A region is defined by the functions \( y = e^{-x} \) for \( x \geq 0 \) and \( y = 0 \). What is the volume generated when this region is revolved around the line \( y = 1 \)?

To find the volume generated when the region defined by \( y = e^{-x} \) for \( x \geq 0 \) and \( y = 0 \) is revolved around the line \( y = 1 \), we can use the method of cylindrical shells or the washer method. Here, I will use the washer method. 1. **Identify the equations**: The curves are \( y = e^{-x} \) and the line \( y = 1 \). The region revolves around the line \( y = 1 \), which is above the curve. 2. **Find outer and inner radii**: - The outer radius \( R \) when revolving around \( y = 1 \) is the distance from the line \( y = 1 \) to the line \( y = 0 \): \[ R = 1 - 0 = 1. \] - The inner radius \( r \) is the distance from the line \( y = 1 \) to the curve \( y = e^{-x} \): \[ r = 1 - e^{-x}. \] 3. **Set up the volume integral**: The volume \( V \) generated by revolving the region can be expressed as: \[ V = \pi \int_{0}^{\infty} \left[ R^2 - r^2 \right] \, dx = \pi \int_{0}^{\infty} \left[ 1^2 - (1 - e^{-x})^2 \right] \, dx. \] 4. **Simplify the integral**: \[ V = \pi \int_{0}^{\infty} \left[ 1 - (1 - 2e^{-x} + e^{-2x}) \right] \, dx = \pi \int_{0}^{\infty} (2e^{-x} - e^{-2x}) \, dx. \] 5. **Separate the integral**: \[ V = \pi \left[ 2 \int_{0}^{\infty} e^{-x} \, dx - \int_{0}^{\infty} e^{-2x} \, dx \right]. \] 6. **Evaluate the integrals**: - The integral of \( e^{-x} \) from \( 0 \) to \( \infty \): \[ \int_{0}^{\infty} e^{-x} \, dx = 1. \] - The integral of \( e^{-2x} \) from \( 0 \) to \( \infty \): \[ \int_{0}^{\infty} e^{-2x} \, dx = \frac{1}{2}. \] 7. **Combine the results**: \[ V = \pi \left[ 2 \cdot 1 - \frac{1}{2} \right] = \pi \left[ 2 - 0.5 \right] = \pi \cdot 1.5 = \frac{3\pi}{2}. \] Thus, the volume generated when the region is revolved around the line \( y = 1 \) is \[ \boxed{\frac{3\pi}{2}}. \]