Figure 3 shows part of the curve C with parametric equations $x = tan \theta$, $y = sin \theta$, $0 \leq \theta < \frac{\pi}{2}$ - Edexcel - A-Level Maths Pure - Question 7 - 2011 - Paper 5

The slope of the tangent at point P can be derived from the parametric equations:

1. Calculate ( \frac{dx}{d\theta} ) and ( \frac{dy}{d\theta} ):

   [ \frac{dx}{d\theta} = sec^2 \theta ]
   [ \frac{dy}{d\theta} = cos \theta ]

2. Then evaluate at (\theta = \frac{\pi}{3}):

   [ \frac{dx}{d\theta} = sec^2(\frac{\pi}{3}) = 4 ] [ \frac{dy}{d\theta} = cos(\frac{\pi}{3}) = \frac{1}{2} ]

3. The slope of the normal, which is the negative reciprocal:
   [ m = -\frac{1}{2} \times \frac{4}{1} = -2 ]

4. Using the point ((\sqrt{3}, \frac{1}{2})) and the slope to find the equation of the normal:

   [ y - \frac{1}{2} = -2(x - \sqrt{3}) ]

   Setting ( y = 0 ) to find Q:

   [ 0 - \frac{1}{2} = -2(x - \sqrt{3}) ] [ x = \frac{\sqrt{3}}{2} ]

Thus, the coordinates of Q: (\left( k \sqrt{3}, 0 \right)) where (k = \frac{1}{2}).

To find the volume of the solid of revolution, we use the formula:

[ V = \pi \int_{0}^{\sqrt{3}} y^2 dx ]

Substituting (y = sin \theta):

1. We change variable with respect to (\theta) by finding ( dx = \sec^2 \theta d\theta ):

   [ V = \pi \int_{0}^{\frac{\pi}{3}} (sin \theta)^2 sec^2 \theta d\theta ]

2. Using the identity ( (sin)^2 = 1 - (cos)^2 ):

   [ V = \pi \int_{0}^{\frac{\pi}{3}} (1 - cos^2 \theta) sec^2 \theta d\theta = \pi \int_{0}^{\frac{\pi}{3}} (sec^2 \theta - 1) d\theta ]

3. The integral evaluates to:

   [ \left[ tan \theta \right]_{0}^{\frac{\pi}{3}} = tan(\frac{\pi}{3}) - tan(0) = \sqrt{3} ]

Therefore, the volume:

[ V = \pi \sqrt{3}(\sqrt{3}) = 3\sqrt{3}\pi ]

So the final volume of the solid of revolution is in the form $p \pi \sqrt{3} + q z^2$ with constants p and q.