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A simple example for newbies, to demonstrate Mathematica's power

Peter Bischet, I'm retired
Posted 9 years ago

I've posted this little mathematics problem maybe two years ago: Find the radius of a circle, bisecting the area of a given circle. The midpoint of the searched circle lies at the circumference of the given circle.

Here is the solution of Ilian Gachevski ( a Wolfram mathematician)

area[R_] = 
  Integrate[
   Boole[x^2 + y^2 <= 1 && (x - 1)^2 + y^2 <= R^2], {x, -1, 
    1}, {y, -1, 1}];
FindRoot[area[R] - Pi/2, {R, 1}]

Have you ever seen a mathematic tool that can be programmed in a shorter or more elegant way? I can only comment it with Tina Turner : ' Mathematica is simply the best'
POSTED BY: Peter Bischet
3 Replies
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no big deal. the point I was trying to make is that Solve can find the answer.
POSTED BY: Frank Kampas

Sorry, Frank, I'm the real Idiot

the original post from Ilian was: 

area[R_] = Integrate[Boole[x^2 + y^2 <= 1 && (x - 1)^2 + y^2 <= R^2],
         {x, -1, 1}, {y, -1, 1}, Assumptions -> R > 0];

FindRoot[area[R] - Pi/2, {R, 1}, WorkingPrecision -> 20]

(* Out[4]= {R -> 1.1587284730181215178} *)
POSTED BY: Peter Bischet 
In[3]:= Solve[area[R] - Pi/2 == 0]

Out[3]= {{R -> 
   Root[{\[Pi] - 2 ArcCos[1 - #1^2/2] - \[Pi] #1^2 + 
       2 ArcTan[#1/Sqrt[4 - #1^2]] #1^2 + #1 Sqrt[4 - #1^2] &, 
     1.15872847301812151783}]}, {R -> 
   Root[{-\[Pi] + 2 ArcCos[1 - #1^2/2] + \[Pi] #1^2 + 
       2 ArcTan[#1/Sqrt[4 - #1^2]] #1^2 + #1 Sqrt[
        4 - #1^2] &, -1.15872847301812151783}]}}
POSTED BY: Frank Kampas
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