Evaluating a definite integral as a semicircle area

The equation $y = \sqrt{9-x^2}$ describes the upper half of the circle $x^2+y^2=9$, which has radius $r=3$ centered at the origin. Since $-3 \le x \le 3$ covers the full diameter, the region under the curve is exactly a semicircle of radius 3.

The area of a full circle of radius 3 is $\pi r^2 = 9\pi$, so the area of the semicircle is half of that: $\frac{9\pi}{2}$. Since $f(x)\ge 0$ on $[-3,3]$, the definite integral equals this area exactly, giving $\int_{-3}^{3}\sqrt{9-x^2}\,dx = \frac{9\pi}{2}$.