A curve $C$ has parametric equations $x = 3\cos\theta - 2\cos 3\theta ,$
$y = 9\sin\theta - \sin 3\theta ,$ for $0 \le \theta \le 2\pi$.

1. Sketch the curve C using a graphing calculator

By plotting the parametric equations $x = 3\cos\theta - 2\cos 3\theta$ and $y = 9\sin\theta - \sin 3\theta$ for $0 \le \theta \le 2\pi$ on a graphing calculator, we obtain a continuous, closed curve (a peanut shape).

It crosses the $x$-axis at $(1, 0)$ and $(-1, 0)$, and the $y$-axis at $(0, 10)$ and $(0, -10)$. Note that the curve extends beyond the $x$-intercepts to a maximum/minimum $x$-value of approximately $\pm 1.84$.

(The student is expected to draw this general shape on a pair of axes, labelling the intercepts.)

2. State the cartesian equations of the lines of symmetry

Observing the shape of the sketched graph (or by verifying that substituting $-\theta$ and $\pi - \theta$ confirms reflectional symmetry), the curve is symmetric about both the $y$-axis and the $x$-axis.

The cartesian equations of these lines of symmetry are $x = 0$ and $y = 0$.

1. Set $x = 0$ for intersections with the $y$-axis and substitute the given identity for $\cos 3\theta$.

The curve meets the $y$-axis when $x = 0$.
$3\cos\theta - 2\cos 3\theta = 0$
Substituting $\cos 3\theta = 4\cos^3\theta - 3\cos\theta$:
$3\cos\theta - 2(4\cos^3\theta - 3\cos\theta) = 0$
$3\cos\theta - 8\cos^3\theta + 6\cos\theta = 0$
$9\cos\theta - 8\cos^3\theta = 0$

2. Solve the equation for $\theta$ within the given domain $0 \le \theta \le 2\pi$.

Factorising the equation:
$\cos\theta(9 - 8\cos^2\theta) = 0$
This gives $\cos\theta = 0$ or $\cos^2\theta = \frac{9}{8}$.
Since $-1 \le \cos\theta \le 1$, we know that $\cos^2\theta \le 1$. Hence, there are no real solutions for $\cos^2\theta = \frac{9}{8}$.
Therefore, we only solve $\cos\theta = 0$.
For $0 \le \theta \le 2\pi$:
$\theta = \frac{\pi}{2}, \frac{3\pi}{2}$

1. Differentiate $y$ with respect to $\theta$.

$y = 9\sin\theta - \sin 3\theta$
$\frac{dy}{d\theta} = 9\cos\theta - 3\cos 3\theta$

2. Set up the integral for the area with respect to $y$ and expand the integrand.

The area enclosed by the axes and the curve in the first quadrant can be found using $\int x \, dy$.
$\int x \, dy = \int x \frac{dy}{d\theta} \, d\theta$
$x \frac{dy}{d\theta} = (3\cos\theta - 2\cos 3\theta)(9\cos\theta - 3\cos 3\theta)$
$= 27\cos^2\theta - 9\cos\theta\cos 3\theta - 18\cos\theta\cos 3\theta + 6\cos^2 3\theta$
$= 27\cos^2\theta - 27\cos\theta\cos 3\theta + 6\cos^2 3\theta$

3. Determine the limits $\theta_1$ and $\theta_2$ bounding the first quadrant.

In the first quadrant, the region is bounded by the x-axis ($y=0$) and y-axis ($x=0$).
For the y-axis, from part (ii), $x = 0$ at $\theta = \frac{\pi}{2}$.
For the x-axis, $y = 9\sin\theta - \sin 3\theta = 0$.
By inspection, at $\theta = 0$, $y = 0$ and $x = 3(1) - 2(1) = 1 > 0$, so the curve starts on the positive x-axis.
Since $y$ increases from $0$ to its y-intercept as $\theta$ increases from $0$ to $\frac{\pi}{2}$, the area is:
$\text{Area} = \int_{0}^{\frac{\pi}{2}} \left( 27\cos^2\theta - 27\cos\theta\cos 3\theta + 6\cos^2 3\theta \right) \, d\theta$
Thus, $\theta_1 = 0$ and $\theta_2 = \frac{\pi}{2}$.

1. Determine the integration limits $\theta_1$ and $\theta_2$ established in part (iii).

In the first quadrant, the curve is bounded by the $x$-axis and $y$-axis.
At the $x$-axis ($y=0$), $9\sin\theta - \sin 3\theta = 0 \implies \theta = 0$.
At the $y$-axis ($x=0$), from part (ii), the curve meets the positive $y$-axis at $\theta = \frac{\pi}{2}$.
Since the curve travels from $(1, 0)$ to $(0, 10)$ as $\theta$ goes from $0$ to $\frac{\pi}{2}$, the corresponding limits for the integral are $\theta_1 = 0$ and $\theta_2 = \frac{\pi}{2}$.

2. Express the integrand using linear trigonometric functions.

Using the identities $\cos^2 A = \frac{1+\cos 2A}{2}$ and $\cos A \cos B = \frac{1}{2}(\cos(A+B) + \cos(A-B))$:
$27\cos^2\theta - 27\cos\theta\cos 3\theta + 6\cos^2 3\theta$
$= 27\left(\frac{1 + \cos 2\theta}{2}\right) - \frac{27}{2}(\cos 4\theta + \cos 2\theta) + 6\left(\frac{1 + \cos 6\theta}{2}\right)$
$= \frac{27}{2} + \frac{27}{2}\cos 2\theta - \frac{27}{2}\cos 4\theta - \frac{27}{2}\cos 2\theta + 3 + 3\cos 6\theta$
$= \frac{33}{2} - \frac{27}{2}\cos 4\theta + 3\cos 6\theta$

3. Integrate the simplified expression.

Let $A_1$ be the area enclosed in the first quadrant. Integrating the simplified expression with respect to $\theta$:
$A_1 = \int_0^{\frac{\pi}{2}} \left( \frac{33}{2} - \frac{27}{2}\cos 4\theta + 3\cos 6\theta \right) \, d\theta$
$= \left[ \frac{33}{2}\theta - \frac{27}{8}\sin 4\theta + \frac{1}{2}\sin 6\theta \right]_0^{\frac{\pi}{2}}$

4. Evaluate the definite integral to find the area in the first quadrant.

Evaluating the definite integral at the upper and lower limits:
$A_1 = \left( \frac{33}{2}\left(\frac{\pi}{2}\right) - \frac{27}{8}\sin(2\pi) + \frac{1}{2}\sin(3\pi) \right) - (0 - 0 + 0)$
$= \frac{33\pi}{4} - 0 + 0 = \frac{33\pi}{4}$

5. Calculate the exact total area enclosed by $C$ using symmetry.

From part (i), the curve $C$ has lines of symmetry at both the $x$-axis and $y$-axis. Therefore, the total area is four times the area in the first quadrant.
$\text{Total Area} = 4 \times A_1$
$= 4 \times \frac{33\pi}{4} = 33\pi$