Figure 1 shows part of the curve with equation $y = \sqrt{\tan x}$ - Edexcel - A-Level Maths Pure - Question 3 - 2007 - Paper 7

Using the trapezium rule, we have:

1. The widths of the intervals are:

   • Width between $0$ and $\frac{\pi}{16}$: $h_1 = \frac{\pi}{16} - 0 = \frac{\pi}{16}$
   • Width between $\frac{\pi}{16}$ and $\frac{\pi}{8}$: $h_2 = \frac{\pi}{8} - \frac{\pi}{16} = \frac{\pi}{16}$
   • Width between $\frac{\pi}{8}$ and $\frac{3\pi}{16}$: $h_3 = \frac{3\pi}{16} - \frac{\pi}{8} = \frac{\pi}{16}$
   • Width between $\frac{3\pi}{16}$ and $\frac{\pi}{4}$: $h_4 = \frac{\pi}{4} - \frac{3\pi}{16} = \frac{\pi}{16}$

2. Calculate the area using:

$A = \frac{h}{2} [f(a) + 2f(x_1) + 2f(x_2) + 2f(x_3) + f(b)]$

• Where:
  • $f(0) = 0$
  • $f(\frac{\pi}{16}) \approx 0.44599$
  • $f(\frac{\pi}{8}) \approx 0.64359$
  • $f(\frac{3\pi}{16}) \approx 0.81714$
  • $f(\frac{\pi}{4}) = 1$

3. Therefore:

$A \approx \frac{\frac{\pi}{16}}{2} [0 + 2(0.44599) + 2(0.64359) + 2(0.81714) + 1] \approx 0.4788$

Thus, the estimated area for region $R$ is approximately 0.4788.

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

$V = \pi \int_{a}^{b} [f(x)]^2 dx$

In this case, $f(x) = \sqrt{\tan x}$, and the limits of integration are from $0$ to $\frac{\pi}{4}$. Therefore:

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

To evaluate this integral, we can use the substitution method or recognize that the integral of $\tan x$ is:

$\int \tan x \, dx = -\ln |\cos x| + C$

After solving:

$V = \pi [ -\ln |\cos(\frac{\pi}{4})| + \ln |\cos(0)| ] = \pi [ -\ln \left(\frac{1}{\sqrt{2}}\right) - 0 ] = \pi \ln(\sqrt{2}) \approx 1.08603$

Thus, the exact value for the volume of the solid generated is approximately $\pi \ln(\sqrt{2})$.