To find the volume of the solid obtained by rotating the region bounded by the curves $x = 3y^2$ and $x = 3$ about the line $x = 3$, we will use the method of cylindrical shells.

Step-by-Step Solution

1. Set up the problem: The region is bounded by the curves $x = 3y^2$ (a parabola) and $x = 3$ (a vertical line). We are rotating this region around the line $x = 3$.

2. Cylindrical shells: For the cylindrical shells method, the radius of each shell is the distance from a point on the curve $x = 3y^2$ to the axis of rotation $x = 3$.

   The height of each shell is given by the difference in the $x$-coordinates: $h(y) = 3 - 3y^2$

3. Volume integral: The formula for the volume using cylindrical shells is: $V = \int_{a}^{b} 2\pi (\text{radius}) \times (\text{height}) \, dy$ Here, the radius is $r(y) = 3 - (3y^2)$ and the height is the $y$-length of the region.

   The bounds are from $y = 0$ to $y = 1$ because $x = 3y^2$ intersects $x = 3$ when $y = 1$.

4. Setup the integral: $V = 2\pi \int_0^1 (3 - 3y^2)(1) \, dy$

5. Evaluate the integral: $V = 2\pi \int_0^1 (3 - 3y^2) \, dy$ First, integrate each term: $\int_0^1 3 \, dy = 3y \Big|_0^1 = 3$ $\int_0^1 3y^2 \, dy = y^3 \Big|_0^1 = 1$

6. Final calculation: $V = 2\pi \left( 3 - 1 \right) = 2\pi \times 2 = 4\pi$

Thus, the volume of the solid is $V = 4\pi$.

Would you like further details or clarifications on any of the steps?

Here are five related questions for further exploration:

1. How would the result change if the rotation were about the $y$-axis?
2. What if the region were rotated around the line $x = 0$ instead?
3. How do you calculate the volume using the disk method for this same problem?
4. What happens if the parabola is $x = ky^2$ for some constant $k$?
5. How can we generalize this technique for other curves?

Tip: The cylindrical shells method is particularly useful when the region is rotated around a vertical line and the slices would be awkward to express in terms of $x$.