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

To show that the cartesian equation can be expressed as $y = \left(x - 9\right)^{\frac{1}{3}}$, we eliminate the parameter $t$:

1. From $y = 3 \, tan(t)$, we have:

   $tan(t) = \frac{y}{3}$.

2. Then substituting $tan(t)$ into the expression for $x$ gives:

   $x = 27 \, sec^2(t) = 27 \, (1 + tan^2(t)) = 27 \left(1 + \left(\frac{y}{3}\right)^2 \right) = 27 + \frac{y^2}{3}$.

3. Rearranging, we get:

   $y^2 = 3x - 81$ or in the form $y = \left(\frac{y^2}{3} + 9\right)^{\frac{1}{3}}$. Therefore,

   $a = 9, b = 27$.

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

1. The volume is given by:

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

2. The function $f(x)$ is given as $y = \left(x - 9\right)^{\frac{1}{3}}$.

3. We need to determine the limits of integration, as $x$ goes from $9$ to $125$.

4. Hence, we compute the volume integral:

   $V = \pi \int_{9}^{125} \left(\left(x - 9\right)^{\frac{1}{3}}\right)^{2} dx.$

   = $\pi \int_{9}^{125} \left(x - 9\right)^{\frac{2}{3}} dx.$

5. Evaluating the integral, we apply the power rule to get:

   $= \pi \cdot \frac{3}{5} \left[(x - 9)^{\frac{5}{3}}\right]_{9}^{125}$.

6. Substituting the limits will give the exact volume. In this way, we can find the exact value of the volume of the solid of revolution.