Find the volume of the solid formed by rotating the area under the curve $$ y = \frac{1}{x} $$ from $$ x = 1 $$ to $$ x = 3 $$ about the y-axis.

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To find the volume of the solid formed by rotating the area under the curve $$ y = \frac{1}{x} $$ from $$ x = 1 $$ to $$ x = 3 $$ about the y-axis, we can use two methods: the **Shell Method** and the **Washer Method**. Both methods should yield the same result when applied correctly.

### 1. Shell Method

The **Shell Method** is particularly useful when dealing with rotations around the y-axis. Here's how to apply it:

- **Radius of the Shell:** $$ x $$ (distance from the y-axis)
- **Height of the Shell:** $$ y = \frac{1}{x} $$ (since we're considering the area under the curve)
- **Volume Element:** The differential volume $$ dV $$ of a cylindrical shell is given by:
  $$
  dV = 2\pi \times \text{radius} \times \text{height} \times dx = 2\pi x \left(\frac{1}{x}\right) dx = 2\pi dx
  $$
  
- **Total Volume $$ V $$:**
  $$
  V = \int_{1}^{3} 2\pi dx = 2\pi (3 - 1) = 4\pi
  $$

### 2. Washer Method

The **Washer Method** involves integrating with respect to $$ y $$:

1. **Express $$ x $$ in terms of $$ y $$:** $$ x = \frac{1}{y} $$
2. **Determine Limits of Integration:**
   - When $$ x = 1 $$, $$ y = 1 $$
   - When $$ x = 3 $$, $$ y = \frac{1}{3} $$

3. **Set Up the Integral:**
   - **Outer Radius $$ R $$:** $$ \frac{1}{y} $$ (from the curve)
   - **Inner Radius $$ r $$:** $$ 1 $$ (from the line $$ x = 1 $$)
   
   However, for $$ y < \frac{1}{3} $$, $$ \frac{1}{y} > 3 $$, which exceeds our domain. Therefore, we have two regions to consider:

- **For $$ y $$ from $$ \frac{1}{3} $$ to $$ 1 $$:**
     $$
     R = \frac{1}{y}, \quad r = 1
     $$
   
   - **For $$ y $$ from $$ 0 $$ to $$ \frac{1}{3} $$:**
     $$
     R = 3, \quad r = 1
     $$

4. **Compute the Integrals:**
   $$
   V = \pi \left( \int_{\frac{1}{3}}^{1} \left( \left(\frac{1}{y}\right)^2 - 1^2 \right) dy + \int_{0}^{\frac{1}{3}} \left(3^2 - 1^2\right) dy \right)
   $$
   
   Simplifying:
   $$
   V = \pi \left( \int_{\frac{1}{3}}^{1} \left(\frac{1}{y^2} - 1\right) dy + \int_{0}^{\frac{1}{3}} 8 \, dy \right)
   $$
   
   Evaluating the integrals:
   $$
   V = \pi \left[ \left(-\frac{1}{y} - y\right) \bigg|_{\frac{1}{3}}^{1} + 8 \left(\frac{1}{3} - 0\right) \right] = \pi \left[ 2 - \frac{2}{3} + \frac{8}{3} \right] = 4\pi
   $$

### Conclusion

Both methods confirm that the volume of the solid is:

$$
\boxed{4\pi}
$$
The volume of the solid is $$ 4\pi $$.

Find the volume of the solid formed by rotating the area under the curve \( y = \frac{1}{x} \) from \( x = 1 \) to \( x = 3 \) about the y-axis.

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