Filling the area enclosed by two arcs and a rectangle

Question

I am trying to fill the area marked gray below.

I found some examples with circles doing this kind a thing using clip. But I could not find the way to do that with arcs.

\documentclass[a4paper,12pt]{article}
\usepackage{mathtools}
\usepackage{tkz-euclide}
\usepackage{tikz}
\usetikzlibrary{quotes,angles,shapes,backgrounds}
\def \firstarc{(4,0) arc (0:90:4cm)}
\def \secondarc{(8,0) arc (0:30:8cm)}
\begin{document}
\begin{tikzpicture}
    \coordinate (A) at (0,0);
    \coordinate (B) at (8,0);
    \coordinate (C) at (8,4);
    \coordinate (D) at (0,4);

    \draw \firstarc;
    \draw \secondarc;

    \draw (A)--(B)--(C)--(D)--(A);
\end{tikzpicture}
\end{document}

Answer 1

All of the answers here are good as they provide the actual area, and do not cheat by filling the background white. (Albeit this would be easier and in most cases an adequate solution.)

However, the solutions does not work when the height and the width of the rectangle changes, as all the solutions hardcode in the 30 degree angle. I am sure you can shorten this code by using the intersection library, and there exists pure tikz solutions; this is my solution.

The result looks like everybody else's attempt. Note that these images are transparent.

However, as mentioned one can also change the dimensions

Code

\documentclass[a4paper,12pt,margin=0.1mm]{standalone}
\usepackage{tkz-euclide}
\usetkzobj{all} % load all objects not neccecary in TexLive2020
\begin{document}
  \begin{tikzpicture}
    \def\width{8} \def\height{3}

    \tkzDefPoints{0/0/A, \width/0/B, \width/\height/C, 0/\height/D}
    \tkzClipPolygon(A,B,C,D)

    % Different waus to define the point E
    % \tkzDefPointBy[rotation= center A angle 90](D) \tkzGetPoint{E}
     \tkzDefPoint(0.5*\width, 0){E}
    % \tkzDefPoint(\height, 0){E}

    \tkzCalcLength[cm](A,E) \tkzGetLength{rAE} % \rAE gives the length

    \tkzInterLC(C,D)(A,B) \tkzGetPoints{J1}{J2}
    \tkzFindSlopeAngle(A,J2)\tkzGetAngle{JAB} % Might have to switch with J1 here

    \tkzDrawCircle(A,B); \tkzDrawCircle(A,E)

    \fill[blue] (E) arc (0:90:\rAE cm) -- (D) -- (\JAB:\width cm) 
                    arc (\JAB:0:\width cm);
    \tkzDrawPolygon[red](A,B,C,D)
  \end{tikzpicture}
\end{document}

Answer 2

As you are already using polar coordinates for the arcs.

\documentclass[a4paper,12pt]{article}
\usepackage{mathtools}
\usepackage{tkz-euclide}
\usepackage{tikz}
\usetikzlibrary{quotes,angles,shapes,backgrounds}
\def \firstarc{(4,0) arc (0:90:4cm)}
\def \secondarc{(8,0) arc (0:30:8cm)}
\begin{document}
\begin{tikzpicture}
    \coordinate (A) at (0,0);
    \coordinate (B) at (8,0);
    \coordinate (C) at (8,4);
    \coordinate (D) at (0,4);

    \draw \firstarc;
    \draw \secondarc;

    \draw[red] (A)--(B)--(C)--(D)-- cycle;
    \fill[blue] (B) -- (4,0) arc (0:90:4cm) -- (30:8cm) arc (30:0:8cm);
\end{tikzpicture}
\end{document}

Answer 3

Simplified version of @BambOo answer (+1):

\documentclass[a4paper,12pt]{article}
\usepackage{tikz}

\begin{document}
    \begin{tikzpicture}
\draw[red]  (0,0) rectangle (8,4);
\fill[blue] (4,0) arc (0:90:4cm) -- (30:8cm) arc (30:0:8cm);
    \end{tikzpicture}
\end{document}

Result is the same:

Answer 4

Hope this help!

\documentclass[a4paper,12pt]{article}
\usepackage{mathtools}
\usepackage{tkz-euclide}
\usepackage{tikz}
\usetikzlibrary{quotes,angles,shapes,backgrounds}
\def \firstarc{(4,0) arc (0:90:4cm)}
\def \secondarc{(8,0) arc (0:30:8cm)}
\begin{document}
\begin{tikzpicture}
    \coordinate (A) at (0,0);
    \coordinate (B) at (8,0);
    \coordinate (C) at (8,4);
    \coordinate (D) at (0,4);

    \draw \firstarc;
    \draw \secondarc;

    \draw (A)--(B)--(C)--(D)--(A);
    \fill[fill=gray](30:8cm)--(0,4) arc (90:0:4cm)--(8,0) arc (0:30:8cm)--cycle;
\end{tikzpicture}
\end{document}