A curve $C$ has parametric equations
$x = 2\cos t + 1, \quad y = \sin t , \quad 0 \le t < 2\pi .$

1. Find the coordinates of the intersections with the $x$-axis.

At the $x$-axis, $y = 0$.
$\sin t = 0$
Given $0 \le t < 2\pi$, $t = 0$ or $t = \pi$.

When $t = 0$:
$x = 2\cos(0) + 1 = 2(1) + 1 = 3$
When $t = \pi$:
$x = 2\cos(\pi) + 1 = 2(-1) + 1 = -1$
The coordinates of the intersections with the $x$-axis are $(-1, 0)$ and $(3, 0)$.

2. Determine the shape of the curve by finding its Cartesian equation to aid the sketch.

Rearrange the parametric equations to make the trigonometric functions the subject:
$\cos t = \frac{x - 1}{2}$
$\sin t = y$
Using the identity $\cos^2 t + \sin^2 t = 1$:
$\left(\frac{x - 1}{2}\right)^2 + y^2 = 1$
$\frac{(x - 1)^2}{4} + y^2 = 1$
This represents an ellipse centred at $(1, 0)$. Because the domain is $0 \le t < 2\pi$, the graph is a complete ellipse.

3. Describe and produce the required sketch.

Draw a set of $x$ and $y$ axes.
Sketch a full oval (ellipse) centred at $(1, 0)$.
The curve intersects the $x$-axis at $(-1, 0)$ and $(3, 0)$.
The highest point on the curve is $(1, 1)$ and the lowest is $(1, -1)$.

1. Set up the parametric integral for the volume of revolution.

Rotating the entire ellipse $C$ by $180^\circ$ about the $x$-axis generates the same complete solid (an ellipsoid) as rotating the upper half of $C$ by $360^\circ$.
The upper half of $C$ corresponds to $y \ge 0$, which implies $0 \le t \le \pi$.
The volume $V$ is given by the standard formula for a full revolution of the upper boundary:
$V = \pi \int_{-1}^{3} y^2 \, \mathrm{d}x$
Using the parametric equations $x = 2\cos t + 1 \implies \frac{\mathrm{d}x}{\mathrm{d}t} = -2\sin t$, and $y = \sin t$:
When $x = -1$, $t = \pi$; when $x = 3$, $t = 0$.
$V = \pi \int_{\pi}^{0} (\sin t)^2 (-2\sin t) \, \mathrm{d}t$

2. Simplify and evaluate the integral to find the numerical volume.

$V = 2\pi \int_{0}^{\pi} \sin^3 t \, \mathrm{d}t$
$V = 2\pi \int_{0}^{\pi} \sin t (1 - \cos^2 t) \, \mathrm{d}t$
$V = 2\pi \left[ -\cos t + \frac{1}{3}\cos^3 t \right]_0^\pi$
$V = 2\pi \left[ \left(-\cos\pi + \frac{1}{3}\cos^3\pi\right) - \left(-\cos 0 + \frac{1}{3}\cos^3 0\right) \right]$
$V = 2\pi \left[ \left(1 - \frac{1}{3}\right) - \left(-1 + \frac{1}{3}\right) \right]$
$V = 2\pi \left( \frac{2}{3} + \frac{2}{3} \right) = \frac{8\pi}{3}$
(Note: Evaluation can also be done directly using a graphing calculator to obtain $\approx 8.38$.)

1. Make $\cos t$ and $\sin t$ the subjects of the parametric equations.

$x = 2\cos t + 1 \implies \cos t = \frac{x-1}{2}$
$y = \sin t \implies \sin t = y$

2. Eliminate the parameter $t$ using the Pythagorean identity.

Since $\cos^2 t + \sin^2 t = 1$,
$\left(\frac{x-1}{2}\right)^2 + y^2 = 1$
$\frac{(x-1)^2}{4} + y^2 = 1$

1. State the cartesian equations of curve $C$ and the target circle.

From part (iii), the cartesian equation of $C$ is $\frac{(x-1)^2}{4} + y^2 = 1$.

The target is a circle with radius $2$ centred at the origin, which has the equation $x^2 + y^2 = 4$. Dividing by $4$, this can be written as:
$\frac{x^2}{4} + \frac{y^2}{4} = 1$

2. Determine the transformation to center the curve at the origin.

To transform $\frac{(x-1)^2}{4} + y^2 = 1$ to $\frac{x^2}{4} + y^2 = 1$, we replace $x$ with $x+1$.
This corresponds to a translation of $-1$ unit in the $x$-direction, which is a translation by the vector $\begin{pmatrix} -1 \\ 0 \end{pmatrix}$.

3. Determine the transformation to adjust the semi-minor axis to match the circle's radius.

To transform $\frac{x^2}{4} + y^2 = 1$ into the target circle $\frac{x^2}{4} + \frac{y^2}{4} = 1$, we replace $y$ with $\frac{y}{2}$.
This corresponds to a stretch (or scaling) parallel to the $y$-axis by a scale factor of $2$.

(Note: As the two transformations affect different axes, they can be performed in either order.)