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The problem is asking for the volume of a solid generated by revolving the given region about the **y-axis**. The equation of the curve is \( y = 3\sqrt{x} \), bounded by \( x = 0 \) and \( y = 6 \). To find the volume, the **shell method** is applied since we are revolving around the y-axis.

### Shell Method Formula:
The volume \( V \) using the shell method is given by:
\[
V = 2\pi \int_{x_1}^{x_2} x \cdot h(x) \, dx
\]
Where:
- \( h(x) \) is the height of the shell (which corresponds to \( y = 3\sqrt{x} \)),
- \( x \) is the radius of the shell (distance to the axis of revolution).

### Step-by-step Solution:
1. **Determine the bounds**: 
   The curve is \( y = 3\sqrt{x} \). We need to find the bounds for \( x \). 
   At \( y = 6 \), we can solve for \( x \):
   \[
   6 = 3\sqrt{x} \quad \Rightarrow \quad x = 4
   \]
   So, we integrate from \( x = 0 \) to \( x = 4 \).

2. **Height of the shell**: 
   The height \( h(x) \) is just the equation of the curve, \( h(x) = 3\sqrt{x} \).

3. **Set up the integral**:
   Using the shell method formula, the volume is:
   \[
   V = 2\pi \int_0^4 x \cdot (3\sqrt{x}) \, dx
   \]
   Simplifying the integrand:
   \[
   V = 6\pi \int_0^4 x^{3/2} \, dx
   \]

4. **Compute the integral**:
   \[
   \int x^{3/2} \, dx = \frac{2}{5}x^{5/2}
   \]
   So:
   \[
   V = 6\pi \left[\frac{2}{5}x^{5/2}\right]_0^4 = 6\pi \cdot \frac{2}{5} \cdot (4^{5/2}) 
   \]
   Since \( 4^{5/2} = (2^2)^{5/2} = 2^5 = 32 \), we get:
   \[
   V = 6\pi \cdot \frac{2}{5} \cdot 32 = 6\pi \cdot \frac{64}{5} = \frac{384\pi}{5}
   \]

Thus, the volume of the solid is \( \frac{384\pi}{5} \).

### Answer:
The correct choice is **(c) \( \frac{384\pi}{5} \)**.

Do you have any questions, or would you like more details on this solution?

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Here are 5 related questions for further exploration:
1. How would the setup change if the region were revolved around the x-axis?
2. How do you apply the washer method to solve a similar problem?
3. What is the difference between the disk and shell methods in calculus?
4. How would the volume change if the upper limit of \( y \) was increased to \( 9 \)?
5. Can this problem be solved using cylindrical coordinates?

**Tip**: Always carefully assess which axis the rotation is occurring around to determine whether to use the shell or disk method.

Use the shell method to set up and evaluate the integral that gives the volume of the solid generated by revolving the plane region about the y-axis. The curve is given by y = 3√x, with bounds x = 0 and y = 6.

This problem demonstrates how to use the shell method to calculate the volume of a solid generated by revolving the curve y = 3√x around the y-axis. The step-by-step solution includes setting up and evaluating the integral, with bounds from x = 0 to x = 4, resulting in the final volume of 384π/5.

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