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.

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: \[ \boxed{4\pi} \]