Figure 1 shows part of the curve with equation $y = \, \sqrt{\tan x}$ - Edexcel - A-Level Maths Pure - Question 8 - 2007 - Paper 8

To apply the trapezium rule:

The formula is given by:

$$\text{Area} = \frac{1}{2} \times h \times (y_0 + 2y_1 + 2y_2 + 2y_3 + y_4)$$

where:

• $h = \frac{\frac{\pi}{4} - 0}{4} = \frac{\pi}{16}$
• $y_0 = 0$
• $y_1 = 0.44600$
• $y_2 = 0.64350$
• $y_3 = 0.81722$
• $y_4 = 1$

Now substituting these values into the formula:

$$\text{Area} = \frac{1}{2} \times \frac{\pi}{16} \times (0 + 2(0.44600) + 2(0.64350) + 2(0.81722) + 1)$$

Calculating:

$$= \frac{1}{2} \times \frac{\pi}{16} \times (0 + 0.89200 + 1.28700 + 1.63444 + 1) \\ = \frac{1}{2} \times \frac{\pi}{16} \times (4.81344) \\ = \frac{\pi \times 4.81344}{32}$$

Thus, the estimated area of the shaded region $R$ is approximately: $\approx 0.4726$ (to 4 decimal places).

To find the volume of the solid generated by rotating the region $R$ about the x-axis, we use the formula for the volume of revolution:

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

Substituting $y = \sqrt{\tan x}$ gives:

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

This integral can be calculated as:

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

Calculating the limits:

1. At $x = \frac{\pi}{4}$:
   $-\ln(\cos(\frac{\pi}{4})) = -\ln\left(\frac{1}{\sqrt{2}}\right) = \frac{1}{2} \ln(2)$

2. At $x = 0$:
   $-\ln(\cos(0)) = -\ln(1) = 0$

Thus, combining these gives: $V = \pi \left( \frac{1}{2} \ln(2) - 0 \right) = \frac{\pi}{2} \ln(2)$

Therefore, the exact volume of the solid generated is: $V = \frac{\pi}{2} \ln(2)$