Drawing Elliptical Arc in tikz or tkz-euclide

Question

How to draw Elliptical Arc in tikz or tkz-euclide

If we know center, X-Radius, Y-Radius , Start Angle - the start angle of the arc and End Angle - the end angle of the arc

MWE

\documentclass{standalone}
\usepackage{tikz}
\usetikzlibrary{calc}
\usepackage{amssymb}
\def\centerarc[#1](#2)(#3:#4:#5)% Syntax: [draw options] (center) (initial angle:final angle:radius)
{ \draw[#1] ($(#2)+({#5*cos(#3)},{#5*sin(#3)})$) arc (#3:#4:#5); }

\tikzset{elliparc/.style args={#1:#2:#3}{%
insert path={(#1:#3) arc (#1:#2:#3)}}}

\begin{document}
\begin{tikzpicture}
\begin{scope}[rotate=0]
\draw[dashed] (0,0) [elliparc=0:180:1cm and .5cm];
\draw[] (0,0) [elliparc=-180:0:1cm and .5cm];

\draw[dashed] (0,3) [elliparc=0:180:1cm and .5cm];
\draw[] (0,3) [elliparc=-180:0:1cm and .5cm];% not working

%\draw (0,3) ellipse ({1cm} and {0.5cm});

\draw[] (1,0) -- ++(0,3);
\draw[] (-1,0) -- ++(0,3);

\path (0,0) -- node[pos=0.5,above] {$r$} ++(1,0);
\draw[thin,color=brown] (0,0) --  ++(0,3.05);
\draw[thin,color=brown] (0,0) -- ++(1,0);
\path (0,0) -- node[pos=0.5,above,sloped] {$h$} ++(0,3.05);
\end{scope}
\end{tikzpicture}
\end{document}

Answer 1

You can fix this simply by adding ++ to the first coordinate in the insert path key of elliparc.

MWE:

\documentclass[tikz,margin=2mm]{standalone}
\usepackage{tikz}
\usepackage{amssymb}

\tikzset{elliparc/.style args={#1:#2:#3}{%
insert path={++(#1:#3) arc (#1:#2:#3)}}}

\begin{document}
\begin{tikzpicture}
\begin{scope}[rotate=0]
\draw[dashed] (0,0) [elliparc=0:180:1cm and .5cm];
\draw (0,0) [elliparc=-180:0:1cm and .5cm];

\draw[dashed] (0,3) [elliparc=0:180:1cm and .5cm];
\draw (0,3) [elliparc=-180:0:1cm and .5cm];% working

\draw (1,0) -- ++(0,3);
\draw (-1,0) -- ++(0,3);

\path (0,0) -- node[pos=0.5,above] {$r$} ++(1,0);
\draw[thin,color=brown] (0,0) -- ++(0,3.05);
\draw[thin,color=brown] (0,0) -- ++(1,0);
\path (0,0) -- node[pos=0.5,above,sloped] {$h$} ++(0,3.05);
\end{scope}
\end{tikzpicture}
\end{document}

Just for fun I also edited your definition of \centerarc such that the optional first argument is really optional, and it does not need the calc library anymore, because that seemed a bit overkill.

\makeatletter
\def\centerarci[#1](#2) (#3:#4:#5)% Syntax: [draw options] (center) (initial angle:final angle:radius)
{\draw[#1] (#2) ++(#3:#5) arc (#3:#4:#5);}
\def\centerarc{\@ifnextchar[{\centerarci}{\centerarci[]}}
\makeatother

MWE with \centerarc and drawing the cylinder with both options:

\documentclass[tikz,margin=2mm]{standalone}
\usepackage{tikz}
\usepackage{amssymb}

\makeatletter
\def\centerarci[#1](#2) (#3:#4:#5)% Syntax: [draw options] (center) (initial angle:final angle:radius)
{\draw[#1] (#2) ++(#3:#5) arc (#3:#4:#5);}
\def\centerarc{\@ifnextchar[{\centerarci}{\centerarci[]}}
\makeatother

\tikzset{elliparc/.style args={#1:#2:#3}{%
insert path={++(#1:#3) arc (#1:#2:#3)}}}

\begin{document}
\begin{tikzpicture}
\begin{scope}[rotate=0]
\draw[dashed] (0,0) [elliparc=0:180:1cm and .5cm];
\draw (0,0) [elliparc=-180:0:1cm and .5cm];

\draw[dashed] (0,3) [elliparc=0:180:1cm and .5cm];
\draw (0,3) [elliparc=-180:0:1cm and .5cm];% working

\draw (1,0) -- ++(0,3);
\draw (-1,0) -- ++(0,3);

\path (0,0) -- node[pos=0.5,above] {$r$} ++(1,0);
\draw[thin,color=brown] (0,0) --  ++(0,3.05);
\draw[thin,color=brown] (0,0) -- ++(1,0);
\path (0,0) -- node[pos=0.5,above,sloped] {$h$} ++(0,3.05);
\end{scope}
\end{tikzpicture}
\begin{tikzpicture}
\begin{scope}[rotate=0]
\centerarc[dashed](0,0) (0:180:1cm and 0.5cm);
\centerarc(0,0) (-180:0:1cm and 0.5cm);

\centerarc[dashed](0,3) (0:180:1cm and 0.5cm);
\centerarc(0,3)  (-180:0:1cm and 0.5cm);

\draw (1,0) -- ++(0,3);
\draw (-1,0) -- ++(0,3);

\path (0,0) -- node[pos=0.5,above] {$r$} ++(1,0);
\draw[thin,color=brown] (0,0) --  ++(0,3.05);
\draw[thin,color=brown] (0,0) -- ++(1,0);
\path (0,0) -- node[pos=0.5,above,sloped] {$h$} ++(0,3.05);
\end{scope}
\end{tikzpicture}
\end{document}