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The definite integral can be evaluated by recognizing that the integrand $y = \sqrt{9-x^2}$ graphs as an upper semicircle of radius 3. Because the function is continuous and nonnegative on $[-3, 3]$, the integral equals the ordinary geometric area of this region.

Conditions: The integrand is recognized as the upper semicircle of $x^2+y^2=9$.; The function is continuous and nonnegative on the interval $[-3, 3]$.; Use the real geometric area formula for a circle.

The definite integral can be evaluated by recognizing that the integrand $y = \sqrt{9-x^2}$ graphs as an upper semicircle of radius 3. Because the function is continuous and nonnegative on $[-3, 3]$, the integral equals the ordinary geometric area of this region.

Conditions: The integrand is recognized as the upper semicircle of $x^2+y^2=9$.; The function is continuous and nonnegative on the interval $[-3, 3]$.; Use the real geometric area formula for a circle.

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.

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.