Figure 1 shows part of the curve with equation $y = \sqrt{\tan x}$ - Edexcel - A-Level Maths Pure - Question 1 - 2006 - Paper 8

To calculate the area using the trapezium rule, we use the formula:

$A \approx \frac{1}{2} \times h \times (y_0 + 2y_1 + 2y_2 + 2y_3 + y_4)$

where $h$ is the width of each section. The width here is $h = \frac{\pi/4 - 0}{4} = \frac{\pi}{16}$. By substituting the values from the table:

$A \approx \frac{1}{2} \times \frac{\pi}{16} \times (0 + 2(0.44600) + 2(0.64354) + 2(0.81719) + 0)$

Calculating gives:

$A \approx \frac{\pi}{32} \times (0 + 0.89200 + 1.28708 + 1.63438) \approx \frac{\pi}{32} \times 3.81346 \approx 0.3757 \text{ (4 decimal places)}$

The volume $V$ of the solid generated by rotating the region $\mathcal{R}$ around the x-axis can be found using the formula:

$V = \pi \int_0^{\frac{\pi}{4}} (\sqrt{\tan x})^2 \, dx = \pi \int_0^{\frac{\pi}{4}} \tan x \, dx$

We know that:

$\int \tan x \, dx = -\ln(\cos x)$

So evaluating the definite integral:

$V = \pi \left[-\ln(\cos x) \right]_0^{\frac{\pi}{4}} = \pi \left[-\ln(\cos(\frac{\pi}{4})) + \ln(\cos(0))\right]$

This simplifies to:

$= \pi \left[-\ln\left(\frac{1}{\sqrt{2}}\right) - 0 \right] = \pi \left[\frac{\ln 2}{2}\right] = \frac{\pi \ln 2}{2}$