Show all your work to receive full credit. Write your answers as complete sentences. 1. Use the method of cylindrical shells to find the volume of the solid of revolution generated by rotating the region between the graphs of \( y=\frac{1}{x^{2}}, x=1, x=10 \), and \( y=0 \) about the \( y \)-axis.

To find the volume of the solid of revolution generated by rotating the region bounded by \( y = \frac{1}{x^{2}} \), \( x = 1 \), \( x = 10 \), and \( y = 0 \) about the \( y \)-axis using the method of cylindrical shells, we will follow these steps: **1. Understanding the Region and Setup:** We are given the function \( y = \frac{1}{x^{2}} \), which is a decreasing function for \( x > 0 \). The region of interest is bounded by: - The curve \( y = \frac{1}{x^{2}} \), - The vertical lines \( x = 1 \) and \( x = 10 \), - The horizontal line \( y = 0 \) (the \( x \)-axis). When we rotate this region about the \( y \)-axis, we can use the method of cylindrical shells to find the volume. **2. Method of Cylindrical Shells:** The formula for the volume \( V \) using cylindrical shells is: \[ V = 2\pi \int_{a}^{b} ( \text{radius} ) \times ( \text{height} ) \, dx \] In this context: - **Radius** is the distance from the \( y \)-axis, which is simply \( x \). - **Height** is the value of the function \( y = \frac{1}{x^{2}} \). **3. Setting Up the Integral:** Given the bounds \( x = 1 \) and \( x = 10 \), the integral becomes: \[ V = 2\pi \int_{1}^{10} x \times \frac{1}{x^{2}} \, dx \] Simplifying the integrand: \[ x \times \frac{1}{x^{2}} = \frac{1}{x} \] So, the integral simplifies to: \[ V = 2\pi \int_{1}^{10} \frac{1}{x} \, dx \] **4. Evaluating the Integral:** The integral of \( \frac{1}{x} \) with respect to \( x \) is \( \ln|x| \). Therefore: \[ V = 2\pi \left[ \ln|x| \right]_{1}^{10} \] Evaluating the definite integral: \[ V = 2\pi \left( \ln(10) - \ln(1) \right) \] Since \( \ln(1) = 0 \): \[ V = 2\pi \ln(10) \] **5. Final Answer:** The volume of the solid of revolution is: \[ V = 2\pi \ln(10) \ \text{cubic units} \] Approximately, since \( \ln(10) \approx 2.302585093 \): \[ V \approx 2\pi \times 2.302585093 \approx 14.4543 \ \text{cubic units} \] **Conclusion:** Using the method of cylindrical shells, we determined that the volume of the solid obtained by rotating the given region about the \( y \)-axis is \( 2\pi \ln(10) \) cubic units, which is approximately \( 14.4543 \) cubic units.