We simplify a bit and determine where this MidPiece area is 1/3 Pi r^2.

Input := 


MidPiece = Expand[MidPiece]/.

			Sqrt[r^2 Sin[t]^2]->r Sin[t]

Output =


    2      2                     2

Pi r  - 2 r  ArcTan[Cot[t]] + 2 r  Cos[t] Sin[t]

Input := 


eq1 = MidPiece == 1/3 Pi r^2

Output =


    2      2                     2

Pi r  - 2 r  ArcTan[Cot[t]] + 2 r  Cos[t] Sin[t] == 

      2

  Pi r

  -----

    3

Input := 


eq2 = Simplify[MidPiece/r^2 == 1/3 Pi r^2/r^2]

Output =


                                    Pi

Pi - 2 ArcTan[Cot[t]] + Sin[2 t] == --

                                    3

Input := 


sol = FindRoot[eq2,{t,1.0}]

Output =


{t -> 0.268133}