We are tasked with finding the volume of the solid generated by revolving the region bounded by $y = \sqrt{x}$, $x = 4$, and $y = 0$ about the line $x = 6$. The most appropriate method here is the shell method because we are revolving around a vertical line that is not the y-axis.

Step-by-Step Solution:

1. General Formula for Shell Method

For revolving around a vertical line, the formula for the shell method is:

$V = 2\pi \int_{a}^{b} \text{(radius)} \times \text{(height)} \, dx$

2. Radius of a Typical Shell

The radius of a shell is the horizontal distance from the line of rotation $x = 6$ to a point on the curve $x = x$. Thus, the radius is:

$\text{radius} = 6 - x$

3. Height of a Typical Shell

The height of the shell is given by the function $y = \sqrt{x}$. So the height is:

$\text{height} = \sqrt{x}$

4. Limits of Integration

We are integrating along the x-axis from $x = 0$ to $x = 4$, as these are the bounds of the region.

Thus, the volume integral is:

$V = 2\pi \int_{0}^{4} (6 - x) \cdot \sqrt{x} \, dx$

5. Simplifying the Integral

We need to expand the integrand before integrating:

$V = 2\pi \int_{0}^{4} (6\sqrt{x} - x\sqrt{x}) \, dx$

This can be broken down into two terms:

$V = 2\pi \left( \int_{0}^{4} 6\sqrt{x} \, dx - \int_{0}^{4} x\sqrt{x} \, dx \right)$

Now, let's solve these two integrals separately.

6. First Integral

The first term involves $6\sqrt{x} = 6x^{1/2}$:

$\int 6x^{1/2} \, dx = 6 \cdot \frac{x^{3/2}}{3/2} = 4x^{3/2}$

Evaluating from $0$ to $4$:

$4(4)^{3/2} - 4(0)^{3/2} = 4 \cdot 8 = 32$

7. Second Integral

The second term involves $x\sqrt{x} = x^{3/2}$:

$\int x^{3/2} \, dx = \frac{x^{5/2}}{5/2} = \frac{2x^{5/2}}{5}$

Evaluating from $0$ to $4$:

$\frac{2(4)^{5/2}}{5} - \frac{2(0)^{5/2}}{5} = \frac{2 \cdot 32}{5} = \frac{64}{5}$

8. Final Volume Calculation

Now substitute these values back into the volume expression:

$V = 2\pi \left( 32 - \frac{64}{5} \right)$

$V = 2\pi \left( \frac{160}{5} - \frac{64}{5} \right) = 2\pi \cdot \frac{96}{5}$

$V = \frac{192\pi}{5}$

Thus, the volume of the solid is:

$V = \frac{192\pi}{5} \, \text{cubic units}$

Would you like any more details or explanations?

Here are some related questions you might find helpful:

1. How does the shell method differ from the disk/washer method?
2. What is the advantage of using the shell method in this problem?
3. How would the setup change if we revolved the region about the y-axis instead of $x = 6$?
4. Can you set up the same volume using the washer method?
5. How would the limits of integration change if the region were bounded by different lines?

Tip: When choosing between the shell method and the washer method, consider the axis of rotation. If the axis is parallel to the y-axis, the shell method is often simpler.