MathGroup Archive 1994

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

Search the Archive

RE: Negative Area: Definitive Answer (not)


I've figured it out!  Your example gives the wrong answer as shown here,

In[10]:=
Integrate[1/(A+B Cos[x])^2,{x,0,2 Pi}]/.{A->2,B->1}
Out[10]=
  -4 Pi
---------
3 Sqrt[3]

but if you reverse the roles of A and B you get the right answer.

In[11]:=
Integrate[1/(B+A Cos[x])^2,{x,0,2 Pi}]/.{A->1, B->2}
Out[11]=
  4 Pi
---------
3 Sqrt[3]

So clearly, you just put your A's and B's in the wrong places!  ;o)

Chip






  • Prev by Date: Question in ParametricPlot
  • Next by Date: RE: Negative Area
  • Previous by thread: Re: Question in ParametricPlot
  • Next by thread: Re: recognition puzzle