Problem source

A  movable  point $P$ has cartesian coordinates $(x,y)$, where  $x$ and $y$ are functions of $t$.  The polar coordinates of  $P$ with respect to the origin $O$ are $r$ and $\theta$. 
Starting with the expression 
\[
\tfrac12 \int r^2 \, \d \theta
\] for the area swept out by $OP$, obtain the equivalent expression
\[
\tfrac12 \int \left( x\frac{\d y}{\d t} - y \frac{\d x}{\d t}\right)\d t
\,.
\tag{$*$}
\]
The ends of a thin  straight rod $AB$ lie on a closed convex curve $\cal C$. The point $P$ on the rod is a fixed distance $a$ from $A$ and a fixed distance $b$ from $B$.  The angle between  $AB$ and the positive $x$ direction  is $t$.  As $A$ and $B$ move anticlockwise round $\cal C$, the angle $t$ increases from $0$ to $2\pi$ and  $P$ traces a closed convex curve $\cal D$ inside $\cal C$, with the origin $O$ lying inside $\cal D$, as shown in the diagram. 
\begin{center}
\begin{tikzpicture}[
    scale=0.8,
    >=stealth,
    line width=0.3pt,
    dot/.style={circle, fill=black, inner sep=0pt, minimum size=3pt}
]
% Bezier curves forming the main shape
\draw (0.04,2.13) .. controls (-0.22,3.77) and (2.63,4.87) .. (3.63,4.35);
\draw (3.63,4.35) .. controls (4.97,3.75) and (7.09,1.81) .. (5.59,0.23);
\draw (0.04,2.13) .. controls (0.05,0.29) and (3.37,-1.84) .. (5.59,0.23);
% Rotated ellipse
\draw[rotate around={11.46:(3,1.95)}] (3,1.95) ellipse (4.32 and 3.27);
% Thick line AB
\draw[line width=1.6pt] (-1.11,0.95) -- (1.98,5.12);
% Coordinate axes
\draw[->] (-2.65,1.27) -- (8.54,1.27);
\draw[->] (3.79,-3.49) -- (3.81,6.8);
% Small arc (approximating the parametric plot)
\draw[line width=0.2pt] (-0.87,1.28) arc (180:126.87:0.87);
\draw[line width=0.2pt] (-0.87,1.28) -- (-0.87,1.28);
% Labels
\node[above left] at (1.31,3.78) {$P$};
\node[above left] at (1.24,4.89) {$b$};
\node[above left] at (-0.65,2.41) {$a$};
\node[above left] at (1.87,5.65) {$B$};
\node[above left] at (-1.6,0.95) {$A$};
\node[above left] at (3.26,1.11) {$O$};
\node[above left] at (5.65,3.17) {$\mathcal{D}$};
\node[above left] at (6.73,4.22) {$\mathcal{C}$};
\node[above left] at (-0.45,1.7) {$t$};
\node[above left] at (3.72,7.25) {$y$};
\node[above left] at (8.6,1.4) {$x$};
% Dot
\fill (1.09,3.93) circle (2.5pt);
\end{tikzpicture}
\end{center}
Let $(x,y)$ be the coordinates of $P$. Write down the coordinates of $A$ and $B$ in terms of $a$, $b$, $x$, $y$ and $t$.
The areas swept out by $OA$, $OB$ and $OP$ are
denoted by $[A]$, $[B]$ and $[P]$, respectively. 
Show, using $(*)$, that 
\[
[A] = [P] +\pi a^2 - af
\]
where 
\[
f = \tfrac12 \int _0^{2\pi} \left(
 \Big(x+\frac{\d y}{\d t}\Big)\cos t
+
  \Big(y- \frac{\d x}{\d t}\Big)\sin t
\right) \d t\,.
\]
Obtain a corresponding expression for $[B]$ involving $b$. Hence show that the area between the curves $\cal C$ and $\cal D$ is $\pi ab$.

Solution source

\begin{align*}
&& \tan \theta &= y/x \\
\Rightarrow && \sec^2 \theta \frac{\d \theta}{\d t} &= \frac{x \frac{\d y}{\d t} - y \frac{\d x}{\d t}}{x^2}
\\
\Rightarrow && \frac{\d \theta}{\d t} &=\left (x \frac{\d y}{\d t} - y \frac{\d x}{\d t} \right) \frac{\cos^2 \theta}{x^2}  \\
&&&=\left (x \frac{\d y}{\d t} - y \frac{\d x}{\d t} \right) \frac{\cos^2 \theta}{r^2 \cos^2 \theta } \\
&&&=\left (x \frac{\d y}{\d t} - y \frac{\d x}{\d t} \right) \frac{1}{r^2 } \\
&& \tfrac12 \int r^2 \, \d \theta &= \tfrac12 \int \left (x \frac{\d y}{\d t} - y \frac{\d x}{\d t} \right) \d t
\end{align*}


$A = (x - a \cos t, y - a \sin t), B = (x + b \cos t , y + b \sin t)$

\begin{align*}
&& [A] &= \tfrac12 \int_0^{2\pi} \left ((x-a \cos t) \frac{\d (y-a \sin t)}{\d t} - (y-a \sin t) \frac{\d (x-a \cos t)}{\d t} \right) \d t \\
&&&= \tfrac12 \int_0^{2\pi} \left ((x - a \cos t) \left ( \frac{\d y}{\d t} - a \cos t \right)  - (y - a \sin t) \left ( \frac{\d x}{\d t} + a \sin t \right)\right) \d t \\
&&&= \tfrac12 \int_0^{2\pi} \left ( x \frac{\d y}{\d t} - y \frac{\d x}{\d t} - a \cos t \frac{\d y}{\d t}-ax \cos t +a^2 \cos^2 t + a \sin t \frac{\d x}{\d t}-y a \sin t + a^2 \sin^2 t \right) \d t \\
&&&= \tfrac12 \int_0^{2\pi} \left ( \underbrace{x \frac{\d y}{\d t} - y \frac{\d x}{\d t}}_{[P]}-a\left ((x + \frac{\d y}{\d x}) \cos t + (y - \frac{\d x}{\d t}) \sin t \right) + \underbrace{a^2}_{\pi a^2}  \right) \d t \\
&&&= [P] + \pi a^2 - af
\end{align*}

\begin{align*}
&& [B] &= \tfrac12 \int_0^{2\pi} \left ((x+b \cos t) \frac{\d (y+b \sin t)}{\d t} - (y+b \sin t) \frac{\d (x+b \cos t)}{\d t} \right) \d t \\
&&&= \tfrac12 \int_0^{2\pi} \left ((x+b \cos t) (\frac{\d y}{\d t} + b \cos t) - (y+b \sin t)(\frac{\d x}{\d t} - b \sin t) \right) \d t \\
&&&=  \tfrac12 \int_0^{2\pi} \left (x \frac{\d y}{\d t} - y \frac{\d x}{\d t} + b^2 + b(\cos t (x + \frac{\d y}{\d t}) +(y - \frac{\d x}{\d t})\sin t\right) \d t \\
&&&= [P] + \pi b^2 + b f
\end{align*}

Since $A$ and $B$ trace out the same area, we must have $\pi a^2 - af = \pi b^2 + bf \Rightarrow \pi (a^2-b^2) = f(b+a) \Rightarrow f = \pi (a-b)$.

In particular the area inbetween is $[A] - [P] = \pi a^2 - a \pi (a-b)$