Can I hatch this region in any way?

Can I hatch this region in any way?

9

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

edited yesterday

asked yesterday

LCarvalho

5,62742885

For what purpose?
– Alex Trounev
yesterday

add a comment |

9

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

edited yesterday

asked yesterday

LCarvalho

5,62742885

For what purpose?
– Alex Trounev
yesterday

add a comment |

9

9

9

1

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

edited yesterday

asked yesterday

LCarvalho

5,62742885

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

edited yesterday

asked yesterday

LCarvalho

5,62742885

edited yesterday

asked yesterday

LCarvalho

5,62742885

edited yesterday

edited yesterday

edited yesterday

asked yesterday

LCarvalho

5,62742885

asked yesterday

LCarvalho

5,62742885

asked yesterday

LCarvalho

5,62742885

5,62742885

For what purpose?
– Alex Trounev
yesterday

add a comment |

For what purpose?
– Alex Trounev
yesterday

For what purpose?
– Alex Trounev
yesterday

For what purpose?
– Alex Trounev
yesterday

add a comment |

3 Answers
3

active

oldest

votes

14

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

edited yesterday

answered yesterday

kglr

177k9198406

add a comment |

5

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

answered yesterday

Henrik Schumacher

49.2k467139

add a comment |

4

   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}];

edited yesterday

answered yesterday

Okkes Dulgerci

4,1151816

add a comment |

Your Answer

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3 Answers
3

active

oldest

votes

3 Answers
3

active

oldest

votes

active

oldest

votes

active

oldest

votes

14

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

edited yesterday

answered yesterday

kglr

177k9198406

add a comment |

14

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

edited yesterday

answered yesterday

kglr

177k9198406

add a comment |

14

14

14

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

edited yesterday

answered yesterday

kglr

177k9198406

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

edited yesterday

answered yesterday

kglr

177k9198406

edited yesterday

edited yesterday

edited yesterday

answered yesterday

kglr

177k9198406

answered yesterday

kglr

177k9198406

answered yesterday

kglr

177k9198406

177k9198406

add a comment |

add a comment |

5

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

answered yesterday

Henrik Schumacher

49.2k467139

add a comment |

5

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

answered yesterday

Henrik Schumacher

49.2k467139

add a comment |

5

5

5

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

answered yesterday

Henrik Schumacher

49.2k467139

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

answered yesterday

Henrik Schumacher

49.2k467139

answered yesterday

Henrik Schumacher

49.2k467139

answered yesterday

Henrik Schumacher

49.2k467139

answered yesterday

Henrik Schumacher

49.2k467139

49.2k467139

add a comment |

add a comment |

4

   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}];

edited yesterday

answered yesterday

Okkes Dulgerci

4,1151816

add a comment |

4

   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}];

edited yesterday

answered yesterday

Okkes Dulgerci

4,1151816

add a comment |

4

4

4

   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}];

edited yesterday

answered yesterday

Okkes Dulgerci

4,1151816

   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}];

edited yesterday

answered yesterday

Okkes Dulgerci

4,1151816

edited yesterday

edited yesterday

edited yesterday

answered yesterday

Okkes Dulgerci

4,1151816

answered yesterday

Okkes Dulgerci

4,1151816

answered yesterday

Okkes Dulgerci

4,1151816

4,1151816

add a comment |

add a comment |

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