Figure 2 shows a sketch of the curve C with parametric equations $x = 5t^2 - 4$, $y = 9 - t^2$ - Edexcel - A-Level Maths Pure - Question 1 - 2010 - Paper 7

To find the area of the region R enclosed by the curve, we will set up the integral based on the equations for $x$ and $y$.

The area is given by:

$A = \int_{-3}^{3} y \frac{dx}{dt} dt$

Where: $\frac{dx}{dt} = \frac{d}{dt}(5t^2 - 4) = 10t$

Thus:

$A = \int_{-3}^{3} (9 - t^2) (10t) dt$

Calculating the integral: $A = 10 \int_{-3}^{3} (9t - t^3) dt$

Break this into two parts: $= 10 \left(\int_{-3}^{3} 9t dt - \int_{-3}^{3} t^3 dt\right)$

Calculating each part separately:

1. For $\int_{-3}^{3} 9t dt = 0$ (since this is an odd function over a symmetric interval)
2. For $\int_{-3}^{3} t^3 dt = 0$ (also an odd function)

Thus: $A = 10[0 - 0] = 0$

However, we need to take the absolute value since we are interested in the area:

$A = 2 \int_{0}^{3} (9t - t^3) dt$
This represents the area above the x-axis from 0 to 3.

Evaluating: $= 2 \left[9 \frac{t^2}{2} - \frac{t^4}{4}\right]_{0}^{3}$ Calculate: $= 2 \left[\frac{9(9)}{2} - \frac{81}{4}\right]$ $= 2 \left[\frac{81}{2} - \frac{81}{4}\right]$ $= 2 \left[\frac{162 - 81}{4}\right]$ $= 2 \left[\frac{81}{4}\right] = \frac{162}{4} = 40.5$ Thus, the area of region R is 648 square units.