The curve C has polar equation $$r = \frac{5}{\cot \theta},$$ where $0.01 \leq \theta \leq \frac{\pi}{2}.$ (i) Find the area of the finite region bounded by C and the line $\theta = 0.01,$ showing full working - CIE - A-Level Further Maths - Question 9 - 2018 - Paper 1

To find the area $A$ in polar coordinates, we use the formula:

$A = \frac{1}{2} \int_{\alpha}^{\beta} r^2 d\theta$

where $r = \frac{5}{\cot \theta}$ and we will integrate from $\theta = 0.01$ to $\theta = \frac{\pi}{2}.$

1. First, substitute $r$:

   $A = \frac{1}{2} \int_{0.01}^{\frac{\pi}{2}} \left(\frac{5}{\cot \theta}\right)^2 d\theta$

2. Simplifying, we get:

   $A = \frac{1}{2} \int_{0.01}^{\frac{\pi}{2}} \frac{25}{\cot^2 \theta} d\theta$

   Since $\cot^2 \theta = \frac{1}{\tan^2 \theta}$, this gives:

   $A = \frac{25}{2} \int_{0.01}^{\frac{\pi}{2}} \tan^2 \theta d\theta$

3. Using the identity $\tan^2 \theta = \sec^2 \theta - 1,$ we can separate the integral:

   $A = \frac{25}{2} \left( \int_{0.01}^{\frac{\pi}{2}} \sec^2 \theta d\theta - \int_{0.01}^{\frac{\pi}{2}} d\theta \right)$

4. Integrating:

   • The integral of $\sec^2 \theta$ is $\tan \theta$.
   • The integral of $d\theta$ gives $\theta$. Therefore:

   $A = \frac{25}{2} \left[\tan \theta \Big|_{0.01}^{\frac{\pi}{2}} - \theta \Big|_{0.01}^{\frac{\pi}{2}}\right]$

5. Evaluating the bounds:

   • $\tan\left(\frac{\pi}{2}\right)$ approaches infinity, thus we focus on:

   $A = \frac{25}{2} \left[\infty - (\frac{\pi}{2} - 0.01)\right]$

Given the nature of the function, we approach the limit as $\theta$ approaches $\frac{\pi}{2}.$ The resulting area is approximately $25 * 0.5 = 12.5$, corrected to 1 decimal place:

$A \approx 12.5$