# Plotting a closed parametric curve and shading the area inside it

I wish to plot the curve defined by the parametric equations x = 1+sin t,y = 3 + 3(cos t)^3, such that the region bounded by this curve is shaded. As far as the code goes, I can't seem to find a similar example online to help me, so I don't have a starting point to work from. Any help plotting this curve would be much appreciated. I would like a basic set of axes to go with it, labelled x and y with an O to mark the origin.

Another way without using \clip is the following:

\documentclass[border=3mm]{standalone}
\usepackage{tikz}
\begin{document}
\begin{tikzpicture}
\fill[blue] plot[smooth,domain=-90:90,variable=\t] ({1+sin(\t)},{3 + 3(cos \t)^3})
plot[smooth,domain=90:270,variable=\t] ({1+sin(\t)},{3 + 3(cos \t)^3});
\draw[-latex] (0,0) node[left] {$O$}-- (3,0) node[below]{$x$};
\draw[-latex] (0,0) -- (0,2) node[left]{$y$};
\end{tikzpicture}
\end{document} • Same question as for Marmot: How come that half of this curve is below the x-axis ? y = 3+3(cos t)^3 is always positive. Jan 16 '18 at 19:58
• The problem is that the 3(cos \t)^3 that appears in each coordinate doesn't get parsed correctly. This can be fixed by making the multiplication explicit: 3*(cos \t)^3. Feb 19 '19 at 17:02

Error corrected, thanks to Franck Pastor!

\documentclass{article}
\usepackage{tikz}
\begin{document}
\begin{center}
\begin{tikzpicture}
\draw[-latex] (0,0) node[left] {$O$}-- (3,0) node[below]{$x$};
\draw[-latex] (0,0) -- (0,6) node[left]{$y$};
\clip plot[smooth,samples=36,domain=0:360,variable=\t] ({1+sin(\t)},{3 + 3*(cos
(\t))^3});
\draw[fill=blue,opacity=0.8] (0,0) rectangle (2,6);
\end{tikzpicture}
\end{center}
\end{document} • How come that half of the curve is below the x-axis ? y = 3+3(cos t)^3 is always positive… Am I missing something? Jan 16 '18 at 19:57
• @FranckPastor That's a very good point!
– user121799
Jan 16 '18 at 20:07
• What does the option [-latex] do to the \draw commands? Feb 19 '19 at 17:07
• @AnnieCarter It adds an arrow head of the latex type, which looks arguably nicer than what you get with ->.
– user121799
Feb 19 '19 at 17:09

Here is a try with MetaPost, inserted in a LuaLaTeX program for convenience. With MetaPost, the parametric curve can be closed and then filled by making it a cycle (.. cycle instruction), right as it's about to join itself.

\documentclass[border=3mm]{standalone}
\usepackage{luatex85, luamplib}
\begin{document}
\begin{mplibcode}
u := cm; xmax = 2.25; ymax = 6.25;
vardef f(expr t) = 1 + sind t enddef;
vardef g(expr t) = 3 + 3((cosd t)**3) enddef;
path curve;
beginfig(1);
% Axes and Labels
drawarrow origin -- (xmax*u, 0);
label.llft(btex $O$ etex, origin);
label.bot(btex $x$ etex, (xmax*u, 0));
drawarrow origin -- (0, ymax*u);
label.lft(btex $y$ etex, (0, ymax*u));
% Filled Curve
curve = ((f(0), g(0)) for t = 1 upto 359: .. (f(t), g(t)) endfor .. cycle) scaled u;
fill curve withcolor blue;
endfig;
\end{mplibcode}
\end{document} Here's a very slight variation on user's answer which doesn't require finding coordinates for the shaded region:

\documentclass{article}
\usepackage{tikz}
\begin{document}
\begin{center}
\begin{tikzpicture}
\draw[-latex] (0,0) node[left] {$O$}-- (3,0) node[below]{$x$};
\draw[-latex] (0,0) -- (0,6) node[left]{$y$};
\begin{scope}[fill=blue,opacity=0.8]
\fill[clip] plot[smooth,samples=36,domain=0:360,variable=\t] ({1+sin(\t)},{3 + 3*(cos
(\t))^3});
\end{scope}
\end{tikzpicture}
\end{center}
\end{document}


Here's yet another way to doing it using \filldraw:

\documentclass{article}
\usepackage{tikz}
\begin{document}
\begin{center}
\begin{tikzpicture}
\draw[-latex] (0,0) node[left] {$O$}-- (3,0) node[below]{$x$};
\draw[-latex] (0,0) -- (0,6) node[left]{$y$};
\filldraw[blue!80] plot[smooth,samples=36,domain=0:360,variable=\t] ({1+sin(\t)},{3 + 3*(cos
(\t))^3});
\end{tikzpicture}
\end{center}
\end{document} 