% [** F1: Using \lim for limit on this page]

\textsc{Third Step}.\quad Applying the Fundamental Theorem,
\[
\lim_{n=\infty} \sum_{i=1}^n \tfrac{1}{2} \rho_{i}^2\, \Delta\theta_{i}
  = \int_\alpha^\beta \tfrac{1}{2}\, \rho^2\, d\theta.
\]
Hence the area swept over by the radius vector of the curve in moving
from the position $OP_{1}$ to the position $OD$ is given by the formula
\[
\tag{A} % Don't worry if the tag is at the right margin
\text{area} = \tfrac{1}{2} \int_\alpha^\beta \rho^2\, d\theta,
\]
the value of~$\rho$ in terms of~$\theta$ being substituted from the equation of
the curve.

% Ideally a PPer would have created a macro; otherwise, hard-code the fonts
% [** F1: Need vertical space]
\textsc{Illustrative Example~1}.\quad Find the entire area of the lemniscate $\rho^2 = a^2 \cos 2\theta$.

\textbf{Solution}.\quad Since the figure is symmetrical with respect to both $OX$ and~$OY$, the
whole area $= 4$~times the area of~$OAB$.

Since $\rho = 0$ when $\theta = \dfrac{\pi}{4}$, we see that if $\theta$ varies
from $0$ to~$\dfrac{\pi}{4}$, the radius vector~$OP$ sweeps over the
area~$OAB$. Hence, substituting in~$(A)$,
\begin{align*}
\text{entire area}
  &= 4 × \text{area}~OAB = 4 \cdot \frac{1}{2} \int_\alpha^\beta \rho^2\, d\theta \\
  &= 2 a^2 \int_0^{\frac{\pi}{4}} \cos 2\theta\, d\theta = a^2;
\end{align*}
% [** Illustration]
that is, the area of both loops equals the area of a square constructed on~$OA$ as
one side.

% Ideally there'd be an examples environment; otherwise, hard-code the formatting
\begin{center}
\textbf{EXAMPLES}
\end{center}

\begin{itemize}

\item[1.] Find the area swept over in one revolution by the radius vector of the spiral
of Archimedes, $\rho = a \theta$, starting with $\theta = 0$. How much additional area is swept over
in the second revolution?
\null\hfill \textit{Ans}.~$\dfrac{4\pi^3 a^2}{3}$; $8\pi^3 a^2$.

\item[2.] Find the area of one loop of the curve $\rho = a \cos 2\theta$.
\null\hfill \textit{Ans}.~$\dfrac{\pi a^2}{8}$.

\item[3.] Show that the entire area of the curve $\rho = a \sin 2\theta$ equals one half the area of
the circumscribed circle.

\item[4.] Find the entire area of the cardioid $\rho = a(1 - \cos \theta)$.\\
\null\hfill \textit{Ans}.~ $\dfrac{3\pi a^2}{2}$; that is, six times the area of the generating circle.

\item[5.] Find the area of the circle $\rho = a \cos \theta$.
\null\hfill \textit{Ans}.~$\dfrac{\pi a^2}{4}$.

\item[6.] Prove that the area of the three loops of $\rho = a \sin 3\theta$ equals one fourth of the % [** F1: Hyphenate one-fourth here and below?]
area of the circumscribed circle.

\item[7.] Prove that the area generated by the radius vector of the spiral $\rho = e^{\theta}$ equals
one fourth of the area of the square described on the radius vector.

\item[8.] Find the area of that part of the parabola $\rho = a \sec^2 \dfrac{\theta}{2}$ which is intercepted
between the curve and the latus rectum.
\null\hfill \textit{Ans}.~$\dfrac{8a^2}{3}$.

\item[9.] Show that the area bounded by any two radii vectors of the hyperbolic spiral
$\rho \theta = a$ is proportional to the difference between the lengths of these
radii.

\end{itemize}
