The problem involves finding the volume of a solid of revolution using the shell method. The region is bounded by $x = 3 - y^2$, $y = \sqrt{3}$, $x = 0$, and $y = 0$, and the solid is formed by rotating the region around the $y$-axis.

Using the shell method, the volume is given by:

$V = 2\pi \int_{y_{\text{min}}}^{y_{\text{max}}} (\text{radius}) \cdot (\text{height}) \, dy$

Key steps:

1. Radius: The distance from the $y$-axis is $x = 3 - y^2$.
2. Height: The height is $x = 3 - y^2$.
3. Limits of integration: $y$ varies from 0 to $\sqrt{3}$.

The volume formula becomes: $V = 2\pi \int_{0}^{\sqrt{3}} y \cdot (3 - y^2) \, dy$

Simplifying:

$V = 2\pi \int_{0}^{\sqrt{3}} (3y - y^3) \, dy$

Solving the integral:

1. Break it into two parts: $\int (3y - y^3) \, dy = \int 3y \, dy - \int y^3 \, dy$
2. Compute each term: $\int 3y \, dy = \frac{3y^2}{2}, \quad \int y^3 \, dy = \frac{y^4}{4}$
3. Substitute back: $V = 2\pi \left[ \frac{3y^2}{2} - \frac{y^4}{4} \right]_0^{\sqrt{3}}$

Evaluate at limits:

At $y = \sqrt{3}$: $\frac{3(\sqrt{3})^2}{2} - \frac{(\sqrt{3})^4}{4} = \frac{3(3)}{2} - \frac{9}{4} = \frac{9}{2} - \frac{9}{4} = \frac{18}{4} - \frac{9}{4} = \frac{9}{4}$

At $y = 0$, the expression is 0.

So: $V = 2\pi \cdot \frac{9}{4} = \frac{9\pi}{2}$

Final Answer:

$V = \frac{9\pi}{2}$

Would you like further clarification or detailed steps?

1. Why is the shell method preferred in this case over the disk method?
2. Can the volume also be computed using the washer method?
3. What does the integral represent geometrically?
4. How does the function $x = 3 - y^2$ affect the bounds?
5. What happens if the region is revolved around a different axis?

Tip: When using the shell method, always identify radius and height clearly before setting up the integral.