To find the volume of the solid generated by revolving the region under the curve $y = \frac{1}{x}$ about the x-axis using the shell method, we follow these steps:

1. Set up the Shell Method Formula:

The general formula for the volume of a solid of revolution using the shell method, when revolving around the x-axis, is:

$V = 2\pi \int_{y_1}^{y_2} (\text{radius}) (\text{height}) \, dy$

2. Identify Variables:

In this case, we revolve the region about the x-axis, and the height of the shells will be given in terms of $y$, while the radius will simply be $y$ because we're revolving around the x-axis. From the curve $y = \frac{1}{x}$, we solve for $x$ in terms of $y$:

$y = \frac{1}{x} \implies x = \frac{1}{y}$

Thus, the radius is $y$, and the height of the shell is the distance between $x = 1$ and $x = \frac{1}{y}$ at each $y$, which is given by $\frac{1}{y} - 1$.

3. Determine the Limits of Integration:

From the given information, the region lies between $x = 1$ and $x = 4$, which corresponds to the values of $y = \frac{1}{x}$, or $y = 1$ to $y = \frac{1}{4}$. Therefore, the limits of integration for $y$ are:

$y_1 = \frac{1}{4}, \quad y_2 = 1$

4. Write the Integral:

Now, we substitute everything into the formula for the volume:

$V = 2\pi \int_{\frac{1}{4}}^{1} y \left( \frac{1}{y} - 1 \right) \, dy$

5. Simplify the Integral:

Distribute the $y$ term inside the parentheses:

$V = 2\pi \int_{\frac{1}{4}}^{1} (1 - y) \, dy$

6. Evaluate the Integral:

First, integrate $1 - y$:

$\int (1 - y) \, dy = y - \frac{y^2}{2}$

Now, evaluate this from $y = \frac{1}{4}$ to $y = 1$:

$V = 2\pi \left[ \left( 1 - \frac{1^2}{2} \right) - \left( \frac{1}{4} - \frac{\left(\frac{1}{4}\right)^2}{2} \right) \right]$

Simplifying the expressions:

$V = 2\pi \left[ \left( 1 - \frac{1}{2} \right) - \left( \frac{1}{4} - \frac{1}{32} \right) \right]$ $V = 2\pi \left[ \frac{1}{2} - \left( \frac{8}{32} - \frac{1}{32} \right) \right]$ $V = 2\pi \left[ \frac{1}{2} - \frac{7}{32} \right]$ $V = 2\pi \left[ \frac{16}{32} - \frac{7}{32} \right]$ $V = 2\pi \cdot \frac{9}{32}$

7. Final Answer:

$V = \frac{9\pi}{16}$

Thus, the volume of the solid generated by revolving the region under the curve $y = \frac{1}{x}$ from $x = 1$ to $x = 4$ about the x-axis is $\frac{9\pi}{16}$.

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Would you like further details or have any questions?

Here are 5 related questions to consider:

1. How would the solution change if we revolved around the y-axis instead?
2. Can you derive the volume using the disk/washer method instead of the shell method?
3. What if the region was bounded between $x = 2$ and $x = 5$?
4. How do you set up the shell method for different types of curves?
5. What is the geometric interpretation of the shell height and radius in this example?

Tip: The shell method is often easier for regions where the function is best described in terms of $y$.