I'm trying to draw the intersection of the following two equations: x^2+y^2=4 (a circle of radius 2) and z=x+4. I got kind of close with tikz-3dplot (which has preferrable output in my opinion), but when it got hard I couldn't find any examples to take code from. I also couldn't find any way to input any function (that would fix everything instantly). The output of the tikz-3dplot figure shows the cylinder nicely:


I also included some vague first attempt with pgfplots. It describes the line enclosing the interval nicely, but about everything is said with that: the colour is random, I wouldn't know how to fill the curve, etc.

enter image description here

I have a strong preference for the nice and clean output of tikz-3dplot as well and I'd still like to show the cylinder in the final picture.

So I can imagine I'm heading in the wrong direction since I've hit a brick wall now, which leads me to the following question:

What's a more accessible way to still get pretty output of the intersection of the cylinder and the plane?


        y domain=0:2*pi%
     ({ cos(deg(y))},%
     { sin(deg(y))},%
     % Is ignored?
%     ({ 2cos(deg(y))},%
%     { 2sin(deg(y))},%
%     0);%
        axis/.style={->,black,very thin},


    %\node[left] at (0,0,.6) {$z=x+4$};

    \draw[axis] (\radius,0,0) -- (\axissize,0,0) node[anchor=north east]{$x$};
    \draw[axis,-,dashed] (0,0,0) -- (\radius,0,0);
    \draw[axis,-,dashed] (0,0,0) -- (0,\radius,0);
    \draw[axis] (0,\radius,0) -- (0,\axissize,0) node[anchor=north west]{$y$};
    \draw[axis,-,dashed] (0,0,0) -- (0,0,\th);
    \draw[axis] (0,0,\th) -- (0,0,\th+.5*\axissize) node[anchor=south]{$z$};

    \draw[axis,dashed,-] (0,0,0) -- (-\radius-.03,0,0) node[anchor=south west]{};
    \draw[axis] (-\radius-.03,0,0) -- (-\axissize,0,0) node[anchor=south west]{};
    \draw[axis,dashed] (0,0,0) -- (0,-\axissize,0) node[anchor=south east]{};
    \foreach \height in {0,\th}{
    \tdplotdrawarc[curve,very thin]{(0,0,\height)}{\radius*\costheta}{-25}{\a}{}{}
    \tdplotdrawarc[curve,very thin,dashed]{(0,0,0)}{\radius*\costheta}{\a}{\a+180}{}{}
    \tdplotdrawarc[curve,very thin]{(0,0,\th)}{\radius*\costheta}{\a}{\a+180}{}{}

    \draw (\radius*\costheta,\radius*\sintheta,0) -- (\radius*\costheta,\radius*\sintheta,\th);

    \draw (\radius*\costheta,\radius*\sintheta,0) -- (\radius*\costheta,\radius*\sintheta,\th);

    % I tried something here but it doesn't work :-(
%   \tdplotsetthetaplanecoords{0}
%   \tdplotsetrotatedthetaplanecoords{45}
%   \tdplotsinandcos{\sintheta}{\costheta}{45}
%   \tdplotdrawarc[draw=none,tdplot_rotated_coords,fill=cyan!50,opacity=.5]{(-4*\radius,0,0)}{\radius}{0}{360}{}{}

  • 1
    Look at this example for cylinder, plane, and intersection. texample.net/tikz/examples/dandelin-spheres It has quite a bit of other items in the figure, but eliminating the unwanted is easier than creating new. – R. Schumacher Apr 11 '15 at 18:09
  • @R.Schumacher Thanks. That example actually worked great! – 1010011010 Apr 11 '15 at 20:28

I settled for the following diagram for now (Special thanks to R. Schumacher). It does kind of get the point across. :-) If somebody has a better rendition I'll surely use that one.

enter image description here


\begin{tikzpicture}[MyPersp,axis/.style={->,black,very thin}]
    \def\h{4}% Heigth of the ellipse center (on the axis of the cylinder)
    \def\a{45}% angle of the section plane with the horizontal




    \draw[axis] (1.4*\radius,0,0) -- (\axissize,0,0) node[anchor=south west]{$x$};
    \draw[axis,-,dashed] (0,0,0) -- (1.4*\radius,0,0);
    \draw[axis,-,dashed] (0,0,0) -- (0,1.4*\radius,0);
    \draw[axis] (0,1.4*\radius,0) -- (0,\axissize,0) node[anchor=south east]{$y$};
    \draw[axis,-,dashed] (0,0,0) -- (0,0,\h+2);
    \draw[axis] (0,0,\h+2) -- (0,0,\h+2+.5*\axissize) node[anchor=south]{$z$};
    \draw[axis,dashed,-] (0,0,0) -- (0,-.7*\axissize,0);
    \draw[axis] (0,-\radius,0) -- (0,-\axissize,0) node[right=.125in,below=-.025in]{$-y$};
    \draw[axis,dashed,-] (0,0,0) -- (-\radius,0,0);
    \draw[axis] (-\radius,0,0) -- (-\axissize,0,0) node[below=.075in,left=-.1in]{$-x$};

    \foreach \q in {1,2,3}
    \draw (\q,0,-.1) -- (\q,0,.1) node[below=.1in] {$\q$};
    \draw (-\q,0,-.1) -- (-\q,0,.1) node[below=.1in] {$\mathllap{-}\q$};
    \draw (0,\q,-.1) -- (0,\q,.1) node[below=.1in] {$\q$};
    \draw (0,-\q,-.1) -- (0,-\q,.1) node[below=.1in] {$\mathllap{-}\q$};

    \foreach \q in {1,2,3}
    \draw (0,-.1,\q) -- (0,.1,\q) node[left=.15in,above=-.03in] {$\q$};

    \foreach \t in {135,315}%
        \draw[\clrcyl, thick] ({2*cos(\t)},{2*sin(\t)},0)
    \draw[\clrcyl,thick] ({2*cos(\bfirst)},{2*sin(\bfirst)},0) % lower circle
    \foreach \t in {\bfirst,\bsecond,...,\blast}
    \draw[\clrcyl,thick,dashed] ({2*cos(\afirst)},{2*sin(\afirst)},0) % lower circle dashed
    \foreach \t in {\afirst,\asecond,...,\alast}
    \fill[\clrplaney,draw=\clrplanex,very thick,opacity=0.5]
     (2,0,\h+2) % elliptical section
        \foreach \t in {5,10,...,360}
    \node[pin=45:{\color{\clrcyl}$x^2+y^2=4$}] at (1.25,-1.25,2) {};
    \node[pin=135:{\color{\clrplanex}$z=x+4$}] at (-1.25,1.25,2) {};
    \draw[thick,->,\clrplanex,opacity=.8] (0,0,4) -- (0,{.5*sqrt(3)},{4+.5}) node[above=.15in,left=-.05in] {$\mathbf{\hat N}$};
    \foreach \q in {4,5,6}
    \draw (0,-.1,\q) -- (0,.1,\q) node[left=.15in,above=-.03in] {$\q$};

    \draw[\clrcyl, thick] (2,0,{\h+2}) % upper circle
        \foreach \t in {10,20,...,360}

  • aka 666 -- Nice conversion of the example. – R. Schumacher Apr 11 '15 at 22:10

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