Let's solve the problem step-by-step. --- ### **Step 1: Understand the Region** **Given curves:** 1. \( y^2 - 4y + x = 0 \) (i.e., \( x = 4y - y^2 \)) 2. \( x = 0 \) (the y-axis) **Region \( R \):** Bounded by the parabola \( x = 4y - y^2 \) and the y-axis. --- ### **Step 2: Find the Intersection Points** Set \( x = 0 \) in the parabola: \[ 0 = 4y - y^2 \implies y(4 - y) = 0 \implies y = 0 \text{ or } y = 4 \] So, the region is between \( y = 0 \) and \( y = 4 \). --- ### **Step 3: Sketch the Region** - The parabola \( x = 4y - y^2 \) opens to the left (since the \( y^2 \) is negative). - The region is to the right of \( x = 0 \) (the y-axis) and to the left of \( x = 4y - y^2 \), for \( y \) in \([0, 4]\). --- ### **Step 4: Set Up the Volume Integral (Washer Method)** We revolve this region about the **x-axis**. - For a small horizontal strip at \( y \), the distance from the x-axis is \( y \). - The thickness is \( dy \). - The length from \( x = 0 \) to \( x = 4y - y^2 \) is \( x \). The volume of a thin disk (washer) is: \[ dV = \pi [\text{outer radius}^2 - \text{inner radius}^2] \, dy \] Here, the inner radius is 0 (since the region starts at \( x = 0 \)), and the outer radius is \( x = 4y - y^2 \). But since we are revolving around the x-axis, and our region is expressed as \( x \) in terms of \( y \), the radius of the disk is \( y \), and the width of the region is \( x \). For a strip at height \( y \), its length is \( x = 4y - y^2 \), and when revolved, it traces a circle of radius \( y \): \[ dV = 2\pi y \cdot x \, dy = 2\pi y (4y - y^2) \, dy \] --- ### **Step 5: Set Up and Compute the Integral** Integrate from \( y = 0 \) to \( y = 4 \): \[ V = \int_{y=0}^{y=4} 2\pi y (4y - y^2) \, dy \] \[ = 2\pi \int_{0}^{4} (4y^2 - y^3) \, dy \] \[ = 2\pi \left[ 4 \int_{0}^{4} y^2 \, dy - \int_{0}^{4} y^3 \, dy \right] \] Calculate the integrals: - \( \int y^2 \, dy = \frac{y^3}{3} \) - \( \int y^3 \, dy = \frac{y^4}{4} \) So, \[ = 2\pi \left[ 4 \cdot \frac{y^3}{3} - \frac{y^4}{4} \right]_{0}^{4} \] \[ = 2\pi \left[ \left( \frac{4 \cdot 64}{3} - \frac{256}{4} \right) - 0 \right] \] \[ = 2\pi \left[ \frac{256}{3} - 64 \right] \] \[ = 2\pi \left( \frac{256 - 192}{3} \right) \] \[ = 2\pi \left( \frac{64}{3} \right) \] \[ = \frac{128\pi}{3} \] --- ### **Step 6: Final Answer** \[ \boxed{ \frac{128\pi}{3} } \] This is the volume of the solid obtained by revolving the region \( R \) about the x-axis.