Can I hatch this region in any way?

Can I hatch this region in any way?

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

edited 10 hours ago

asked 10 hours ago

LCarvalho

5,61242885

For what purpose?
– Alex Trounev
10 hours ago

add a comment |

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

edited 10 hours ago

asked 10 hours ago

LCarvalho

5,61242885

For what purpose?
– Alex Trounev
10 hours ago

add a comment |

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

edited 10 hours ago

asked 10 hours ago

LCarvalho

5,61242885

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 10 hours ago

asked 10 hours ago

LCarvalho

5,61242885

edited 10 hours ago

asked 10 hours ago

LCarvalho

5,61242885

edited 10 hours ago

edited 10 hours ago

edited 10 hours ago

asked 10 hours ago

LCarvalho

5,61242885

asked 10 hours ago

LCarvalho

5,61242885

asked 10 hours ago

LCarvalho

5,61242885

5,61242885

For what purpose?
– Alex Trounev
10 hours ago

add a comment |

For what purpose?
– Alex Trounev
10 hours ago

For what purpose?
– Alex Trounev
10 hours ago

For what purpose?
– Alex Trounev
10 hours ago

add a comment |

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

edited 5 hours ago

answered 9 hours ago

kglr

177k9198406

add a comment |

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

answered 9 hours ago

Henrik Schumacher

49k467139

add a comment |

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}];

edited 45 mins ago

answered 8 hours ago

Okkes Dulgerci

4,0851816

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

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

edited 5 hours ago

answered 9 hours ago

kglr

177k9198406

add a comment |

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

edited 5 hours ago

answered 9 hours ago

kglr

177k9198406

add a comment |

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

edited 5 hours ago

answered 9 hours ago

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 5 hours ago

answered 9 hours ago

kglr

177k9198406

edited 5 hours ago

edited 5 hours ago

edited 5 hours ago

answered 9 hours ago

kglr

177k9198406

answered 9 hours ago

kglr

177k9198406

answered 9 hours ago

kglr

177k9198406

177k9198406

add a comment |

add a comment |

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

answered 9 hours ago

Henrik Schumacher

49k467139

add a comment |

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

answered 9 hours ago

Henrik Schumacher

49k467139

add a comment |

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

answered 9 hours ago

Henrik Schumacher

49k467139

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 9 hours ago

Henrik Schumacher

49k467139

answered 9 hours ago

Henrik Schumacher

49k467139

answered 9 hours ago

Henrik Schumacher

49k467139

answered 9 hours ago

Henrik Schumacher

49k467139

49k467139

add a comment |

add a comment |

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}];

edited 45 mins ago

answered 8 hours ago

Okkes Dulgerci

4,0851816

add a comment |

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}];

edited 45 mins ago

answered 8 hours ago

Okkes Dulgerci

4,0851816

add a comment |

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}];

edited 45 mins ago

answered 8 hours ago

Okkes Dulgerci

4,0851816

   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 45 mins ago

answered 8 hours ago

Okkes Dulgerci

4,0851816

edited 45 mins ago

edited 45 mins ago

edited 45 mins ago

answered 8 hours ago

Okkes Dulgerci

4,0851816

answered 8 hours ago

Okkes Dulgerci

4,0851816

answered 8 hours ago

Okkes Dulgerci

4,0851816

4,0851816

add a comment |

add a comment |

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