Circular Segment

Author

Eric W. Weisstein
July 22, 2014

This notebook downloaded from http://mathworld.wolfram.com/notebooks/Laminae/CircularSegment.nb.

For more information, see Eric's MathWorld entry http://mathworld.wolfram.com/CircularSegment.html.

©2014 Wolfram Research, Inc. except for portions noted otherwise

Deifnitions

ineq=x^2+y^2≤a^2&&y≥a Cos[q/2];

reg=RegionIntersection[ImplicitRegion[y>a Cos[q/2],{x,y}],Disk[{0,0},a,Pi/2+{-1,1} q/2]];

assum=a>0&&0<q<π;

Plot

Diagram

LaminaData["CircularSegment","Diagram"]

DiscretizeRegion

Show[DiscretizeRegion[Block[{a=1,q=π/2},reg]],ImageSize->Small]

RegionPlot

Block[{a=1,q=π/2},RegionPlot[ineq,{x,-a,a},{y,0,a},AspectRatio->Automatic]]

Graphics

Options[Segment]:=
{
Fills->Yellow,
Curves->None,
Axes->False
};
Segment[d_,R_:1,opts___]:=Module[
{
x,
y
},
InequalityPlot[x^2+y^2<R&&y>d,{x},{y},opts,Sequence@@Options[Segment],
PlotRange->R{{-1,1},{-1,1}},
DisplayFunction->Identity][[1]]
]

With[{d=.6,R=1},
c=Sqrt[R^2-d^2];
Show[Graphics[{
Segment[d,R] ,
Circle[{0,0},R],
PointSize[.04],Point[{0,0}],
Line[{{-c,d},{0,0},{c,d}}] ,
Dashing[{.04}],Line[{{-c,d},{c,d}}]
}],ImageSize->200,AspectRatio->Automatic,
PlotRange->All
]
]

Properties

Area

As a function of R and θ

v5.0

Integrate[(Sqrt[R^2-x^2]-R Cos[θ/2]),{x,-R Sin[θ/2],R Sin[θ/2]},Assumptions->0<θ≤π&&R>0]

Integrate[1,{x,-R Sin[θ/2],R Sin[θ/2]},{y,R Cos[θ/2],Sqrt[R^2-x^2]},Assumptions->0<θ≤π&&R>0]

As a function of R and h

A[R_,h_]:=R^2 ArcCos[(R-h)/R]-(R-h)Sqrt[2R h-h^2]

A[1,x]

FindRoot[A[1,x]==Pi/2,{x,.5}]

Assuming[a>0&&0<q<π,Area[ineq,{x,y}]//FullSimplify]//Timing

RegionMeasure

Assuming[assum,Simplify@RegionMeasure[reg]]//Timing

AreaMomentOfInertia

Assuming[a>0&&0<q<π,FullSimplify[(1/((1/2) (a^2) (q-Sin[q])))Integrate[AreaInertiaTensor[{x,y},"CoordinateSystem"->"Cartesian"]Boole[ineq[x,y]],{x,-∞,∞},{y,-∞,∞}]]]

Assuming[a>0&&0<q<π,AreaInertiaTensor[ineq,{x,y}]//FullSimplify]//Timing

Centroid

Assuming[a>0&&0<q<π,Centroid[ineq,{x,y}]//FullSimplify]//Timing

RegionCentroid

Assuming[assum,RegionCentroid[EntityValue[Entity["Lamina","CircularSector"],"ImplicitRegion"][q,a]]//FullSimplify]//Timing

Direct

Integrate[x,{x,-R Sin[θ/2],R Sin[θ/2]},{y,R Cos[θ/2],Sqrt[R^2-x^2]},Assumptions->0<θ≤π&&R>0]

Integrate[y,{x,-R Sin[θ/2],R Sin[θ/2]},{y,R Cos[θ/2],Sqrt[R^2-x^2]},Assumptions->0<θ≤π&&R>0]

FullSimplify[%]

ybar=(2/3) (R^3) ((Sin[(θ/2)])^3)/((1/2) (R^2) (θ-Sin[θ]))

ybar/.θ->Pi

Limit[ybar,θ->0]

Generalized Diameter

ap/.ap->2a Sin[q/2]

Perimeter

FullSimplify[s+ap/.{s->a q,ap->2a Sin[q/2]},a>0&&0<q<π]

RadiiOfGyration

Assuming[a>0&&0<q<π,FullSimplify[RadiiOfGyration[ineq[x,y],{x,y}]]]

SectionModuli

(* Roark, p. 12: The term pertains to the cross section of a beam.  The section modulus with respect to either principal
		central axis is the moment of inertia with respect to that axis divided by the distance from that axis to the most remote
		point of the section.  The section modulus largely determines the flexural strength of a beam of given material.
		*)

Tr[{{(a^2 (-2 q+Sin[2 q]))/(8 (-q+Sin[q])),0},{0,(a^2 (6 q-8 Sin[q]+Sin[2 q]))/(24 (q-Sin[q]))}},List]/{2a Sin[q/2],a}

Half of a Semicircle

Plot[A[1,x]-π/4,{x,0,2}]

Solve[A[1,x]==Pi/4,x]

FindRoot[-(1-x) (Sqrt[2 x-(x^2)])+ArcCos[1-x]==(π/4),{x,.5},WorkingPrecision->30]

Reduce[-(1-x) (Sqrt[2 x-(x^2)])+ArcCos[1-x]==(π/4)&&1/2<x<6/10,x,Reals]//Timing

Reduce[-(1-x) (Sqrt[2 x-(x^2)])+ArcCos[1-x]==(π/4)&&1/100<x<1,x,Reals]//Timing

Reduce[-(1-x) (Sqrt[2 x-(x^2)])+ArcCos[1-x]==(π/4)&&0<x<1,x,Reals]//Timing

...hangs...

With[{d=1-.596027,R=1},
c=Sqrt[R^2-d^2];
Show[GraphicsArray[
{Graphics[{
Segment[d,R],
Circle[{0,0},R],
PointSize[.04],Point[{0,0}],
Line[{{-c,d},{0,0},{c,d}}],
Dashing[{.04}],Line[{{-c,d},{c,d}}]
},
AspectRatio->Automatic,
PlotRange->All
],
Graphics[{
{Yellow,Disk[{0,0},R,{0,Pi/2}]},
Circle[{0,0},R],
Dashing[{.04}],
Line[{{1,0},{0,0},{0,1}}]
},AspectRatio->Automatic,
PlotRange->All]
}
]
]
]

<<Calculus`Integration`

Integrate[Boole[x^2+y^2<1&&y>1-.596027],{x},{y}]

Pi/4.