Figure 1 shows part of the curve with equation $y = \, \sqrt{\tan x}$ - Edexcel - A-Level Maths Pure - Question 7 - 2007 - Paper 8

To use the trapezium rule, we calculate:

Area $= \frac{1}{2} \cdot (b_1 + b_2) \cdot h$, where:

• $b_1 = 0$ for $x=0$
• $b_2 = 0.81719$ for $x=\frac{3\pi}{16}$
• $h = \frac{\pi/4 - 0}{4} = \frac{\pi}{16}$

So,

$\text{Area} = \frac{1}{2} \times (0 + 0.81719) \times \frac{\pi}{16}$

Calculating this yields an area estimate of approximately $0.4726$.

Using the formula for the volume of revolution, we have:

$V = \int_{0}^{\frac{\pi}{4}} \pi (\sqrt{\tan x})^2 \, dx = \int_{0}^{\frac{\pi}{4}} \pi \tan x \, dx$

This integral simplifies to:

$V = \pi \left[ \ln|\sec x| \right]_{0}^{\frac{\pi}{4}}$

Calculating the limits:

• At $x = \frac{\pi}{4}$: $\ln|\sec(\frac{\pi}{4})| = \ln|\sqrt{2}|$
• At $x = 0$: $\ln|\sec(0)| = \ln|1| = 0$

Therefore:

$V = \pi \left( \ln|\sqrt{2}| - \ln|1| \right) = \pi \ln|\sqrt{2}| = \frac{\pi}{2} \ln(2)$

Final answer: Volume = $\frac{\pi}{2} \ln(2)$.