Determine the volume generated by revolving the area between \( y = \sqrt{x} \) and the x-axis from \( x = 0 \) to \( x = 4 \) about the y-axis with the washer method.

To find the volume generated by revolving the area between \( y = \sqrt{x} \) and the x-axis from \( x = 0 \) to \( x = 4 \) about the y-axis using the washer method, we first express \( x \) in terms of \( y \). The equation \( y = \sqrt{x} \) can be rewritten as \( x = y^2 \). The limits for \( y \) when \( x \) goes from \( 0 \) to \( 4 \) are \( y = 0 \) (when \( x = 0 \)) to \( y = 2 \) (when \( x = 4 \)). When revolving around the y-axis, we will take the differential volume element as \( dV = \pi [R^2 - r^2] \, dy \). Here, \( R \) is the outer radius, which is the distance from the y-axis to the curve, and \( r \) is the inner radius. In this case: - The outer radius \( R = y^2 \) (since \( x = y^2 \)) - The inner radius \( r = 0 \) (since it's revolved around the x-axis, there’s no inner volume) Thus, the volume can be computed as follows: \[ V = \int_{y=0}^{y=2} \pi (R^2 - r^2) \, dy = \int_{0}^{2} \pi ((y^2)^2 - 0^2) \, dy = \int_{0}^{2} \pi y^4 \, dy \] Now we compute the integral: \[ V = \pi \cdot \left[ \frac{y^5}{5} \right]_{0}^{2} = \pi \cdot \frac{2^5}{5} = \pi \cdot \frac{32}{5} = \frac{32\pi}{5} \] Therefore, the volume generated by revolving the area between \( y = \sqrt{x} \) and the x-axis from \( x = 0 \) to \( x = 4 \) about the y-axis is \[ \frac{32\pi}{5}. \]