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To solve the problem, we are given the function \( y = 3\sqrt{x} \), the boundary lines \( y = 6 \), and \( x = 0 \). We need to compute the volume of the solid generated by revolving the region about different lines.

### 1. **Revolving about the x-axis:**
We use the disk method, where the volume is given by:
\[
V = \pi \int_{a}^{b} [f(x)]^2 \, dx
\]
Here, \( f(x) = 3\sqrt{x} \) and the boundaries are from \( x = 0 \) to \( x = 4 \) (since \( y = 6 \) gives \( x = 4 \) when \( 6 = 3\sqrt{x} \)).

\[
V = \pi \int_0^4 (3\sqrt{x})^2 \, dx = \pi \int_0^4 9x \, dx
\]

Calculating the integral:
\[
V = \pi \left[ \frac{9x^2}{2} \right]_0^4 = \pi \left( \frac{9(4)^2}{2} - 0 \right) = \pi \times \frac{144}{2} = 72\pi
\]

So, the volume when revolving about the x-axis is \( 72\pi \).

### 2. **Revolving about the y-axis:**
For revolution about the y-axis, we use the shell method:
\[
V = 2\pi \int_{a}^{b} x f(x) \, dx
\]
Here, \( f(x) = 3\sqrt{x} \), and the limits are from \( x = 0 \) to \( x = 4 \).

\[
V = 2\pi \int_0^4 x (3\sqrt{x}) \, dx = 6\pi \int_0^4 x^{3/2} \, dx
\]

Integrating:
\[
V = 6\pi \left[ \frac{2}{5}x^{5/2} \right]_0^4 = 6\pi \times \frac{2}{5} \times (4^{5/2}) = 6\pi \times \frac{2}{5} \times 32 = \frac{384\pi}{5}
\]

So, the volume when revolving about the y-axis is \( \frac{384\pi}{5} \).

### 3. **Revolving about the line \( y = 6 \):**
We use the washer method. The volume is given by:
\[
V = \pi \int_{a}^{b} \left( R(x)^2 - r(x)^2 \right) \, dx
\]
Here, \( R(x) = 6 \) and \( r(x) = 3\sqrt{x} \), with limits from \( x = 0 \) to \( x = 4 \).

\[
V = \pi \int_0^4 \left( 6^2 - (3\sqrt{x})^2 \right) dx = \pi \int_0^4 \left( 36 - 9x \right) dx
\]

Calculating the integral:
\[
V = \pi \left[ 36x - \frac{9x^2}{2} \right]_0^4 = \pi \left( 36(4) - \frac{9(4)^2}{2} \right) = \pi \left( 144 - 72 \right) = 72\pi
\]

So, the volume when revolving about \( y = 6 \) is \( 72\pi \).

### 4. **Revolving about the line \( x = 4 \):**
This is also a shell method problem. The distance from the line \( x = 4 \) to \( x \) is \( 4 - x \). The volume is:
\[
V = 2\pi \int_0^4 (4 - x) (3\sqrt{x}) \, dx
\]
Simplifying:
\[
V = 6\pi \int_0^4 (4 - x) x^{1/2} \, dx = 6\pi \left[ \int_0^4 4x^{1/2} \, dx - \int_0^4 x^{3/2} \, dx \right]
\]

Integrating:
\[
V = 6\pi \left[ \frac{4 \times 2}{3}x^{3/2} \right]_0^4 - 6\pi \left[ \frac{2}{5}x^{5/2} \right]_0^4
\]
\[
V = 6\pi \left[ \frac{8}{3} \times 8 - \frac{2}{5} \times 32 \right] = 6\pi \left( \frac{64}{3} - \frac{64}{5} \right) = 6\pi \times \frac{64}{15} = \frac{384\pi}{15} = \frac{128\pi}{5}
\]

So, the volume when revolving about \( x = 4 \) is \( \frac{128\pi}{5} \).

---

If you have any further questions or need additional clarifications, feel free to ask! Here are some related questions you might want to explore:

1. How does the shell method differ from the disk/washer method for volume calculation?
2. What are the typical applications of these volume of revolution methods in real life?
3. How would you approach the problem if the region was bounded by a different function?
4. Can you explain why the washer method was appropriate for revolving about \( y = 6 \)?
5. How do different axis rotations affect the setup of the integral?

**Tip:** Always sketch the region and axis of rotation to visualize the solid before setting up your integral.

Find the volume of the solid generated by revolving the region bounded by y = 3√x and the lines y = 6 and x = 0 about the following lines.

Explore how to find the volume of a solid generated by revolving the region bounded by y = 3√x, y = 6, and x = 0 about the x-axis, y-axis, y = 6, and x = 4. Learn detailed solutions using the disk, washer, and shell methods.

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