Edge and shading of a cylinder

Question

I am trying to create a wedge from a cylinder like

I have been able to get the basics:

\documentclass[11pt,tikz,border=3.14pt]{standalone}
\usepackage{tikz-3dplot}
\usepackage{pgfplots}

\begin{document}
\tdplotsetmaincoords{70}{130}
\begin{tikzpicture}[tdplot_main_coords,scale=3]

\def\pos{0.81} %value 0 < \pos < 1

\coordinate (O) at (0,0,0);

\coordinate (A) at (0,0,0);
\coordinate (B) at (1,0,0);
\coordinate (C) at (1,2,0);
\coordinate (D) at (0,2,0);

\coordinate (AT) at (0,0,1);
\coordinate (BT) at (1,0,1);
\coordinate (CT) at (1,2,1);
\coordinate (DT) at (0,2,1);

        \draw[thick,black] (A) -- (B) -- (C) -- (D) -- cycle;
        \draw[thick,black] (AT) -- (BT) -- (CT) -- (DT) -- cycle;

        \draw[thick,black] (A) -- (AT);
        \draw[thick,black] (B) -- (BT);
        \draw[thick,black] (C) -- (CT);
        \draw[thick,black] (D) -- (DT);

        \draw[thin,black] (1,1,0) -- (0,1,0) -- (0,1,1) -- cycle;

% The circle/ellipse:
\begin{scope}[canvas is xy plane at z=0]
        \draw[thick] (1,0) arc (270:90:1);
\end{scope}

\begin{scope}[
                plane origin={(1,0,0)},
                plane x={(1,1,0)},
                plane y={(0,0,1)},
                canvas is plane
             ]
        \draw[thick] (0,0) arc (180:0:1);
\end{scope}

% The manually computed right-edge:
        \draw[thin] ({1-sqrt(1-\pos^2)},1+\pos,0) -- ({1-sqrt(1-\pos^2)},1+\pos,{sqrt(1-\pos^2)}) ;

% The axis:
    \draw[->] (O) --++ (1.5,0,0) node[below] {$x$};
    \draw[->] (O) --++ (0,2.5,0) node[below] {$y$};
    \draw[->] (O) --++ (0,0,1.25) node[right] {$z$};
\end{tikzpicture}

\end{document}

that renders as:

But I have a couple of roadblocks:

I would like to compute the right-edge of the cylinder automatically, much like this answer here, so I can experiment with the best angle view of the image, but I guess, I did not manage to understand how that solution fully works.

I also would like to shade the whole solid surface, but I am having problems with the edge computed by hand.

Answer 1

I was thinking about something like this; in this image the longitude (defining the observer equals 63 degrees and the angle theta equals 60 degrees.

In the code, the point of interest (the "tangency" point) is denoted P. The code uses tikz-3dplot and the library math to have all the parameters: latitude, longitude, angle theta, radius.

Below is the drawing with longitude=45 and theta=39.

The code

\documentclass[11pt, border=10pt]{standalone}
\usepackage{tikz}
\usetikzlibrary{math}
\usepackage{tikz-3dplot}
\begin{document}

\tikzset{%
  view/.style 2 args={%  observer longitude and latitude (y upwards)
                      %  Remark. lomg=0 means x=0
    z={({-sin(#1)}, {-cos(#1)*sin(#2)})},
    x={({cos(#1)}, {-sin(#1)*sin(#2)})},
    y={(0, {cos(#2)})},
    evaluate={%
      \toxx={sin(#1)*cos(#2)};
      \toyy={sin(#2)};
      \tozz={cos(#1)*cos(#2)};
    },
    longitude = #1,
    latitude = #2
  },
  c arc/.style args={#1:#2:#3}{insert path={++(#1:#3) arc (#1:#2:#3)}},
}
\pgfkeys{/tikz/.cd,
  latitude/.store in=\toLatit,  % observer's latitude
  latitude=0
}
\pgfkeys{/tikz/.cd,
  longitude/.store in=\toLongit,  % observer's longitude
  longitude=0  % corresponds to x=0
}

\begin{tikzpicture}[view={45}{20}, every node/.style={scale=.8}]
  \tikzmath{%
    \r = 2;
    \ang = 39;
    \h = \r*tan(\ang);
    \tanAng = \toLongit +90;
    \rPrime = \r/cos(\ang);
  }

  % the prism
  \draw[gray, dashed, very thin, canvas is yz plane at x=-\r]
  (0, 0) -- ++(\h, 0) -- ++(0, -\r) -- ++(-\h, 0) -- cycle;
  \draw[gray, dashed, very thin, canvas is xy plane at z=-\r]
  (-\r, 0) -- ++(2*\r, 0) -- ++(0, \h) -- ++(-2*\r, 0) -- cycle;
  \draw[gray, dashed, very thin, canvas is zx plane at y=0]
  (0, -\r) -- ++(0, 2*\r) -- ++(-\r, 0) -- ++(0, -2*\r) -- cycle;
  \draw[canvas is yz plane at x=\r]
  (0, 0) -- ++(\h, 0) -- ++(0, -\r) -- ++(-\h, 0) -- cycle;
  \draw[canvas is xy plane at z=0]
  (-\r, 0) -- ++(2*\r, 0) -- ++(0, \h) -- ++(-2*\r, 0) -- cycle;
  \draw[canvas is zx plane at y=\h]
  (0, -\r) -- ++(0, 2*\r) -- ++(-\r, 0) -- ++(0, -2*\r) -- cycle;

  % the angle seen in x=0
  \draw[blue, very thin, canvas is yz plane at x=0]
  (\h, -\r) -- (0, 0) -- (0, -\r);
  \draw[blue, very thin, canvas is yz plane at x=0]
  (-90: 1/3*\r)  arc (-90:{-90+\ang}:1/3*\r)
  node[pos=.6, right] {$\theta$};
  
  % the arc in the horizontal plane y = 0
  \draw[red, dashed, canvas is zx plane at y=0]
  (90:\r) arc (90:270:\r);

  % the arc in the slanted plane
  \draw[red, thick, fill=red!60, fill opacity=.5,
  rotate around x=\ang, canvas is zx plane at y=0]
  (90:\r) arc (90:270:\rPrime cm and \r cm) -- cycle;

  % the tangency point
  \path[canvas is zx plane at y=0] (\tanAng: \r) coordinate (P);

  % the cylinder (visible part)
  \fill[red, opacity=.5] (0, 0, 0)
  {[rotate around x=\ang, canvas is zx plane at y=0]
    -- (90:\r) arc (90:\tanAng:\rPrime cm and \r cm)}
  {[canvas is zx plane at y=0] -- (P) arc (\tanAng:90:\r)}
  -- cycle;

  % the visible arc in the plane y = 0
  \draw[red, thick, canvas is zx plane at y=0]
  (P) arc (\tanAng:90:\r);

  % z axis
  \draw[very thick, ->] (-\r, 2/3*\h, 0) -- ++(0, 0, 2/3*\r)
  node[pos=.6, above] {$y$};
\end{tikzpicture}
\end{document}