Figure 4 shows a sketch of part of the curve C with parametric equations $x = 3 heta ext{sin} heta$, $y = ext{sec} heta$, $0 < heta < \frac{ ext{π}}{2}$ - Edexcel - A-Level Maths Pure - Question 1 - 2005 - Paper 5

The area of region R can be obtained using the formula:

$\text{Area} = \int_{a}^{b} (f(x)) dx$ where f(x) is defined by the curve.

Given our parametric equations, we can express the area using:

$\text{Area} = \int_{0}^{k} y(x) \, dx$ Substituting for y from the parametric equation, we have:

$y = \text{sec}\theta$. We also differentiate x with respect to \theta:

$\frac{dx}{d\theta} = 3 \text{sin}\theta + 3\theta\text{cos}\theta$. Now, we can express the area in the required form:

$\int_{0}^{\alpha} \text{sec}^2\theta d\theta + \tan\theta\text{sec}^2\theta d\theta$ showing that constants \lambda, \alpha, and \beta can be determined later through integration where necessary.

To find the exact area, we evaluate the integral obtained from part (b):

$\text{Area} = \lambda \int_{0}^{\alpha} ( \text{sec}^2\theta + \tan\theta \text{sec}^2\theta ) d\theta$. Using integration techniques, we can work out the integral:

$\int ( \text{sec}^2\theta + \tan\theta \text{sec}^2\theta ) d\theta$ This integrates to:

$\text{Area} = \left[ \tan\theta + \frac{1}{2} (\text{sec}^2\theta) \right]_{0}^{\alpha}$ Evaluating this for the limits and substituting values will yield the exact area of region R.

By substituting the limits after finding \alpha, through attention to the parameters set in the problem, we can calculate the final area.