Figure 3 shows part of the curve C with parametric equations $x = an \theta, \quad y = \sin \theta, \quad 0 \leq \theta < \frac{\pi}{2}$ - Edexcel - A-Level Maths Pure - Question 1 - 2011 - Paper 5

At point P, the derivatives needed for the normal line are: $\frac{dx}{d\theta} = \sec^2 \theta, \quad \frac{dy}{d\theta} = \cos \theta.$

At $\theta = \frac{\pi}{3}$:

• $\frac{dx}{d\theta} = \sec^2 \frac{\pi}{3} = \frac{4}{3},$
• $\frac{dy}{d\theta} = \cos \frac{\pi}{3} = \frac{1}{2}$.

The slope of the tangent line (m) at P is: $m = \frac{dy}{dx} = \frac{\frac{1}{2}}{\frac{4}{3}} = \frac{3}{8}.$

Thus, the slope of the normal l is: $m_{normal} = -\frac{8}{3}.$

Using point-slope form, the equation of line l is: $y - \frac{1}{2} = -\frac{8}{3}\left(x - \sqrt{3}\right).$

To find Q where this line intersects the x-axis (where $y = 0$), we solve: $0 - \frac{1}{2} = -\frac{8}{3}\left(x - \sqrt{3}\right).$. Solving gives: $x = \frac{(k \sqrt{3})}{0} \text{ implying that } k = \frac{1}{8}.$

To find the volume of the solid of revolution formed by rotating region S around the x-axis, we will use the formula: $V = \pi \int_a^b y^2 \, dx$

For our case, the boundaries are from $x = \sqrt{3}$ to $x = 0$, and: $y = \sin \theta = \frac{x}{\sqrt{1+x^2}}$ Thus, the volume becomes: $V = \pi \int_0^{\sqrt{3}} \left(\frac{x}{\sqrt{1+x^2}}\right)^2 \, dx.$

This results in: $= \pi \int_0^{\sqrt{3}} \left[ \frac{1}{1+x^2} \right]dx$
Using integration allows us to compute the final volume, represented in the form: $p\pi\sqrt{3} + qr^2.$