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

To estimate the area using the trapezium rule, we can calculate:

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

Where:

• $h = \frac{\pi/4 - 0}{4} = \frac{\pi}{16}$
• $y_0 = 0$, $y_1 = 0.44610$, $y_2 = 0.64359$, $y_3 = 0.81742$, and $y_4 = 0.81742$.

Thus, substituting these values:

$\text{Area} = \frac{\pi/16}{2} \times \left(0 + 2(0.44610) + 2(0.64359) + 2(0.81742) + 0.81742\right)$

Carrying out this calculation gives:

$\approx 0.4726$

To find the volume $V$ of the solid generated when the region $R$ is rotated around the $x$-axis, we use the formula:

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

Where $y = \sqrt{\tan x}$, so:

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

Evaluating this integral:

1. The integral of $\tan x$ is $-\ln(\cos x)$, leading to: $V = \pi \left[-\ln(\cos x)\right]_0^{\frac{\pi}{4}}$

2. Evaluating the limits: $= \pi \left[-\ln(\cos(\frac{\pi}{4})) + \ln(\cos(0))\right] = \pi \left[-\ln(\frac{\sqrt{2}}{2}) + 0\right] = \pi \left[\ln(2^{1/2})\right] = \frac{\pi}{2} \ln 2$

Therefore, the exact volume is:

$V = \frac{\pi}{2} \ln 2$