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The real domain of the function $y=\sqrt{9-x^2}$ is the closed interval $[-3, 3]$. This restriction exists because, over the real numbers, the expression inside a square root (the radicand) must be nonnegative for the principal square root to be real-valued.

Conditions: Real-valued interpretation of the square root.; Using the principal square root convention.; The radicand is $9 - x^2$.

The exact value of the definite integral $\int_{-3}^{3} \sqrt{9-x^2}\,dx$ is $\frac{9\pi}{2}$. This is derived by interpreting the integral as the geometric area of the upper semicircle of a circle with radius 3.

Conditions: Interpret the integral as area under the graph on $[-3,3]$.; Recognize the graph as the upper semicircle of radius 3.; Exact equality for the displayed definite integral.