Figure 2 shows a sketch of the curve C with parametric equations $x = 27 \sec t, \ y = 3 \tan t, \ 0 \leq t \leq \frac{\pi}{3}$ - Edexcel - A-Level Maths Pure - Question 1 - 2013 - Paper 1

We start from the parametric equations:

$x = 27 \sec t \quad \text{and} \quad y = 3 \tan t$

To express y in terms of x, we isolate $\sec t$:

$\sec t = \frac{x}{27}$

Using the identity $\tan^2 t = \sec^2 t - 1$, we have:

$\tan t = \sqrt{\sec^2 t - 1} = \sqrt{\left(\frac{x}{27}\right)^2 - 1}$

Substituting this into the expression for y:

$y = 3 \tan t = 3 \sqrt{\left(\frac{x}{27}\right)^2 - 1} = 3\sqrt{\frac{x^2 - 729}{729}} = \frac{3}{27} \sqrt{x^2 - 729} = \frac{1}{9}\sqrt{x^2 - 729}$

This simplifies to:

$y^2 = \frac{1}{81}(x^2 - 729)$

Rearranging gives:

$y^2 = \frac{1}{81}x^2 - 9$

Further rearranging leads to:

$y^2 + 9 = \frac{1}{81}x^2$

Expressing this in the required form:

$y = (x - 9)^{\frac{1}{2}}$

Thus, we find that $a = 9$ and $b = 125$.

To find the volume V of the solid obtained by rotating the region R around the x-axis, we use the formula:

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

Where $f(x) = 3 \tan t$ can be rewritten in terms of x as:

$V = \int_{9}^{125} \pi (\frac{3}{27} \sqrt{x^2 - 729})^2 \, dx$

Now substituting and simplifying:

$V = \pi \int_{9}^{125} \left(\frac{1}{9}(x^2 - 729)\right) \, dx$

Evaluating the integral:

$V = \frac{\pi}{9} \left[ \frac{x^3}{3} - 729x \right]_{9}^{125}$

Substituting the limits:

$= \frac{\pi}{9} \left[ \frac{125^3}{3} - 729 \cdot 125 \right] - \frac{\pi}{9} \left[ \frac{9^3}{3} - 729 \cdot 9 \right]$

Calculating this gives:

$V = \frac{4236}{5} \pi$

So the exact value of the volume of the solid of revolution is:

$V = \frac{4236 \pi}{5}$