1. The problem is to evaluate the definite integral $$\int_{-3}^{3} \sqrt{9 - x^2} \, dx$$. 2. This integral represents the area under the curve of the function $$y = \sqrt{9 - x^2}$$ from $$x = -3$$ to $$x = 3$$. 3. The function $$y = \sqrt{9 - x^2}$$ describes the upper half of a circle centered at the origin with radius 3, since $$x^2 + y^2 = 9$$. 4. The area of a full circle with radius $$r$$ is given by $$\pi r^2$$. 5. Since the integral covers the upper semicircle, the area is half the area of the full circle: $$\text{Area} = \frac{1}{2} \pi r^2 = \frac{1}{2} \pi (3)^2 = \frac{1}{2} \pi 9 = \frac{9\pi}{2}$$. 6. Therefore, the value of the integral is $$\boxed{\frac{9\pi}{2}}$$.