Step 1: Use a trigonometric substitution:
Let $x = 3\sin\theta$, so $dx = 3\cos\theta\,d\theta$. Then $\sqrt{9 - x^2} = \sqrt{9 - 9\sin^2\theta} = 3\cos\theta$, since $\cos\theta \ge 0$ on the range we need.
Step 2: Change the limits:
When $x = 0$, $\sin\theta = 0$ so $\theta = 0$. When $x = 3$, $\sin\theta = 1$ so $\theta = \frac{\pi}{2}$.
Step 3: Integrate:
The integral becomes $\int_0^{\pi/2} 3\cos\theta\cdot 3\cos\theta\,d\theta = 9\int_0^{\pi/2}\cos^2\theta\,d\theta$. Use $\cos^2\theta = \frac{1+\cos 2\theta}{2}$:
\[ 9\cdot\frac{1}{2}\left[\theta + \frac{\sin 2\theta}{2}\right]_0^{\pi/2} = \frac{9}{2}\cdot\frac{\pi}{2} = \frac{9\pi}{4} \]
Step 4: Compare with the options:
The value $\frac{9\pi}{4}$ is option (B). The smaller values in (A), (C) and (D) would come from using the wrong upper limit or forgetting the factor $9$ from $3\cdot 3$.
Final Answer:
Both the substitution and the geometry give $\frac{9\pi}{4}$, option (B).
\[ \boxed{\frac{9\pi}{4}} \]