# LATEX Removing part of shaded area in polar plot

I've been working on creating the following picture using TikZ.

The following code is as far as I've gotten.

\documentclass[width=5cm, height=5cm, border=2mm]{standalone}
\usepackage{tikz}
\begin{document}
\begin{tikzpicture}[>=latex]

% cartesian coordinates
\draw[-] (-2cm,0cm) -- (2cm,0cm) node[right,fill=white] {$x$};
\draw[-] (0cm,-1cm) -- (0cm,3.1cm) node[above,fill=white] {$y$};

\draw [thick, domain=0:2*pi, samples=200, smooth] plot (xy polar cs:angle=\x r,radius=
{1+1*sin(\x r)});
\filldraw [thick, even odd rule, fill=gray!50, fill opacity=0.3, domain=pi/6:5*pi/6,
samples=200, smooth] plot (xy polar cs:angle=\x r,radius= {1+1*sin(\x r)})
(0,1.5)circle(1.5cm);

\end{tikzpicture}
\end{document}


I was attempting with this code to first plot the sine graph(the one with a little dimple at the bottom) and then plot the shaded circle graph. The sine graph will only exist between pi/6 and 5pi/6, thus it'll "even odd rule" with the existing shaded circle and remove the shading within that region but as seen below it doesn't quite work out the way I was expecting. Pardon my limp explanation of the even odd rule, it's new to me and I'm still learning how to competently use it.

The above code produces the following polar plot.

How can I remove the obvious gray semicircle so it'll be like the sketch above?

This solution does not use the even odd rule option, just exploits the fact that you know the intersection points

\documentclass[width=5cm, height=5cm, border=2mm]{standalone}
\usepackage{tikz}
\begin{document}
\begin{tikzpicture}[>=latex]

% cartesian coordinates
\draw[-] (-2cm,0cm) -- (2cm,0cm) node[right,fill=white] {$x$};
\draw[-] (0cm,-1cm) -- (0cm,3.1cm) node[above,fill=white] {$y$};