The volume of the solid formed by rotating the region about the x-axis is given by the integral:

\(\text{Vol} = \pi \int_{1}^{2} y^2 \, dx\)

Substitute \(y = x^2 + \frac{2}{x}\) into the equation:

\(\text{Vol} = \pi \int_{1}^{2} \left(x^2 + \frac{2}{x}\right)^2 \, dx\)

Expand the integrand:

\(\left(x^2 + \frac{2}{x}\right)^2 = x^4 + 4 + \frac{4x}{x^2}\)

\(= x^4 + 4x + \frac{4}{x^2}\)

Integrate term by term:

\(\int x^4 \, dx = \frac{x^5}{5}\)

\(\int 4x \, dx = 2x^2\)

\(\int \frac{4}{x^2} \, dx = -\frac{4}{x}\)

Thus, the integral becomes:

\(\text{Vol} = \pi \left[ \frac{x^5}{5} - \frac{4}{x} + 2x^2 \right]_{1}^{2}\)

Evaluate the definite integral:

\(\text{Vol} = \pi \left( \left[ \frac{32}{5} - 2 + 8 \right] - \left[ \frac{1}{5} - 4 + 2 \right] \right)\)

\(= \pi \left( \frac{71}{5} \right)\)

Therefore, the volume is \(\frac{71\pi}{5}\) or approximately 44.6.

(i) Differentiate \(y = \frac{18}{x}\) to find the gradient of the tangent: \(\frac{dy}{dx} = -\frac{18}{x^2}\). At \(P(6, 3)\), the gradient is \(-\frac{1}{2}\). The gradient of the normal is the negative reciprocal, \(2\). The equation of the normal is \(y - 3 = 2(x - 6)\), simplifying to \(y = 2x - 9\). Setting \(y = 0\) to find \(R\), we get \(0 = 2x - 9\), so \(x = 4.5\). Thus, \(R\) is \(\left(4\frac{1}{2}, 0\right)\).

(ii) The volume of the solid is given by \(\pi \int_{4.5}^{6} \left(\frac{18}{x}\right)^2 \, dx\). This simplifies to \(\pi \int_{4.5}^{6} \frac{324}{x^2} \, dx = \pi \left[-\frac{324}{x}\right]_{4.5}^{6}\). Evaluating, we get \(\pi \left(-54 + 72\right) = 18\pi\).

(b) To find the volume of revolution, use the formula:

\(V = \pi \int_{1}^{2} y^2 \, dx\)

Substitute \(y = \frac{1}{(3x - 2)^{\frac{3}{2}}}\) into the integral:

\(V = \pi \int_{1}^{2} \left( \frac{1}{(3x - 2)^{\frac{3}{2}}} \right)^2 \, dx = \pi \int_{1}^{2} (3x - 2)^{-3} \, dx\)

Integrate:

\(V = \pi \left[ \frac{(3x - 2)^{-2}}{-2 \times 3} \right]_{1}^{2}\)

\(V = \pi \left[ -\frac{1}{6(3x - 2)^2} \right]_{1}^{2}\)

Evaluate the definite integral:

\(V = \pi \left[ -\frac{1}{6 \times 16} + \frac{1}{6} \right] = \frac{5\pi}{32}\)

(c) Differentiate \(y = \frac{1}{(3x - 2)^{\frac{3}{2}}}\) to find \(\frac{dy}{dx}\):

\(\frac{dy}{dx} = -\frac{3}{2} \times 3(3x - 2)^{-\frac{5}{2}}\)

At \(x = 1\), \(\frac{dy}{dx} = -\frac{9}{2}\).

The gradient of the normal is \(-\frac{1}{\left( -\frac{9}{2} \right)} = \frac{2}{9}\).

The equation of the normal is:

\(y - 1 = \frac{2}{9}(x - 1)\)

Substitute \(x = 0\) to find the y-intercept:

\(y - 1 = \frac{2}{9}(0 - 1)\)

\(y = 1 - \frac{2}{9} = \frac{7}{9}\)

(i) To find the equation of the tangent at B, first find the derivative of \(y = \frac{8}{3x + 2}\). The derivative is \(\frac{dy}{dx} = -\frac{24}{(3x + 2)^2}\). At B (2, 1), \(\frac{dy}{dx} = -\frac{3}{8}\).

The equation of the tangent line at B is \(y - 1 = -\frac{3}{8}(x - 2)\).

To find where this line crosses the x-axis, set \(y = 0\):

\(0 - 1 = -\frac{3}{8}(x - 2)\)

\(-1 = -\frac{3}{8}x + \frac{3}{4}\)

\(-\frac{7}{4} = -\frac{3}{8}x\)

\(x = \frac{7}{4} \times \frac{8}{3} = \frac{14}{3}\)

The area of triangle BDC is \(\frac{1}{2} \times 2 \times \frac{2}{3} \times 1 = \frac{4}{3}\).

(ii) The volume of the solid formed by rotating the region ODBA about the x-axis is given by:

\(V = \pi \int_0^2 y^2 \, dx = \pi \int_0^2 \left(\frac{8}{3x + 2}\right)^2 \, dx\)

\(= \pi \int_0^2 \frac{64}{(3x + 2)^2} \, dx\)

Let \(u = 3x + 2\), then \(du = 3 \, dx\), so \(dx = \frac{1}{3} \, du\).

Change the limits: when \(x = 0, u = 2\); when \(x = 2, u = 8\).

\(V = \pi \int_2^8 \frac{64}{u^2} \times \frac{1}{3} \, du\)

\(= \frac{64\pi}{3} \int_2^8 u^{-2} \, du\)

\(= \frac{64\pi}{3} \left[ -u^{-1} \right]_2^8\)

\(= \frac{64\pi}{3} \left( -\frac{1}{8} + \frac{1}{2} \right)\)

\(= \frac{64\pi}{3} \times \frac{3}{8}\)

\(= 8\pi\).

The curve intersects \(y = 1\) at \((3, 1)\).

Volume = \(\pi \int (x-2) \,[dx]\)

\(\left[ \pi \left( \frac{1}{2} x^2 - 2x \right) \right] \text{ or } \left[ \pi \left( \frac{1}{2} (x-2)^2 \right) \right]\)

= \(\left[ \pi \left( \frac{5^2}{2} - 2 \times 5 \right) - \left( \text{their } 3^2/2 - 2 \times \text{their } 3 \right) \right]\)

= \(\left[ \pi \left( \frac{5}{2} + \frac{3}{2} \right) \right] \text{ as a minimum requirement for their values}\)

Volume of cylinder = \(\pi \times 1^2 \times (5 - \text{their } 3) = [2\pi]\)

\([Volume of solid = 4\pi - 2\pi =] 2\pi \text{ or } 6.28\)