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I want to measure the region covered by the inequalities given below, however, I am unable to do so by using the 'RegionMeasure' command. It will be very helpful if someone can guide me on the correct usage of this command. The two commands that I have used are:

`In[360]:= Reduce[γ > 0 && 
  0.47497220901624654` < θ < 
   1.` && (8.948725914798513`*^9 (4.195206175959918`*^16 γ + 
       3.1817826487031024`*^16 γ θ))/(
    4.212658842590404`*^24 + 2.0744515411908167`*^25 θ + 
     1.2131943826821323`*^25 θ^2) + 
    2.6275268714928966`*^-8 Sqrt[(
     1.7547286214020834`*^68 γ^2 + 
      1.6847605658382023`*^68 γ^2 θ + 
      3.486052528664758`*^67 γ^2 \
θ^2)/(4.212658842590404`*^24 + 
       2.0744515411908167`*^25 θ + 
       1.2131943826821323`*^25 θ^2)^2] < 
   A <= -((732.` γ)/(-25.` + 25.` θ)) && δ > 0 &&
   0.8765495860065979` < θ < 
   0.9999999999999997` && (
    5.5165693397017784`*^7 (-5.112505890383376`*^15 δ + 
       6.9456700794325384`*^16 δ \
θ))/(-6.857951394460775`*^23 - 
     2.052441539265766`*^23 θ + 
     1.3917822876389182`*^24 θ^2) + 
    2.339212652763844`*^-9 Sqrt[-((
      1.` (-1.5438780163828724`*^66 δ^2 - 
         6.271842145024957`*^64 δ^2 θ + 
         4.2067188049000795`*^65 δ^2 \
θ^2))/(-6.857951394460775`*^23 - 
        2.052441539265766`*^23 θ + 
        1.3917822876389182`*^24 θ^2)^2)] < 
   A < -((1.1333846813411608`*^16 δ)/(-3.269644300141837`*^15 \
+ 3.269644300141838`*^15 θ)) && A > 0]`

`Out[360]= δ > 0 && 
 0.87655 < θ < 
  1. && (20. (-4.20265*10^14 δ + 
      5.70957*10^15 δ θ))/(-2.04383*10^16 - 
    6.11675*10^15 θ + 4.14783*10^16 θ^2) + 
   3.37116*10^8 Sqrt[-((
     1. (-6.60228*10^16 δ^2 - 
        2.68211*10^15 δ^2 θ + 
        1.79897*10^16 δ^2 θ^2))/(-2.04383*10^16 - 
       6.11675*10^15 θ + 4.14783*10^16 θ^2)^2)] < 
  A < -((1.13338*10^16 δ)/(-3.26964*10^15 + 
    3.26964*10^15 θ)) && 
 0.00136612 (25. A - 25. A θ) <= γ < (
   4.29852*10^21 (4.28668*10^31 A + 2.43602*10^32 A θ + 
      2.83549*10^32 A θ^2 + 9.36294*10^31 A θ^3))/(
   2.30619*10^54 + 1.13564*10^55 θ + 
    6.64155*10^54 θ^2) - 
   2.54329*10^29 Sqrt[(
    4.51194*10^47 A^2 + 4.87686*10^48 A^2 θ + 
     1.78959*10^49 A^2 θ^2 + 2.66799*10^49 A^2 θ^3 + 
     1.87189*10^49 A^2 θ^4 + 6.13523*10^48 A^2 θ^5 + 
     7.43423*10^47 A^2 θ^6)/(2.30619*10^54 + 
      1.13564*10^55 θ + 6.64155*10^54 θ^2)^2]`

`In[361]:= RegionMeasure[{δ > 0 && 
   0.8765495860065979` < θ < 
    0.9999999999999997` && (
     20.` (-4.20264799057449`*^14 δ + 
        5.709569246155256`*^15 δ \
θ))/(-2.0438287837686464`*^16 - 
      6.116752443509597`*^15 θ + 
      4.147834442969913`*^16 θ^2) + 
     3.371160714024831`*^8 Sqrt[-((
       1.` (-6.602276286200451`*^16 δ^2 - 
          2.682105336398663`*^15 δ^2 θ + 
          1.7989711307229038`*^16 δ^2 \
θ^2))/(-2.0438287837686464`*^16 - 
         6.116752443509597`*^15 θ + 
         4.147834442969913`*^16 θ^2)^2)] < 
    A < -((1.1333846813411608`*^16 δ)/(-3.269644300141837`*^15 \
+ 3.269644300141838`*^15 θ)) && 
   0.001366120218579235` (25.` A - 25.` A θ) <= γ < (
     4.2985242006630915`*^21 (4.286680224396589`*^31 A + 
        2.4360180583882686`*^32 A θ + 
        2.835489371677133`*^32 A θ^2 + 
        9.362943190468041`*^31 A θ^3))/(
     2.3061908420688815`*^54 + 1.1356440968450611`*^55 θ + 
      6.641548436593824`*^54 θ^2) - 
     2.54329256250521`*^29 √((4.511942384755368`*^47 A^2 + 
           4.876860487023246`*^48 A^2 θ + 
           1.7895890107057892`*^49 A^2 θ^2 + 
           2.667989225957878`*^49 A^2 θ^3 + 
           1.8718881217671476`*^49 A^2 θ^4 + 
           6.135226535329527`*^48 A^2 θ^5 + 
           7.43422522803339`*^47 A^2 \
θ^6)/(2.3061908420688815`*^54 + 
           1.1356440968450611`*^55 θ + 
           6.641548436593824`*^54 θ^2)^2)}, {{A, 1, 
   100}, {θ, 0.374626, 1}, {δ, 0, 1}, {γ, 0, 1}}]`

`Out[361]= 61.912 √Det[
   SymbolicTensors`Tensor[{{1}}, \
{SymbolicTensors`CotangentBasis[{SymbolicTensors`SymbolicTensorsDump`\
x22}], SymbolicTensors`CotangentBasis[{SymbolicTensors`\
SymbolicTensorsDump`x22}]}]]`

The first 'Reduce' command gives me a common feasible region, and in the second command I have tried to measure the feasible region so obtained by putting the restrictions on the values of the four parameters.

Please let me know if you want me to share any other information or re-format my question for better readability. Thank you so much.

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1 Answer 1

1
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Normally you would run this where ineq is your inequality, however Mathematica v12.1 will give up after 5-10 minutes because your problem is too complex.

implr = ImplicitRegion[
   ineq, {{A, 1, 100}, {θ, 0.374626, 1}, {δ, 0, 1}, {γ, 0, 1}}];
measure = RegionMeasure[implr,4]

Instead, I recommend using NIntegrate to find the 4-volume. It may complain about convergence and precision though:

NIntegrate[If[ineq, 1, 0], {A, 1, 100}, {θ, 0.374626, 1}, {δ, 0, 1}, {γ, 0, 1}]
(* output: 2.68304 *)

... or you could even do Monte-Carlo integration if you only need an estimate:

BoxWhiskerChart[
 ParallelTable[
  NIntegrate[
   If[ineq, 1, 0], {A, 1, 100}, {θ, 0.374626, 1}, {δ, 0,1}, {γ, 0, 1},
   Method -> "AdaptiveMonteCarlo"]
 , 30]]

The results of that seem to suggest 2.68 is not an unreasonable figure: box whisker for monte-carlo integration

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  • $\begingroup$ Note, RegionMember has a Method option. There's an example in the docs: RegionMeasure[{t, t^2}, {{t, 0, 1}}, Method -> "NIntegrate"]. Weirdly using the Method on RegionMeasure causes red syntax highlighting suggesting it's an invalid option but it's not. $\endgroup$
    – flinty
    Jun 6, 2020 at 12:25
  • $\begingroup$ Thank you so much. I will just try using your suggestions. Thanks a lot. $\endgroup$ Jun 6, 2020 at 17:49

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