# How to obtain the area enclosed by such a curve?

The desired curve is defined as curve02 below :

ClearAll["Global*"];
R = 48.78;
r = 8.13;
z1 = R/r;
z2 = 1 - z1;
e = 7.05;
f = r/e;
re = 12.6;
θ = ArcTan[Sin[z1 τ]/(f + Cos[z1 τ])] - τ;
φ = ArcSin[f Sin[θ + τ]] - θ;
ψ = z1/(z1 - 1) φ;
curve01 = {(R - r) Sin[τ] + e Sin[z2 τ] -
re Sin[θ], (R - r) Cos[τ] - e Cos[z2 τ] +
re Cos[θ]} // FullSimplify;
curve02 = {curve01[] Cos[φ - ψ] -
curve01[] Sin[φ - ψ] - e Sin[ψ],
curve01[] Sin[φ - ψ] +
curve01[] Cos[φ - ψ] - e Cos[ψ]} //
Simplify;
ParametricPlot[Evaluate[curve02], {τ, 0, 5 π},
Exclusions -> None, MaxRecursion -> 15, PlotPoints -> 1000]


which can be visualized as: How to obtain the area of its enclosed region? update

Green's theorem can solve another similar problem with high accuracy but does not suit this one, below is an example:

I tried to rewrite your original curve as below, just in order to make sure the derivatives of the parametric form can be obtained by Mathematica by avoiding Abs or Sign:

ncurve={(Cos[t]^2 )^(1/4),(Sin[t]^2)^(1/4)}


Then the numerical result of the closed area can be obtained by applying Green's theorem:

4*Quiet@NIntegrate[ncurve[] D[ncurve[], t], {t, 0, Pi/2}] //
NumberForm[#, 15] &


which gives:

3.708149351621483

Do you wan the entire area enclosed by the outer envelope? A bit brute force, but note the 10-fold symmetry, so that only two arc segments define the outer boundary:

base = Line@Table[ curve02, {\[Tau], 0, 5 Pi, Pi/1000}];
r1 = FindRoot[ (curve02 /.  \[Tau] -> x) == (curve02 /.  \[Tau] ->
y), {x, .5}, {y, 5.5}];
p1 = y /. FindRoot[ (ArcTan @@ (curve02 /. \[Tau] -> y)) ==
3 Pi/ 10 , {y, .55}]
top = x /. FindRoot[ curve02[] /.  \[Tau] -> x , {x, 5}];
arc = Join[Table[ curve02 , {\[Tau], top, y /. r1 , .0001}],
Table[ curve02 , {\[Tau], x /. r1, p1 , .0001}]];
Graphics[{base, {Red,
Line[{curve02 /. \[Tau] -> top, {0, 0}, curve02 /. \[Tau] -> p1}],
Line@arc }}] now the area of the polygon slice: ( by 10 gives the total ) (https://mathematica.stackexchange.com/a/22587/2079 )

PolygonSignedArea[pts_?MatrixQ] := Total[Det /@ Partition[pts, 2, 1, 1]]/2;
area = 10 PolygonSignedArea[Reverse@Join[{{0, 0}}, arc]]


7936.86

as noted in the comments, if we set the increment to 10^-6 this converges to the more sophisticated NIntegrate result of 7945.5

• I got 7945.52, but I used NIntegrate. Jul 17, 2015 at 23:23
• @LCFactorization Using george's setup, 5 (NIntegrate[-First[curve02] D[Last@curve02, \[Tau]] + Last[curve02] D[First@curve02, \[Tau]], {\[Tau], top, y /. r1}] + NIntegrate[-First[curve02] D[Last@curve02, \[Tau]] + Last[curve02] D[First@curve02, \[Tau]], {\[Tau], x /. r1, p1}]). My brute-force approach was similar and not distinctive enough to justify a separate answer. Jul 18, 2015 at 0:34
• @LCFactorization Each arc segment spans a convex wedge and any polygonalization (with vertices on the arc and origin) will underestimate the area. Jul 18, 2015 at 0:58
• @MichaelE2 thank you for instructive supplemental code snippet. Jul 18, 2015 at 2:05
• bugger. base is the original graphic, something like Line@Table[ curve02 , {\[Tau], 0,5 Pi , .01}] (or use parametric plot ). Jul 18, 2015 at 13:54

I may have missed the point but I post out of interest.

p = ParametricPlot[curve02, {\[Tau], 0, 5 Pi}]
c[t_] := curve02 /. \[Tau] -> t
point = SortBy[c /@ Range[0, 5 Pi, 0.001], Norm];
Manipulate[
ListPlot[point[[1 ;; n]], AspectRatio -> Automatic], {n, 1, 15000,
1}] The manipulate allows to get "interior"

Getting desired points:

points = point[[1 ;; 9473]];
pnts = DeleteCases[
points, {x_, y_} /;
Norm[{x, y}] >
45 && (Pi/5 < ArcTan[x, y] < Pi/2.5 ||
Pi/5 < ArcTan[-x, y] < Pi/2.5 || 0 < ArcTan[-x, -y] < Pi/6 ||
Pi/2.5 < ArcTan[-x, -y] < Pi/2 ||
Pi/2.5 < ArcTan[x, -y] < Pi/2 || 0 < ArcTan[x, -y] < Pi/6)]


then

pg = Polygon[pnts[[Last@FindShortestTour[pnts]]]];
Show[p, Graphics[{Red, pg}]]
Area[pg]
`

where area yields: 5242.29 • thanks. Two issues: (1). The desired area is the enclosed region by the profile of the photo, not the red; (2). how do you obtain the magic number 9473, and the logic expression for selecting desired points? Jul 17, 2015 at 14:06
• @LCFactorization re:(1) I am sorry I am not sure I understand your desired region (2) I used manipulate to get as close to 'interior' for step size I chose then deleted 'outside' to allow polygon...in my Timezone it is midnight so am off to sleep. Good luck :) Jul 17, 2015 at 14:12