Thread #16918610
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Well?
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>>16918686
~4.82
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>>16918716
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>>16918716
>>16918717
approximation, no algebraic solution
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>>16918610
Difference of integrals maybe? The area of a convex Jordan curve [math]J[/math] can be found by integrating a CW parameterization:
[eqn]\int_{0}^1|\gamma\left(t\right)|dt[/eqn]
where [math]\gamma:[0,1]\rightarrow J[/math].
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>>16918655
yes
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>>16918610
ContourPlot[{2 π Sin[y Cos[y] – 2 π Sin[x]] = Sin[x] y Cos[y], (x – (3 π)/2)^2 + (y – 2 π)^2 = (π/2)^2}, {x, π, 2 π}, {y, (3 π)/2, (5 π)/2}]
The area of the depicted yellow circle is π^3/4.
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>>16920217
ContourPlot[{2 Pi Sin[x Cos[x] + 2 Pi Cos[y]] = (–x) Cos[x] Cos[y], (x – 2 Pi)^2 + y^2 = (Pi/2)^2}, {x, (3 Pi)/2, (5 Pi)/2}, {y, 0, Pi}]
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>>16920537
ContourPlot[{2 Pi Sin[x Cos[x] + 2 Pi Cos[y]] = (–x) Cos[x] Cos[y], 2 Pi (1 – Cos[y]) = x Cos[x] + ArcSin[(x Cos[x] Cos[1.5814630598732267])/(2 Pi)]}, {x, (3 Pi)/2, (5 Pi)/2}, {y, 0, Pi}]
The depicted yellow curve can be easily solved for y.
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>>16920601
>The depicted yellow curve can be easily solved for y.
correction: The equation of the depicted [...] for y.
A(n) = area of region enclosed by n-th shortest blue curve
Cos[y] = 1 – (x Cos[x] + ArcSin[(x Cos[x] Cos[1.5814630598732267])/(2 Pi)])/(2 Pi)
N = 2 Integrate[y, {x, (3 Pi)/2, (5 Pi)/2}] ≈ 7.27796177
A(1) < A(2) < N < A(3) ≈ N
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>>16920956
>7.27796177
This is a lower bound.
>>16918717
>7.4377073357
I don't know how to approximate the exact value.
>>16920217
>π^3/4
≈ 7.75156917
This is an upper bound.
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>>16921825
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>>16922001
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I can't get into these threads unless they're in the form of an agartha meme
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>>16921825
ds = 0.002000
4850 points
area = 7.437716013334791
ds = 0.001000
9701 points
area = 7.437717586974413
ds = 0.000500
19403 points
area = 7.437717980011843
ds = 0.000250
38807 points
area = 7.437718078224901
ds = 0.000125
77615 points
area = 7.437718102772335
et cetera
area ≈ 7.437718110951834