Step 1: Compare the given ellipse $\frac{x^2}{4} + \frac{y^2}{9} = 1$ with the standard form $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$. We get $a^2 = 4$ and $b^2 = 9$, so $a = 2$ and $b = 3$.

Step 2: Since the ellipse is symmetrical about both axes, the total area = $4 \times$ (area in the first quadrant).

Step 3: From the equation, $y = \frac{b}{a}\sqrt{a^2 - x^2} = \frac{3}{2}\sqrt{4 - x^2}$ (taking positive value in first quadrant).

Step 4: Area = $4 \int_{0}^{a} y \, dx = 4 \int_{0}^{2} \frac{3}{2}\sqrt{4 - x^2} \, dx = \frac{12}{2} \left[ \frac{x}{2}\sqrt{4 - x^2} + \frac{4}{2}\sin^{-1}\frac{x}{2} \right]_{0}^{2}$

Step 5: Substituting limits: $= 6 \left[ \left(0 + 2 \sin^{-1}1\right) - 0 \right] = 6 \times 2 \times \frac{\pi}{2} = 6\pi$ square units.

Step 6: Alternatively, using the formula Area = $\pi ab = \pi \times 2 \times 3 = 6\pi$ square units.