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}
]
EDIT
I want to define a region of a circle, because I want to determine the area of this region...
graphics
asked yesterday
LCarvalho
5,62742885
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}
]
EDIT
I want to define a region of a circle, because I want to determine the area of this region...
graphics
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}
]
EDIT
I want to define a region of a circle, because I want to determine the area of this region...
graphics
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}
]
EDIT
I want to define a region of a circle, because I want to determine the area of this region...
graphics
graphics
asked yesterday
LCarvalho
5,62742885
asked yesterday
LCarvalho
5,62742885
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]]
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
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}]
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}]
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}]}]]
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}];
answered yesterday
Okkes Dulgerci
4,1151816
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
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]]
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
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]]
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
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]]
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
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]]
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
answered yesterday
kglr
177k9198406
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}]
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}]
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}]
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}]
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}]
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}]}]]
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}];
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}]
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}]}]]
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}];
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}]
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}]}]]
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}];
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}]
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}]}]]
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}];
answered yesterday
Okkes Dulgerci
4,1151816
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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yesterday