1. **Problem:** Find the volume of the solid formed by revolving the region under the curve $y=\sqrt{3 - x}$ from $x=0$ to $x=3$ about the x-axis. 2. **Formula:** The volume $V$ of a solid of revolution about the x-axis is given by the disk method: $$V = \pi \int_a^b [f(x)]^2 \, dx$$ where $f(x)$ is the function defining the radius of the disks. 3. **Apply the formula:** Here, $f(x) = \sqrt{3 - x}$, so $$[f(x)]^2 = (\sqrt{3 - x})^2 = 3 - x$$ 4. **Set up the integral:** $$V = \pi \int_0^3 (3 - x) \, dx$$ 5. **Calculate the integral:** $$\int_0^3 (3 - x) \, dx = \left[3x - \frac{x^2}{2}\right]_0^3 = (3 \times 3 - \frac{3^2}{2}) - (0) = 9 - \frac{9}{2} = \frac{9}{2}$$ 6. **Find the volume:** $$V = \pi \times \frac{9}{2} = \frac{9\pi}{2}$$ 7. **Answer:** The volume of the solid is $$\boxed{\frac{9\pi}{2}}$$