We determine the volume at the various extrema - even though we note that the solution t = 11.4134 is out of range (for we only consider cutting for t in the interval [0, 2 Pi]). Solution 1 is t = 2 Pi, which says cut all the circle out and obtain a (minimal!) 0 volume.

Input := 


V1 = V[t]/.sol1[[1]]

Output =


                      2

((-6.28319 r + 2 Pi r)  

                                   2

          2   (-6.28319 r + 2 Pi r)

    Sqrt[r  - ----------------------]) / (12 Pi)

                          2

                      4 Pi

Input := 


V2 = V[t]/.sol1[[2]]//N

Output =


          2       2

0.403067 r  Sqrt[r ]

Input := 


V3 = V[t]/.sol1[[3]]//N

Output =


          2       2

0.403067 r  Sqrt[r ]