Mathematica 9 is now available
Services & Resources / Wolfram Forums
-----
 /
MathGroup Archive
2003
*January
*February
*March
*April
*May
*June
*July
*August
*September
*October
*November
*December
*Archive Index
*Ask about this page
*Print this page
*Give us feedback
*Sign up for the Wolfram Insider

MathGroup Archive 2003

[Date Index] [Thread Index] [Author Index]

Search the Archive

RE: Error?

  • To: mathgroup at smc.vnet.net
  • Subject: [mg44717] RE: [mg44680] Error?
  • From: "Florian Jaccard" <florian.jaccard at eiaj.ch>
  • Date: Sat, 22 Nov 2003 02:17:12 -0500 (EST)
  • Reply-to: <florian.jaccard at eiaj.ch>
  • Sender: owner-wri-mathgroup at wolfram.com

Hello !

Very interesting question !

As


Integrate[Sqrt[r^2 - x^2], x] = (1/2)*(x*Sqrt[r^2 - x^2] +
r^2*ArcTan[x/Sqrt[r^2 - x^2]])

you will have to use complex numbers using Newton-Leibnitz's formula with
R<d<2R  !
As (I suppose) you want to compute a surface under a circle, it is not good
!

so I would suggest that you should inetgrate only with  d<R !

Indeed :

In[34]:=
Integrate[Sqrt[r^2 - x^2], {x, -r, -r + d}, Assumptions -> {d > 0, r > 0, d
< r}]

Out[34]=
(1/4)*(2*Sqrt[(-d)*(d - 2*r)]*(d - r) + Pi*r^2 + 2*r^2*ArcSin[(d - r)/r])

So if d>R, just compute for x from 0 to d-R and add Pi r^2/4 !

Meilleures salutations

Florian Jaccard


-----Message d'origine-----
De : Baris Altunkaynak [mailto:altunkai at boun.edu.tr]
Envoyé : ven., 21. novembre 2003 11:13
À : mathgroup at smc.vnet.net
Objet : [mg44680] Error?



This integral below give 0 on Mathematica 5.0
Integrate[Sqrt[R^2 - x^2], {x, -R, -R + d},
Assumptions -> R > 0 && d > 0 && d < 2R]

This is a mistake I think, isn't it?

Thanks,
Baris Altunkaynak





  • Prev by Date: Re: Error?
  • Next by Date: Re: filled plot on part of x-interval
  • Previous by thread: Re: Re: Error?
  • Next by thread: solution of linear equations, please help!!