Step 1: Rewrite the equation

The given equation is $4x^2 + y^2 = 36$. We can rewrite this equation in the standard form of an ellipse:

Step 2: Divide by 36

Divide both sides of the equation by 36:

$\frac{4x^2}{36} + \frac{y^2}{36} = 1$

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

Step 3: Identify the semi-major and semi-minor axes

This is an ellipse with $a^2 = 9$ and $b^2 = 36$, so $a = 3$ and $b = 6$.

Step 4: Express y in terms of x

Solve for $y$ in terms of $x$:

$\frac{y^2}{36} = 1 - \frac{x^2}{9}$

$y^2 = 36(1 - \frac{x^2}{9})$

$y^2 = 36 - 4x^2$

$y = \sqrt{36 - 4x^2} = 2\sqrt{9 - x^2}$

Since we are finding the area, we consider the positive square root.

Step 5: Set up the integral for the area

The area of the ellipse can be found by integrating $y$ with respect to $x$ from $-a$ to $a$ and multiplying by 2 (since the ellipse is symmetric about the x-axis):

$Area = 2 \int_{-3}^{3} 2\sqrt{9 - x^2} \, dx = 4 \int_{-3}^{3} \sqrt{9 - x^2} \, dx$

Step 6: Evaluate the integral

The integral $\int \sqrt{a^2 - x^2} \, dx = \frac{x}{2}\sqrt{a^2 - x^2} + \frac{a^2}{2}\sin^{-1}(\frac{x}{a}) + C$. In our case, $a = 3$.

So, $\int \sqrt{9 - x^2} \, dx = \frac{x}{2}\sqrt{9 - x^2} + \frac{9}{2}\sin^{-1}(\frac{x}{3}) + C$

Now, evaluate the definite integral:

$4 \int_{-3}^{3} \sqrt{9 - x^2} \, dx = 4 \left[ \frac{x}{2}\sqrt{9 - x^2} + \frac{9}{2}\sin^{-1}(\frac{x}{3}) \right]_{-3}^{3}$

$= 4 \left[ (\frac{3}{2}\sqrt{9 - 9} + \frac{9}{2}\sin^{-1}(1)) - (\frac{-3}{2}\sqrt{9 - 9} + \frac{9}{2}\sin^{-1}(-1)) \right]$

$= 4 \left[ (0 + \frac{9}{2} \cdot \frac{\pi}{2}) - (0 + \frac{9}{2} \cdot (-\frac{\pi}{2})) \right]$

$= 4 \left[ \frac{9\pi}{4} + \frac{9\pi}{4} \right] = 4 \left[ \frac{18\pi}{4} \right] = 18\pi$