The curve $C$ (as shown in the diagram below) has equation $y = x^2 \sin x$, $-\pi \le x \le \pi$.

1. Formulate the definite integral for the total area of region $R$, utilising the symmetry of the curve.

Since $y = x^2 \sin x \le 0$ for $x \in [-\pi, 0]$ and $y \ge 0$ for $x \in [0, \pi]$, the exact area $A$ is given by:
$A = \int_{-\pi}^0 -x^2 \sin x \,dx + \int_0^\pi x^2 \sin x \,dx$
Because $f(x) = x^2 \sin x$ is an odd function ($f(-x) = -f(x)$), the regions below and above the $x$-axis are symmetric in area.
Thus, we can compute the area as:
$A = 2 \int_0^\pi x^2 \sin x \,dx$

2. Apply integration by parts for the first time on $x^2 \sin x$.

Using integration by parts, $\int u \,dv = uv - \int v \,du$:
Let $u = x^2$ and $dv = \sin x \,dx$.
Then $du = 2x \,dx$ and $v = -\cos x$.
$\int x^2 \sin x \,dx = -x^2 \cos x - \int (-\cos x)(2x) \,dx$
$= -x^2 \cos x + 2 \int x \cos x \,dx$

3. Apply integration by parts a second time to evaluate $\int x \cos x \,dx$.

For $\int x \cos x \,dx$, let $u = x$ and $dv = \cos x \,dx$.
Then $du = dx$ and $v = \sin x$.
$\int x \cos x \,dx = x \sin x - \int \sin x \,dx$
$= x \sin x + \cos x$
Substituting this back gives the integral:
$\int x^2 \sin x \,dx = -x^2 \cos x + 2(x \sin x + \cos x)$
$= -x^2 \cos x + 2x \sin x + 2\cos x$

4. Evaluate the definite integral using the limits from $0$ to $\pi$ to find the exact area.

$A = 2 \left[ -x^2 \cos x + 2x \sin x + 2\cos x \right]_0^\pi$
$= 2 \Big( \big(-\pi^2 \cos \pi + 2\pi \sin \pi + 2\cos \pi \big) - \big(-0 + 0 + 2\cos 0 \big) \Big)$
Using $\cos \pi = -1$, $\sin \pi = 0$, and $\cos 0 = 1$:
$= 2 \Big( (-\pi^2(-1) + 0 + 2(-1)) - (2(1)) \Big)$
$= 2 (\pi^2 - 2 - 2)$
$= 2\pi^2 - 8$

1. Identify the type of curve and its orientation from its equation.

The equation is $(y + 1)^2 - 4(x + 2)^2 = 1$, which can be rewritten as:
$\frac{(y + 1)^2}{1^2} - \frac{(x + 2)^2}{(\frac{1}{2})^2} = 1$
This is a hyperbola centred at $(-2, -1)$ that opens vertically (upwards and downwards).

2. Find the coordinates of the turning points (vertices).

The turning points lie on the vertical axis of symmetry, $x = -2$.
Substitute $x = -2$ into the equation:
$(y + 1)^2 - 4(0)^2 = 1$
$(y + 1)^2 = 1$
$y + 1 = \pm 1$
So, $y = 0$ or $y = -2$.
The coordinates of the turning points are $(-2, 0)$ and $(-2, -2)$.

3. Determine the equations of the asymptotes.

The equations of the asymptotes are found by setting the constant to 0:
$(y + 1)^2 - 4(x + 2)^2 = 0$
$(y + 1)^2 = 4(x + 2)^2$
$y + 1 = \pm 2(x + 2)$

For the positive case:
$y + 1 = 2x + 4 \implies y = 2x + 3$

For the negative case:
$y + 1 = -2x - 4 \implies y = -2x - 5$

The asymptotes are $y = 2x + 3$ and $y = -2x - 5$.

4. Describe the features to be shown on the sketch.

Sketch the following features on the Cartesian axes:

• Centre at $(-2, -1)$ (not required to be explicitly marked, but guides the drawing).
• The two straight line asymptotes (drawn as dashed lines) $y = 2x + 3$ and $y = -2x - 5$ intersecting at the centre.
• The upper branch of the hyperbola with its minimum turning point at $(-2, 0)$ (which lies on the $x$-axis).
• The lower branch of the hyperbola with its maximum turning point at $(-2, -2)$.
• The curves approaching the asymptotes as $x \to \pm\infty$.

1. Identify the relevant branch of the hyperbola and deduce the bounded region.

From the sketch in (ii), the hyperbola $(y+1)^2 - 4(x+2)^2 = 1$ has an upper branch $y = -1 + \sqrt{1 + 4(x+2)^2} \ge 0$ and a lower branch $y = -1 - \sqrt{1 + 4(x+2)^2}$.
Since Curve $C$ ($y = x^2 \sin x$) is strictly below the $x$-axis for $x \in (-\pi, 0)$ and the upper branch of the hyperbola does not intersect it to form a closed region, the bounded region must lie between Curve $C$ and the lower branch.

2. Find the $x$-coordinates of the intersection points using a graphing calculator.

Equating the two curves to find their points of intersection:
$x^2 \sin x = -1 - \sqrt{1 + 4(x+2)^2}$
Using a graphing calculator, the $x$-coordinates of the intersections are:
$x \approx -2.76212 \quad \text{and} \quad x \approx -1.53745$

3. Set up the integral for the volume of the solid of revolution.

For $-2.76212 < x < -1.53745$, both curves lie below the $x$-axis. Curve $C$ is further from the $x$-axis (more negative) than the lower branch, making it the outer boundary. The outer radius is $R = |x^2 \sin x|$ and the inner radius is $r = |-1 - \sqrt{1 + 4(x+2)^2}|$.
The volume $V$ generated by rotating the region through 4 right angles ($360^\circ$) about the $x$-axis is:
$V = \pi \int_{-2.76212}^{-1.53745} \left[ \left(x^2 \sin x\right)^2 - \left(-1 - \sqrt{1 + 4(x+2)^2}\right)^2 \right] \mathrm{d}x$

4. Evaluate the definite integral using a graphing calculator.

Evaluating the integral directly with a graphing calculator:
$V \approx 26.748$
$V = 26.7 \text{ units}^3 \text{ (to 3 s.f.)}$