Can I hatch this region in any way?

Can I hatch this region in any way?

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 1 hour ago

asked 2 hours ago

LCarvalho

5,58542885

For what purpose?
– Alex Trounev
2 hours ago

add a comment |

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 1 hour ago

asked 2 hours ago

LCarvalho

5,58542885

For what purpose?
– Alex Trounev
2 hours ago

add a comment |

1

1

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 1 hour ago

asked 2 hours ago

LCarvalho

5,58542885

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 1 hour ago

asked 2 hours ago

LCarvalho

5,58542885

edited 1 hour ago

asked 2 hours ago

LCarvalho

5,58542885

edited 1 hour ago

edited 1 hour ago

edited 1 hour ago

asked 2 hours ago

LCarvalho

5,58542885

asked 2 hours ago

LCarvalho

5,58542885

asked 2 hours ago

LCarvalho

5,58542885

5,58542885

For what purpose?
– Alex Trounev
2 hours ago

add a comment |

For what purpose?
– Alex Trounev
2 hours ago

For what purpose?
– Alex Trounev
2 hours ago

For what purpose?
– Alex Trounev
2 hours ago

add a comment |

3 Answers
3

active

oldest

votes

3

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

N @ Area @ impreg

52.3599

Show[Graphics[{Gray, Circle[{0, 0}, 10]}], 

 RegionPlot[impreg, 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, -12, 12}, {y, -12, 12}, 

 MeshFunctions -> {# + #2 &, # - #2 &}, Mesh -> {50, 50}, 

 MeshShading -> None, PlotStyle -> None, BoundaryStyle -> Red, 

 PlotPoints -> 90, Epilog -> {Gray, Scale[Circle, 10]}, 

 Frame -> False]

same picture

edited 1 hour ago

answered 1 hour ago

kglr

177k9198405

add a comment |

1

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 1 hour ago

Henrik Schumacher

49k467139

add a comment |

0

   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]}];



    Area@RegionIntersection[reg1, reg2]

$frac{50 pi }{3}$

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 33 mins ago

answered 42 mins ago

Okkes Dulgerci

4,0451816

add a comment |

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3 Answers
3

active

oldest

votes

3 Answers
3

active

oldest

votes

active

oldest

votes

active

oldest

votes

3

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

N @ Area @ impreg

52.3599

Show[Graphics[{Gray, Circle[{0, 0}, 10]}], 

 RegionPlot[impreg, 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, -12, 12}, {y, -12, 12}, 

 MeshFunctions -> {# + #2 &, # - #2 &}, Mesh -> {50, 50}, 

 MeshShading -> None, PlotStyle -> None, BoundaryStyle -> Red, 

 PlotPoints -> 90, Epilog -> {Gray, Scale[Circle, 10]}, 

 Frame -> False]

same picture

edited 1 hour ago

answered 1 hour ago

kglr

177k9198405

add a comment |

3

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

N @ Area @ impreg

52.3599

Show[Graphics[{Gray, Circle[{0, 0}, 10]}], 

 RegionPlot[impreg, 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, -12, 12}, {y, -12, 12}, 

 MeshFunctions -> {# + #2 &, # - #2 &}, Mesh -> {50, 50}, 

 MeshShading -> None, PlotStyle -> None, BoundaryStyle -> Red, 

 PlotPoints -> 90, Epilog -> {Gray, Scale[Circle, 10]}, 

 Frame -> False]

same picture

edited 1 hour ago

answered 1 hour ago

kglr

177k9198405

add a comment |

3

3

3

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

N @ Area @ impreg

52.3599

Show[Graphics[{Gray, Circle[{0, 0}, 10]}], 

 RegionPlot[impreg, 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, -12, 12}, {y, -12, 12}, 

 MeshFunctions -> {# + #2 &, # - #2 &}, Mesh -> {50, 50}, 

 MeshShading -> None, PlotStyle -> None, BoundaryStyle -> Red, 

 PlotPoints -> 90, Epilog -> {Gray, Scale[Circle, 10]}, 

 Frame -> False]

same picture

edited 1 hour ago

answered 1 hour ago

kglr

177k9198405

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

N @ Area @ impreg

52.3599

Show[Graphics[{Gray, Circle[{0, 0}, 10]}], 

 RegionPlot[impreg, 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, -12, 12}, {y, -12, 12}, 

 MeshFunctions -> {# + #2 &, # - #2 &}, Mesh -> {50, 50}, 

 MeshShading -> None, PlotStyle -> None, BoundaryStyle -> Red, 

 PlotPoints -> 90, Epilog -> {Gray, Scale[Circle, 10]}, 

 Frame -> False]

same picture

edited 1 hour ago

answered 1 hour ago

kglr

177k9198405

edited 1 hour ago

edited 1 hour ago

edited 1 hour ago

answered 1 hour ago

kglr

177k9198405

answered 1 hour ago

kglr

177k9198405

answered 1 hour ago

kglr

177k9198405

177k9198405

add a comment |

add a comment |

1

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 1 hour ago

Henrik Schumacher

49k467139

add a comment |

1

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 1 hour ago

Henrik Schumacher

49k467139

add a comment |

1

1

1

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 1 hour 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 1 hour ago

Henrik Schumacher

49k467139

answered 1 hour ago

Henrik Schumacher

49k467139

answered 1 hour ago

Henrik Schumacher

49k467139

answered 1 hour ago

Henrik Schumacher

49k467139

49k467139

add a comment |

add a comment |

0

   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]}];



    Area@RegionIntersection[reg1, reg2]

$frac{50 pi }{3}$

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 33 mins ago

answered 42 mins ago

Okkes Dulgerci

4,0451816

add a comment |

0

   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]}];



    Area@RegionIntersection[reg1, reg2]

$frac{50 pi }{3}$

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 33 mins ago

answered 42 mins ago

Okkes Dulgerci

4,0451816

add a comment |

0

0

0

   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]}];



    Area@RegionIntersection[reg1, reg2]

$frac{50 pi }{3}$

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 33 mins ago

answered 42 mins ago

Okkes Dulgerci

4,0451816

   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]}];



    Area@RegionIntersection[reg1, reg2]

$frac{50 pi }{3}$

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 33 mins ago

answered 42 mins ago

Okkes Dulgerci

4,0451816

edited 33 mins ago

edited 33 mins ago

edited 33 mins ago

answered 42 mins ago

Okkes Dulgerci

4,0451816

answered 42 mins ago

Okkes Dulgerci

4,0451816

answered 42 mins ago

Okkes Dulgerci

4,0451816

4,0451816

add a comment |

add a comment |

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