Step 1 (Formula / Governing Law): [1 Mark]

The given equation is $4x^2 + y^2 = 36$. Dividing by 36, we get the standard form of an ellipse:

$$ \frac{x^2}{9} + \frac{y^2}{36} = 1 \implies \frac{x^2}{3^2} + \frac{y^2}{6^2} = 1 $$

This is an ellipse with semi-major axis $b = 6$ (along the y-axis) and semi-minor axis $a = 3$ (along the x-axis). The area of the ellipse is given by $4 \times \int_{0}^{a} y \, dx$.

Step 2 (Step-by-Step Working): [3 Marks]

From the equation $y^2 = 36 - 4x^2$, we have $y = \sqrt{36 - 4x^2} = 2\sqrt{9 - x^2}$ (considering the first quadrant).

The area $A$ is:

$$ A = 4 \int_{0}^{3} 2\sqrt{3^2 - x^2} \, dx = 8 \int_{0}^{3} \sqrt{3^2 - x^2} \, dx $$

Using the standard integration formula $\int \sqrt{a^2 - x^2} \, dx = \frac{x}{2}\sqrt{a^2 - x^2} + \frac{a^2}{2}\sin^{-1}\left(\frac{x}{a}\right)$:

$$ A = 8 \left[ \frac{x}{2}\sqrt{9 - x^2} + \frac{9}{2}\sin^{-1}\left(\frac{x}{3}\right) \right]_{0}^{3} $$

Substituting the limits:

$$ A = 8 \left[ \left( \frac{3}{2}\sqrt{9 - 9} + \frac{9}{2}\sin^{-1}(1) \right) - (0 + 0) \right] $$ $$ A = 8 \left[ 0 + \frac{9}{2} \cdot \frac{\pi}{2} \right] $$

Step 3 (Final Answer): [1 Mark]

$$ A = 8 \cdot \frac{9\pi}{4} = 18\pi $$

The area of the region bounded by the curve is $18\pi$ square units.