I have very limited LaTeX knowledge. I am trying to plot the following figure using "tikz" and I appreciate if anyone can help me to finish plotting my figure. I did the following so far:

\documentclass{article}
\usepackage{tikz}
\begin{document}

\begin{tikzpicture}

\draw (0,0) arc (0:180:6);
\draw   (-12,0)  - -(0,0);
\draw (-8,0) arc (0:180:2);
\draw (0,0) arc (0:180:4);
\draw (-6,-0.1) - - (-6,0.1);

\end{tikzpicture}

\end{document}


There is a BIG semi-circle and two inner SMALL semi-circles. I want to label the right endpoint of the BIG semi-circle by "A" and  the left endpoint of the BIG semi-circle by "B". I want to label the center of the BIG semi-circle by "O", and I want to label the point where the two SMALL semicircles meet by "P". I also want to shade the area inside the BIG semi-circle and outside the two SMALL semi-circles. I also want to mention on the figure that the radius of the big circle is: 6. and the distance between the point "O" and the point "P" is "x"

Any help is much appreciated


Question: How do I find the area between two polar curves?
Answer: We can use any one of the line integrals that by Green's Theorem yield the area:

dA = RandomChoice[{x Dt[y], -y Dt[x], (x Dt[y] - y Dt[x])/2}]
param = {x -&gt; r Cos[θ], y -&gt; r Sin[θ]};
boundary1 = {r -&gt; 6};
boundary2 = {r -&gt; 4 + 4 Cos[θ]};
θ0 = θ /. Solve[{Equal @@ (r /. {boundary1, boundary2}), -Pi &lt; θ &lt;  Pi}, {θ}];
Integrate[dA /. param /. boundary2 /. Dt[θ] -&gt; 1, Flatten[{θ, θ0}]] -
 Integrate[dA /. param /. boundary1 /. Dt[θ] -&gt; 1, Flatten[{θ, θ0}]] // Simplify
(*
  1/2 (-y Dt[x] + x Dt[y])

  18 Sqrt[3] - 4 π
*)




Now this is an interesting bug result (the double integral in polar coordinates):

Integrate[r, {r, θ} ∈ ImplicitRegion[6 &lt;= r &lt;= 4 + 4 Cos[θ], {r, θ}]]




As Michael Siefert astutely pointed out, the problem is that the range of θ is assumed to be all reals.  Restricting θ to one period of length 2 Pi gives the right answer:

Integrate[r, {r, θ} ∈ ImplicitRegion[6 &lt;= r &lt;= 4 + 4 Cos[θ] &amp;&amp; 0 &lt;= θ &lt;= 2 Pi, {r, θ}]]
(*  2 (9 Sqrt[3] - 2 π)  *)