \[Pi]\)\/2\) > In[5]:= > \!\(\[Integral]\_0\%2\(\[Integral]\_\(R\)\%R\( Rz\^2\/\@\(R\^2  x\^2\)\) > \\[DifferentialD]x \[DifferentialD]z\)\) > Out[5]= > \!\(\(2\)\ \[Pi]\ Rz\^2\) > > In both cases, the sign is wrong and, with the first integral, it is not > obvious. Interestingly, Mathematica 5.0 computes these integrals correctly. > So beware!
More than not obvious. I assume that the first integral is
Integrate[2 y, {x, 3, 3}, {y, Sqrt[9  x^2], Sqrt[9  x^2]}, {z, 0, y  3}]
for which Mathematica returns.
(81 Pi)/2
As far as I can see, this integral _is_ correct. It certainly agrees with the numerical integral:
NIntegrate[2 y, {x, 3, 3}, {y, Sqrt[9  x^2], Sqrt[9  x^2]}, {z, 0, y  3}]
127.23450236818807
For the second integral, there is no space between R and z^2 in your input expression. Moreover, telling Mathematica that R > 0, yields the correct answer:
Assuming[R > 0, Integrate[(R z^2)/Sqrt[R^2  x^2], {z, 0, 2}, {x, R, R}]]
(8 Pi R)/3
Cheers, Paul
_______________________________________________________________________ Paul Abbott Phone: 61 8 6488 2734 School of Physics, M013 Fax: +61 8 6488 1014 The University of Western Australia (CRICOS Provider No 00126G) AUSTRALIA http://physics.uwa.edu.au/~paul

