\(% Differentiation % https://tex.stackexchange.com/a/60546/ \newcommand{\Diff}{\mathop{}\!\mathrm{d}} \newcommand{\DiffFrac}[2]{\frac{\Diff #1}{\Diff #2}} \newcommand{\DiffOp}[1]{\frac{\Diff}{\Diff #1}} \newcommand{\Ndiff}[1]{\mathop{}\!\mathrm{d}^{#1}} \newcommand{\NdiffFrac}[3]{\frac{\Ndiff{#1} #2}{\Diff {#3}^{#1}}} \newcommand{\NdiffOp}[2]{\frac{\Ndiff{#1}}{\Diff {#2}^{#1}}} % Evaluation \newcommand{\LEvalAt}[2]{\left.#1\right\vert_{#2}} \newcommand{\SqEvalAt}[2]{\left[#1\right]_{#2}} % Epsilon & Phi \renewcommand{\epsilon}{\varepsilon} \renewcommand{\phi}{\varphi} % Sets \newcommand{\NN}{\mathbb{N}} \newcommand{\ZZ}{\mathbb{Z}} \newcommand{\QQ}{\mathbb{Q}} \newcommand{\RR}{\mathbb{R}} \newcommand{\CC}{\mathbb{C}} \newcommand{\PP}{\mathbb{P}} \renewcommand{\emptyset}{\varnothing} % Probabililty \DeclareMathOperator{\Cov}{Cov} \DeclareMathOperator{\Corr}{Corr} \DeclareMathOperator{\Var}{Var} \DeclareMathOperator{\Expt}{E} \DeclareMathOperator{\Prob}{P} % Distribution \DeclareMathOperator{\Binomial}{B} \DeclareMathOperator{\Poisson}{Po} \DeclareMathOperator{\Normal}{N} \DeclareMathOperator{\Exponential}{Exp} \DeclareMathOperator{\Geometric}{Geo} \DeclareMathOperator{\Uniform}{U} % Complex Numbers \DeclareMathOperator{\im}{Im} \DeclareMathOperator{\re}{Re} % Missing Trigonometric & Hyperbolic functions \DeclareMathOperator{\arccot}{arccot} \DeclareMathOperator{\arcsec}{arcsec} \DeclareMathOperator{\arccsc}{arccsc} \DeclareMathOperator{\sech}{sech} \DeclareMathOperator{\csch}{csch} \DeclareMathOperator{\arsinh}{arsinh} \DeclareMathOperator{\arcosh}{arcosh} \DeclareMathOperator{\artanh}{artanh} \DeclareMathOperator{\arcoth}{arcoth} \DeclareMathOperator{\arsech}{arsech} \DeclareMathOperator{\arcsch}{arcsch} % UK Notation \DeclareMathOperator{\cosec}{cosec} \DeclareMathOperator{\arccosec}{arccosec} \DeclareMathOperator{\cosech}{cosech} \DeclareMathOperator{\arcosech}{arcosech} % Paired Delimiters \DeclarePairedDelimiter{\ceil}{\lceil}{\rceil} \DeclarePairedDelimiter{\floor}{\lfloor}{\rfloor} \DeclarePairedDelimiter{\abs}{\lvert}{\rvert} \DeclarePairedDelimiter{\ang}{\langle}{\rangle} % Vectors \newcommand{\vect}[1]{\mathbf{#1}} \newcommand{\bvect}[1]{\overrightarrow{#1}} % https://tex.stackexchange.com/a/28213 % \DeclareMathSymbol{\ii}{\mathalpha}{letters}{"10} % \DeclareMathSymbol{\jj}{\mathalpha}{letters}{"11} % \newcommand{\ihat}{\vect{\hat{\ii}}} % \newcommand{\jhat}{\vect{\hat{\jj}}} \newcommand{\ihat}{\textbf{\^{ı}}} \newcommand{\jhat}{\textbf{\^{ȷ}}} \newcommand{\khat}{\vect{\hat{k}}} % Other Functions \DeclareMathOperator{\sgn}{sgn} \DeclareMathOperator{\tr}{tr} % Other Math Symbols \DeclareMathOperator{\modulo}{mod} \newcommand{\divides}{\mid} \newcommand{\notdivides}{\nmid} \newcommand{\LHS}{\text{LHS}} \newcommand{\RHS}{\text{RHS}} \newcommand{\degree}{^{\circ}}\)
Since we have \[ \tan \theta = \frac {y}{x} \implies \theta = \arctan \frac {y}{x} + k\pi \] for some \(k \in \ZZ \), differentiating with respect to \(t\) gives us \[ \DiffFrac {\theta }{t} = \frac {1}{1 + \left (\frac {y}{x}\right )^2} \cdot \DiffFrac {\frac {y}{x}}{t} = \frac {x^2}{x^2 + y^2} \cdot \frac {x\DiffFrac {y}{t} - y\DiffFrac {x}{t}}{x^2} = \frac {x\DiffFrac {y}{t} - y\DiffFrac {x}{t}}{r^2}. \]
Hence, \[ \frac {1}{2} \int r^2 \Diff \theta = \frac {1}{2} \int r^2 \cdot \frac {x\DiffFrac {y}{t} - y\DiffFrac {x}{t}}{r^2} \cdot \Diff t = \frac {1}{2} \int \left (x\DiffFrac {y}{t} - y\DiffFrac {x}{t}\right ) \Diff t, \] as desired.
The coordinates of \(A\) and \(B\) are \[ A \left (x - a \cos t, y - a \sin t\right ), B \left (x + b \cos t, y + b \sin t\right ). \]
Hence, we have \begin {align*} [A] & = \frac {1}{2} \int _{0}^{2\pi } \left (x_A \DiffFrac {y_A}{t} - y_A \DiffFrac {x_A}{y}\right ) \Diff t \\ & = \frac {1}{2} \int _{0}^{2\pi } \left [(x - a \cos t) \left (\DiffFrac {y}{t} - a \cos t\right ) - (y - a \sin t) \left (\DiffFrac {x}{t} + a \sin t\right )\right ] \Diff t \\ & = \frac {1}{2} \int _{0}^{2\pi } \left (x\DiffFrac {y}{t} - y\DiffFrac {x}{t}\right ) \Diff t - \frac {a}{2} \int _{0}^{2\pi } \left [\cos t \left (\DiffFrac {y}{t} + x\right ) + \sin t \left (y - \DiffFrac {x}{t}\right )\right ] \Diff t \\ & \phantom {=} + \frac {a^2}{2} \int _{0}^{2\pi } \left (\cos ^2 t + \sin ^2 t\right ) \Diff t \\ & = [P] - af + \frac {a^2}{2} \int _{0}^{2\pi } \Diff t \\ & = [P] - af + 2\pi \cdot \frac {a^2}{2} \\ & = [P] + \pi a^2 - af, \end {align*}
as desired.
Similarly, \begin {align*} [B] & = \frac {1}{2} \int _{0}^{2\pi } \left (x_B \DiffFrac {y_B}{t} - y_B \DiffFrac {x_B}{y}\right ) \Diff t \\ & = \frac {1}{2} \int _{0}^{2\pi } \left [(x + b \cos t) \left (\DiffFrac {y}{t} + b \cos t\right ) - (y + b \sin t) \left (\DiffFrac {x}{t} - b \sin t\right )\right ] \Diff t \\ & = \frac {1}{2} \int _{0}^{2\pi } \left (x \DiffFrac {y}{t} - y \DiffFrac {x}{t}\right ) \Diff t + \frac {b}{2} \int _{0}^{2\pi } \left [\cos t \left (\DiffFrac {y}{t} + x\right ) + \sin t \left (y - \DiffFrac {x}{t}\right )\right ] \Diff t \\ & \phantom {=} + \frac {b^2}{2} \int _{0}^{2\pi } \left (\cos ^2 t + \sin ^2 t\right ) \Diff t \\ & = [P] + bf + \frac {b^2}{2} \int _{0}^{2\pi } \Diff t \\ & = [P] + bf + 2\pi \cdot \frac {b^2}{2} \\ & = [P] + \pi b^2 + bf. \end {align*}
Since over \(t \in [0, 2\pi ]\), \(A\) and \(B\) both trace over \(\mathcal {D}\), we must have \[ [A] = [B], \] and hence \[ \pi a^2 - af = \pi b^2 + bf, \] which means \[ \pi (a + b) (a - b) = (a + b)f, \] and hence \[ f = (a - b)\pi , \] and therefore \[ [A] = [B] = [P] + ab\pi . \]
The area between the curves \(\mathcal {C}\) and \(\mathcal {D}\) is represented as \([A] - [P]\) or \([B] - [P]\), and hence this area is \(\pi ab\), as desired.