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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 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 equation $x^2 + y^2 = 9$ describes a full circle centered at the origin with radius 3, which includes both upper and lower branches. However, the original function is defined using the principal square root, $y = \sqrt{9-x^2}$.

Conditions: Working over the real numbers.; Using the principal (nonnegative) square root convention.; The underlying relation is the circle $x^2 + y^2 = 9$.

The equation $x^2 + y^2 = 9$ describes a full circle centered at the origin with radius 3, which includes both upper and lower branches. However, the original function is defined using the principal square root, $y = \sqrt{9-x^2}$.

Conditions: Working over the real numbers.; Using the principal (nonnegative) square root convention.; The underlying relation is the circle $x^2 + y^2 = 9$.

To evaluate the definite integral $\int_{-3}^{3} \sqrt{9-x^2}\,dx$ geometrically, recognize that the integrand $y = \sqrt{9-x^2}$ represents the upper semicircle of a circle centered at the origin with radius 3. Because the function is nonnegative and continuous on the interval $[-3, 3]$, the definite integral equals the ordinary geometric area of this shaded region.

Conditions: The integrand is $\sqrt{9-x^2}$ and the limits of integration are -3 and 3.; The square root denotes the principal (nonnegative) root, restricting the graph to $y \ge 0$.; The function is continuous and nonnegative on the closed interval $[-3, 3]$, ensuring the definite integral equals the ordinary area under the curve.

To evaluate the definite integral $\int_{-3}^{3} \sqrt{9-x^2}\,dx$ geometrically, recognize that the integrand $y = \sqrt{9-x^2}$ represents the upper semicircle of a circle centered at the origin with radius 3. Because the function is nonnegative and continuous on the interval $[-3, 3]$, the definite integral equals the ordinary geometric area of this shaded region.

Conditions: The integrand is $\sqrt{9-x^2}$ and the limits of integration are -3 and 3.; The square root denotes the principal (nonnegative) root, restricting the graph to $y \ge 0$.; The function is continuous and nonnegative on the closed interval $[-3, 3]$, ensuring the definite integral equals the ordinary area under the curve.