Can I hatch this region in any way?

Multi tool use
Multi tool use












6














Can I hatch this region in any way?



Graphics[{
Circle[{0,0},10,{ArcCos[5/10],ArcCos[5Sqrt[3]/10]}],
Line[{{5Sqrt[3],5},{5Sqrt[3],-5}}],
Circle[{0,0},10,{-ArcCos[5/10],-ArcCos[5Sqrt[3]/10]}],
Line[{{5,-5Sqrt[3]},{5,5Sqrt[3]}}]
},
Axes->True,
AxesOrigin->{0,0}
]


enter image description here



EDIT



I want to define a region of a circle, because I want to determine the area of this region...



enter image description here










share|improve this question
























  • For what purpose?
    – Alex Trounev
    10 hours ago
















6














Can I hatch this region in any way?



Graphics[{
Circle[{0,0},10,{ArcCos[5/10],ArcCos[5Sqrt[3]/10]}],
Line[{{5Sqrt[3],5},{5Sqrt[3],-5}}],
Circle[{0,0},10,{-ArcCos[5/10],-ArcCos[5Sqrt[3]/10]}],
Line[{{5,-5Sqrt[3]},{5,5Sqrt[3]}}]
},
Axes->True,
AxesOrigin->{0,0}
]


enter image description here



EDIT



I want to define a region of a circle, because I want to determine the area of this region...



enter image description here










share|improve this question
























  • For what purpose?
    – Alex Trounev
    10 hours ago














6












6








6







Can I hatch this region in any way?



Graphics[{
Circle[{0,0},10,{ArcCos[5/10],ArcCos[5Sqrt[3]/10]}],
Line[{{5Sqrt[3],5},{5Sqrt[3],-5}}],
Circle[{0,0},10,{-ArcCos[5/10],-ArcCos[5Sqrt[3]/10]}],
Line[{{5,-5Sqrt[3]},{5,5Sqrt[3]}}]
},
Axes->True,
AxesOrigin->{0,0}
]


enter image description here



EDIT



I want to define a region of a circle, because I want to determine the area of this region...



enter image description here










share|improve this question















Can I hatch this region in any way?



Graphics[{
Circle[{0,0},10,{ArcCos[5/10],ArcCos[5Sqrt[3]/10]}],
Line[{{5Sqrt[3],5},{5Sqrt[3],-5}}],
Circle[{0,0},10,{-ArcCos[5/10],-ArcCos[5Sqrt[3]/10]}],
Line[{{5,-5Sqrt[3]},{5,5Sqrt[3]}}]
},
Axes->True,
AxesOrigin->{0,0}
]


enter image description here



EDIT



I want to define a region of a circle, because I want to determine the area of this region...



enter image description here







graphics






share|improve this question















share|improve this question













share|improve this question




share|improve this question








edited 10 hours ago

























asked 10 hours ago









LCarvalho

5,61242885




5,61242885












  • For what purpose?
    – Alex Trounev
    10 hours ago


















  • For what purpose?
    – Alex Trounev
    10 hours ago
















For what purpose?
– Alex Trounev
10 hours ago




For what purpose?
– Alex Trounev
10 hours ago










3 Answers
3






active

oldest

votes


















11














reg = ImplicitRegion[x^2 + y^2 <= 100 && 5 <= x <= 5 Sqrt[3], {x, y}];
N[Area[reg]]



52.3599




Show[Graphics[{Gray, Circle[{0, 0}, 10]}, Axes -> True], 
RegionPlot[reg, MeshFunctions -> {# + #2 &, # - #2 &},
Mesh -> {50, 50}, MeshShading -> None, PlotStyle -> None,
BoundaryStyle -> Red]]


enter image description here



RegionPlot[x^2 + y^2 <= 100 && 5 <= x <= 5 Sqrt[3],{x, -10, 10}, {y, -10, 10}, 
MeshFunctions -> {# + #2 &, # - #2 &}, Mesh -> {50, 50},
MeshShading -> None, PlotStyle -> None, BoundaryStyle -> Red,
PlotPoints -> 90, Axes -> True, Epilog -> {Gray, Scale[Circle, 10]},
Frame -> False]



same picture







share|improve this answer































    4














    poly = MeshPrimitives[
    BoundaryDiscretizeRegion[
    RegionIntersection[
    Disk,
    HalfPlane[{{5/10, 0}, {5/10, 1}}, {1, 0}],
    HalfPlane[{{5 Sqrt[3]/10, 0}, {5 Sqrt[3]/10, 1}}, {-1, 0}]
    ],
    MaxCellMeasure -> {1 -> 0.001}
    ],
    2
    ][[1]];
    Area[poly]



    0.523599




    Graphics[{Circle, Gray, EdgeForm[{Thick, Black}], poly}]


    enter image description here






    share|improve this answer





























      3














         Graphics[{Red, Opacity@0.7, Disk[{0, 0}, 10], Opacity@1, Blue, Thick, 
      Circle[{0, 0}, 10, {π/6, π/3}],
      Circle[{0, 0}, 10, {-(π/6), -(π/3)}], Green, Opacity@0.6,
      Rectangle[{10 Cos[π/3], -10 Sin[π/3]}, {10 Cos[π/6],
      10 Sin[π/3]}]}, Axes -> True, AxesOrigin -> {0, 0}]


      enter image description here



      Therefore we can use the following.



      reg1 = Disk[{0, 0}, 10];
      reg2 = Rectangle[{10 Cos[π/3], -10 Sin[π/3]}, {10 Cos[π/ 6], 10 Sin[π/3]}];
      reg=RegionIntersection[reg1, reg2];
      Area@reg



      $frac{50 pi }{3}$




      Show[Graphics[{Circle[{0, 0}, 10]}, Axes -> True], 
      Region[reg, BaseStyle -> {LightBlue, EdgeForm[{Red, Thick}]}]]


      enter image description here



      You can also choose reg2 as



      reg2 = Rectangle[{5, -5 Sqrt[3]}, {5 Sqrt[3], 5 Sqrt[3]}];


      Or



      reg2 = Rectangle[{5, -10}, {5 Sqrt[3], 10}];





      share|improve this answer























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        3 Answers
        3






        active

        oldest

        votes








        3 Answers
        3






        active

        oldest

        votes









        active

        oldest

        votes






        active

        oldest

        votes









        11














        reg = ImplicitRegion[x^2 + y^2 <= 100 && 5 <= x <= 5 Sqrt[3], {x, y}];
        N[Area[reg]]



        52.3599




        Show[Graphics[{Gray, Circle[{0, 0}, 10]}, Axes -> True], 
        RegionPlot[reg, MeshFunctions -> {# + #2 &, # - #2 &},
        Mesh -> {50, 50}, MeshShading -> None, PlotStyle -> None,
        BoundaryStyle -> Red]]


        enter image description here



        RegionPlot[x^2 + y^2 <= 100 && 5 <= x <= 5 Sqrt[3],{x, -10, 10}, {y, -10, 10}, 
        MeshFunctions -> {# + #2 &, # - #2 &}, Mesh -> {50, 50},
        MeshShading -> None, PlotStyle -> None, BoundaryStyle -> Red,
        PlotPoints -> 90, Axes -> True, Epilog -> {Gray, Scale[Circle, 10]},
        Frame -> False]



        same picture







        share|improve this answer




























          11














          reg = ImplicitRegion[x^2 + y^2 <= 100 && 5 <= x <= 5 Sqrt[3], {x, y}];
          N[Area[reg]]



          52.3599




          Show[Graphics[{Gray, Circle[{0, 0}, 10]}, Axes -> True], 
          RegionPlot[reg, MeshFunctions -> {# + #2 &, # - #2 &},
          Mesh -> {50, 50}, MeshShading -> None, PlotStyle -> None,
          BoundaryStyle -> Red]]


          enter image description here



          RegionPlot[x^2 + y^2 <= 100 && 5 <= x <= 5 Sqrt[3],{x, -10, 10}, {y, -10, 10}, 
          MeshFunctions -> {# + #2 &, # - #2 &}, Mesh -> {50, 50},
          MeshShading -> None, PlotStyle -> None, BoundaryStyle -> Red,
          PlotPoints -> 90, Axes -> True, Epilog -> {Gray, Scale[Circle, 10]},
          Frame -> False]



          same picture







          share|improve this answer


























            11












            11








            11






            reg = ImplicitRegion[x^2 + y^2 <= 100 && 5 <= x <= 5 Sqrt[3], {x, y}];
            N[Area[reg]]



            52.3599




            Show[Graphics[{Gray, Circle[{0, 0}, 10]}, Axes -> True], 
            RegionPlot[reg, MeshFunctions -> {# + #2 &, # - #2 &},
            Mesh -> {50, 50}, MeshShading -> None, PlotStyle -> None,
            BoundaryStyle -> Red]]


            enter image description here



            RegionPlot[x^2 + y^2 <= 100 && 5 <= x <= 5 Sqrt[3],{x, -10, 10}, {y, -10, 10}, 
            MeshFunctions -> {# + #2 &, # - #2 &}, Mesh -> {50, 50},
            MeshShading -> None, PlotStyle -> None, BoundaryStyle -> Red,
            PlotPoints -> 90, Axes -> True, Epilog -> {Gray, Scale[Circle, 10]},
            Frame -> False]



            same picture







            share|improve this answer














            reg = ImplicitRegion[x^2 + y^2 <= 100 && 5 <= x <= 5 Sqrt[3], {x, y}];
            N[Area[reg]]



            52.3599




            Show[Graphics[{Gray, Circle[{0, 0}, 10]}, Axes -> True], 
            RegionPlot[reg, MeshFunctions -> {# + #2 &, # - #2 &},
            Mesh -> {50, 50}, MeshShading -> None, PlotStyle -> None,
            BoundaryStyle -> Red]]


            enter image description here



            RegionPlot[x^2 + y^2 <= 100 && 5 <= x <= 5 Sqrt[3],{x, -10, 10}, {y, -10, 10}, 
            MeshFunctions -> {# + #2 &, # - #2 &}, Mesh -> {50, 50},
            MeshShading -> None, PlotStyle -> None, BoundaryStyle -> Red,
            PlotPoints -> 90, Axes -> True, Epilog -> {Gray, Scale[Circle, 10]},
            Frame -> False]



            same picture








            share|improve this answer














            share|improve this answer



            share|improve this answer








            edited 5 hours ago

























            answered 9 hours ago









            kglr

            177k9198406




            177k9198406























                4














                poly = MeshPrimitives[
                BoundaryDiscretizeRegion[
                RegionIntersection[
                Disk,
                HalfPlane[{{5/10, 0}, {5/10, 1}}, {1, 0}],
                HalfPlane[{{5 Sqrt[3]/10, 0}, {5 Sqrt[3]/10, 1}}, {-1, 0}]
                ],
                MaxCellMeasure -> {1 -> 0.001}
                ],
                2
                ][[1]];
                Area[poly]



                0.523599




                Graphics[{Circle, Gray, EdgeForm[{Thick, Black}], poly}]


                enter image description here






                share|improve this answer


























                  4














                  poly = MeshPrimitives[
                  BoundaryDiscretizeRegion[
                  RegionIntersection[
                  Disk,
                  HalfPlane[{{5/10, 0}, {5/10, 1}}, {1, 0}],
                  HalfPlane[{{5 Sqrt[3]/10, 0}, {5 Sqrt[3]/10, 1}}, {-1, 0}]
                  ],
                  MaxCellMeasure -> {1 -> 0.001}
                  ],
                  2
                  ][[1]];
                  Area[poly]



                  0.523599




                  Graphics[{Circle, Gray, EdgeForm[{Thick, Black}], poly}]


                  enter image description here






                  share|improve this answer
























                    4












                    4








                    4






                    poly = MeshPrimitives[
                    BoundaryDiscretizeRegion[
                    RegionIntersection[
                    Disk,
                    HalfPlane[{{5/10, 0}, {5/10, 1}}, {1, 0}],
                    HalfPlane[{{5 Sqrt[3]/10, 0}, {5 Sqrt[3]/10, 1}}, {-1, 0}]
                    ],
                    MaxCellMeasure -> {1 -> 0.001}
                    ],
                    2
                    ][[1]];
                    Area[poly]



                    0.523599




                    Graphics[{Circle, Gray, EdgeForm[{Thick, Black}], poly}]


                    enter image description here






                    share|improve this answer












                    poly = MeshPrimitives[
                    BoundaryDiscretizeRegion[
                    RegionIntersection[
                    Disk,
                    HalfPlane[{{5/10, 0}, {5/10, 1}}, {1, 0}],
                    HalfPlane[{{5 Sqrt[3]/10, 0}, {5 Sqrt[3]/10, 1}}, {-1, 0}]
                    ],
                    MaxCellMeasure -> {1 -> 0.001}
                    ],
                    2
                    ][[1]];
                    Area[poly]



                    0.523599




                    Graphics[{Circle, Gray, EdgeForm[{Thick, Black}], poly}]


                    enter image description here







                    share|improve this answer












                    share|improve this answer



                    share|improve this answer










                    answered 9 hours ago









                    Henrik Schumacher

                    49k467139




                    49k467139























                        3














                           Graphics[{Red, Opacity@0.7, Disk[{0, 0}, 10], Opacity@1, Blue, Thick, 
                        Circle[{0, 0}, 10, {π/6, π/3}],
                        Circle[{0, 0}, 10, {-(π/6), -(π/3)}], Green, Opacity@0.6,
                        Rectangle[{10 Cos[π/3], -10 Sin[π/3]}, {10 Cos[π/6],
                        10 Sin[π/3]}]}, Axes -> True, AxesOrigin -> {0, 0}]


                        enter image description here



                        Therefore we can use the following.



                        reg1 = Disk[{0, 0}, 10];
                        reg2 = Rectangle[{10 Cos[π/3], -10 Sin[π/3]}, {10 Cos[π/ 6], 10 Sin[π/3]}];
                        reg=RegionIntersection[reg1, reg2];
                        Area@reg



                        $frac{50 pi }{3}$




                        Show[Graphics[{Circle[{0, 0}, 10]}, Axes -> True], 
                        Region[reg, BaseStyle -> {LightBlue, EdgeForm[{Red, Thick}]}]]


                        enter image description here



                        You can also choose reg2 as



                        reg2 = Rectangle[{5, -5 Sqrt[3]}, {5 Sqrt[3], 5 Sqrt[3]}];


                        Or



                        reg2 = Rectangle[{5, -10}, {5 Sqrt[3], 10}];





                        share|improve this answer




























                          3














                             Graphics[{Red, Opacity@0.7, Disk[{0, 0}, 10], Opacity@1, Blue, Thick, 
                          Circle[{0, 0}, 10, {π/6, π/3}],
                          Circle[{0, 0}, 10, {-(π/6), -(π/3)}], Green, Opacity@0.6,
                          Rectangle[{10 Cos[π/3], -10 Sin[π/3]}, {10 Cos[π/6],
                          10 Sin[π/3]}]}, Axes -> True, AxesOrigin -> {0, 0}]


                          enter image description here



                          Therefore we can use the following.



                          reg1 = Disk[{0, 0}, 10];
                          reg2 = Rectangle[{10 Cos[π/3], -10 Sin[π/3]}, {10 Cos[π/ 6], 10 Sin[π/3]}];
                          reg=RegionIntersection[reg1, reg2];
                          Area@reg



                          $frac{50 pi }{3}$




                          Show[Graphics[{Circle[{0, 0}, 10]}, Axes -> True], 
                          Region[reg, BaseStyle -> {LightBlue, EdgeForm[{Red, Thick}]}]]


                          enter image description here



                          You can also choose reg2 as



                          reg2 = Rectangle[{5, -5 Sqrt[3]}, {5 Sqrt[3], 5 Sqrt[3]}];


                          Or



                          reg2 = Rectangle[{5, -10}, {5 Sqrt[3], 10}];





                          share|improve this answer


























                            3












                            3








                            3






                               Graphics[{Red, Opacity@0.7, Disk[{0, 0}, 10], Opacity@1, Blue, Thick, 
                            Circle[{0, 0}, 10, {π/6, π/3}],
                            Circle[{0, 0}, 10, {-(π/6), -(π/3)}], Green, Opacity@0.6,
                            Rectangle[{10 Cos[π/3], -10 Sin[π/3]}, {10 Cos[π/6],
                            10 Sin[π/3]}]}, Axes -> True, AxesOrigin -> {0, 0}]


                            enter image description here



                            Therefore we can use the following.



                            reg1 = Disk[{0, 0}, 10];
                            reg2 = Rectangle[{10 Cos[π/3], -10 Sin[π/3]}, {10 Cos[π/ 6], 10 Sin[π/3]}];
                            reg=RegionIntersection[reg1, reg2];
                            Area@reg



                            $frac{50 pi }{3}$




                            Show[Graphics[{Circle[{0, 0}, 10]}, Axes -> True], 
                            Region[reg, BaseStyle -> {LightBlue, EdgeForm[{Red, Thick}]}]]


                            enter image description here



                            You can also choose reg2 as



                            reg2 = Rectangle[{5, -5 Sqrt[3]}, {5 Sqrt[3], 5 Sqrt[3]}];


                            Or



                            reg2 = Rectangle[{5, -10}, {5 Sqrt[3], 10}];





                            share|improve this answer














                               Graphics[{Red, Opacity@0.7, Disk[{0, 0}, 10], Opacity@1, Blue, Thick, 
                            Circle[{0, 0}, 10, {π/6, π/3}],
                            Circle[{0, 0}, 10, {-(π/6), -(π/3)}], Green, Opacity@0.6,
                            Rectangle[{10 Cos[π/3], -10 Sin[π/3]}, {10 Cos[π/6],
                            10 Sin[π/3]}]}, Axes -> True, AxesOrigin -> {0, 0}]


                            enter image description here



                            Therefore we can use the following.



                            reg1 = Disk[{0, 0}, 10];
                            reg2 = Rectangle[{10 Cos[π/3], -10 Sin[π/3]}, {10 Cos[π/ 6], 10 Sin[π/3]}];
                            reg=RegionIntersection[reg1, reg2];
                            Area@reg



                            $frac{50 pi }{3}$




                            Show[Graphics[{Circle[{0, 0}, 10]}, Axes -> True], 
                            Region[reg, BaseStyle -> {LightBlue, EdgeForm[{Red, Thick}]}]]


                            enter image description here



                            You can also choose reg2 as



                            reg2 = Rectangle[{5, -5 Sqrt[3]}, {5 Sqrt[3], 5 Sqrt[3]}];


                            Or



                            reg2 = Rectangle[{5, -10}, {5 Sqrt[3], 10}];






                            share|improve this answer














                            share|improve this answer



                            share|improve this answer








                            edited 45 mins ago

























                            answered 8 hours ago









                            Okkes Dulgerci

                            4,0851816




                            4,0851816






























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