Can I hatch this region in any way?
11
$begingroup$
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 Jan 3 at 19:06
LCarvalhoLCarvalho
5,86142986
$endgroup$
add a comment |
11
$begingroup$
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 Jan 3 at 19:06
LCarvalhoLCarvalho
5,86142986
$endgroup$
$begingroup$
For what purpose?
$endgroup$
– Alex Trounev
Jan 3 at 19:10
add a comment |
11
11
11
2
$begingroup$
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 Jan 3 at 19:06
LCarvalhoLCarvalho
5,86142986
$endgroup$
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 Jan 3 at 19:06
LCarvalhoLCarvalho
5,86142986
asked Jan 3 at 19:06
LCarvalhoLCarvalho
5,86142986
edited Jan 3 at 19:16
LCarvalho
asked Jan 3 at 19:06
LCarvalhoLCarvalho
5,86142986
asked Jan 3 at 19:06
LCarvalhoLCarvalho
5,86142986
asked Jan 3 at 19:06
LCarvalhoLCarvalho
5,86142986
5,86142986
$begingroup$
For what purpose?
$endgroup$
– Alex Trounev
Jan 3 at 19:10
add a comment |
$begingroup$
For what purpose?
$endgroup$
– Alex Trounev
Jan 3 at 19:10
$begingroup$
For what purpose?
$endgroup$
– Alex Trounev
Jan 3 at 19:10
$begingroup$
For what purpose?
$endgroup$
– Alex Trounev
Jan 3 at 19:10
add a comment |
3 Answers
3
active
oldest
votes
16
$begingroup$
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 Jan 3 at 19:57
kglrkglr
190k10206425
$endgroup$
add a comment |
5
$begingroup$
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 Jan 3 at 19:30
Henrik SchumacherHenrik Schumacher
59.1k582164
$endgroup$
add a comment |
4
$begingroup$
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 Jan 3 at 20:29
Okkes DulgerciOkkes Dulgerci
5,4241919
$endgroup$
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
16
$begingroup$
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 Jan 3 at 19:57
kglrkglr
190k10206425
$endgroup$
add a comment |
16
$begingroup$
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 Jan 3 at 19:57
kglrkglr
190k10206425
$endgroup$
add a comment |
16
16
16
$begingroup$
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 Jan 3 at 19:57
kglrkglr
190k10206425
$endgroup$
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 Jan 3 at 19:57
kglrkglr
190k10206425
edited Jan 4 at 0:10
answered Jan 3 at 19:57
kglrkglr
190k10206425
answered Jan 3 at 19:57
kglrkglr
190k10206425
answered Jan 3 at 19:57
kglrkglr
190k10206425
190k10206425
add a comment |
add a comment |
5
$begingroup$
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 Jan 3 at 19:30
Henrik SchumacherHenrik Schumacher
59.1k582164
$endgroup$
add a comment |
5
$begingroup$
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 Jan 3 at 19:30
Henrik SchumacherHenrik Schumacher
59.1k582164
$endgroup$
add a comment |
5
5
5
$begingroup$
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 Jan 3 at 19:30
Henrik SchumacherHenrik Schumacher
59.1k582164
$endgroup$
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 Jan 3 at 19:30
Henrik SchumacherHenrik Schumacher
59.1k582164
answered Jan 3 at 19:30
Henrik SchumacherHenrik Schumacher
59.1k582164
answered Jan 3 at 19:30
Henrik SchumacherHenrik Schumacher
59.1k582164
answered Jan 3 at 19:30
Henrik SchumacherHenrik Schumacher
59.1k582164
59.1k582164
add a comment |
add a comment |
4
$begingroup$
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 Jan 3 at 20:29
Okkes DulgerciOkkes Dulgerci
5,4241919
$endgroup$
add a comment |
4
$begingroup$
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 Jan 3 at 20:29
Okkes DulgerciOkkes Dulgerci
5,4241919
$endgroup$
add a comment |
4
4
4
$begingroup$
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 Jan 3 at 20:29
Okkes DulgerciOkkes Dulgerci
5,4241919
$endgroup$
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 Jan 3 at 20:29
Okkes DulgerciOkkes Dulgerci
5,4241919
edited Jan 4 at 4:42
answered Jan 3 at 20:29
Okkes DulgerciOkkes Dulgerci
5,4241919
answered Jan 3 at 20:29
Okkes DulgerciOkkes Dulgerci
5,4241919
answered Jan 3 at 20:29
Okkes DulgerciOkkes Dulgerci
5,4241919
5,4241919
add a comment |
add a comment |
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$begingroup$
For what purpose?
$endgroup$
– Alex Trounev
Jan 3 at 19:10