Step 1: Identify the Curve

The equation $\frac{x^{2}}{9}+\frac{y^{2}}{4}=1$ represents an ellipse centered at the origin with semi-major axis $a = 3$ along the x-axis and semi-minor axis $b = 2$ along the y-axis.

Step 2: Express y in terms of x

We need to find the area bounded by the curve and the x-axis. First, solve the equation for $y$:

$\frac{y^{2}}{4} = 1 - \frac{x^{2}}{9}$

$y^{2} = 4(1 - \frac{x^{2}}{9})$

$y^{2} = \frac{4}{9}(9 - x^{2})$

$y = \pm \frac{2}{3}\sqrt{9 - x^{2}}$

Since we are finding the area bounded by the curve and the x-axis, we consider the positive part of $y$, i.e., $y = \frac{2}{3}\sqrt{9 - x^{2}}$.

Step 3: Determine the Limits of Integration

The ellipse intersects the x-axis at $x = -3$ and $x = 3$. These will be our limits of integration.

Step 4: Set up the Integral

The area $A$ of the region bounded by the curve and the x-axis is given by the integral:

$A = \int_{-3}^{3} \frac{2}{3}\sqrt{9 - x^{2}} \, dx$

Step 5: Evaluate the Integral

We can use the formula $\int \sqrt{a^{2} - x^{2}} \, dx = \frac{x}{2}\sqrt{a^{2} - x^{2}} + \frac{a^{2}}{2}\sin^{-1}(\frac{x}{a}) + C$. In our case, $a = 3$.

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

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

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

$A = \frac{2}{3} \left[ \frac{9\pi}{4} + \frac{9\pi}{4} \right] = \frac{2}{3} \cdot \frac{18\pi}{4} = \frac{36\pi}{12} = 3\pi$