# Viviani curve with no self intersection

I was trying to make the following in TiKz which I made using Mathematica, which is the intersection of a sphere and cylinder similar to the Viviani's curve.

I want to highlight the shaded region of the intersection of a solid ball with a cylinder (the intersection is the blue curve and the region is the region bounded by the blue curve) which I was unable to make into TiKz. I don't have a good amount of knowledge of TiKz (still in learning process). I tried to change the code but unable to get the required one. Any help will be appreciated.

Edit

I am not getting how the parameterization is being used in the code. I have the following parametrization: • An idea could be to use the points of the curve calculate with Mathematica and pass them to TikZ. Oct 21, 2020 at 15:31
• @vipa could you please elaborate yourself? Oct 22, 2020 at 5:14
• @Á.Iborra I have changed it. There was a typo. Oct 27, 2020 at 1:02

Compile on your computer with Asymptote.

settings.outformat="pdf";
settings.prc=false;
settings.render=0;
import x11colors;
import graph3;
size(8cm,0);

/*
real h = 5;
triple[] axes = new triple[] {X, Y, Z};
for(int i=0; i<axes.length; ++i){
draw(O--h*axes[i],Arrow3());
}
*/
currentprojection = orthographic(0,-1,0.5);
surface sphere=surface(new triple(pair p) {
real t=p.x, s=p.y;
return (1+cos(t)*sin(s),sin(t)*sin(s),cos(s));},  (0,0),(2*pi,2*pi),Spline);
draw(sphere,lightblue+opacity(.6),MidnightBlue+0.1bp);

surface cylinder=surface(new triple(pair p) {
real t=p.x, s=p.y;
return (cos(t),sin(t),s);}, (0,-2),(2*pi,2),Spline);
draw(cylinder,  lightblue+opacity(.6),MidnightBlue+0.1bp);

draw(graph(new triple(real t) {return
(cos(t),
sin(t),
sqrt(2*cos(t)-1)); },
-pi/3,pi/3,300),red+1.5bp);
draw(graph(new triple(real t) {return
(cos(t),
sin(t),
-sqrt(2*cos(t)-1)); },
-pi/3,pi/3,300),red+1.5bp);
shipout(bbox(2mm,invisible)); Compile the code above without

settings.outformat="pdf";
settings.prc=false;
settings.render=0;


and without shipout(bbox(2mm,invisible));

in http://asymptote.ualberta.ca/ to get an interaction. • Can you answer my question here with Asymptote? Oct 27, 2020 at 8:56
• @MoneyOrientedProgrammer can you guess, :-))
– user213378
Oct 27, 2020 at 9:00
• I cannot guess because you said that you haven't learnt Asymptote 3D yet. :-) Oct 27, 2020 at 9:03

This is what I got. You might want to fine-tune the z-buffer.

\documentclass[tikz]{standalone}
\usepackage{pgfplots}
\pgfplotsset{compat=1.17}
\begin{document}

\def\CylinderLower[#1]{
domain=-180:180, samples=60,
domain y=-1.2:0, samples y=2,
#1
]({cos(x)}, {sin(x)}, {min(y, -sqrt(1-4*sin(x/2)^2) )});
}

\def\CylinderInner[#1]{
domain=-60:60, samples=30,
domain y=-1:1, samples y=2,
#1
]({cos(x)}, {sin(x)}, {y * sqrt(1-4*sin(x/2)^2)});
}

\def\CylinderUpper[#1]{
domain=-180:180, samples=60,
domain y=0:1.2, samples y=2,
#1
]({cos(x)}, {sin(x)}, {max(y, sqrt(1-4*sin(x/2)^2) )});
}

\pgfmathdeclarefunction{dual}1{\pgfmathparse{sqrt(1-(#1)^2)}}

\def\SphereInner[#1]{
domain=-1:1, samples=10,
domain y=-1:1, samples y=20,
#1
]({1 - cos(x*acos(dual(y)/2)) * dual(y)},
{sin(x*acos(dual(y)/2)) * dual(y)},
{y});
}

\def\SphereOuter[#1]{
domain=-1:1, samples=10,
domain y=-1:1, samples y=20,
#1
]({1 + cos(x*acos(-dual(y)/2)) * dual(y)},
{sin(x*acos(-dual(y)/2)) * dual(y)},
{y});
}

\tikz[cap=round,join=round]{
\begin{axis}[axis equal,]
\CylinderLower[point meta=x]
\SphereInner[point meta=-z^2,opacity=.9]
\CylinderInner[point meta=x,opacity=.5]
\end{axis}
}

\end{document} • Thanks for this wonderful illustration. Just a small comment, I want to highlight the intersection. Is it possible to do? Oct 27, 2020 at 6:55