Quelques solutions avec d`autres outils proposées par des savants

···
π
···
π

0≤y≤1
[
BAE θ =π
2y
θ=
\
AF A0=
[
BAE ρ = 2θ/πsin(θ)
2/π lim
θ→0
2θ
πsinθ =2 π
M(Oy)P
π=OP/OM

3
√2
x y a
a/x =x/y =y/2a x3= 2a3
a/x =x/y =y/2a
x2=ay y2= 2ax x2=ay xy = 2a2
x2+y2=ax (cylindre)
x2+y2+z2=ax2+y2(tore)
x2+y2+z2=a2x2/b2(cˆone)

KO =3
√2
θ
3t=θ cosθ = 4cos3t−3cost
cost =x4x3−3x=cosθ

\
ODC
θ
θ/3