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 the volume as an integral. Since we are revolving around the y-axis, it's helpful to express \( x \) in terms of \( y \). From the equation \( y = \sqrt{x} \), we can solve for \( x \): \[ x = y^2 \] The bounds for \( y \) when \( x \) ranges from \( 0 \) to \( 4 \) can be found by substituting these \( x \) values back into the equation for \( y \): - When \( x = 0 \), \( y = \sqrt{0} = 0 \). - When \( x = 4 \), \( y = \sqrt{4} = 2 \). Next, the volume \( V \) of the solid of revolution can be calculated using the following formula for revolving a region around the y-axis: \[ V = \pi \int_{a}^{b} [R(y)^2 - r(y)^2] \, dy \] Here, \( R(y) \) represents the outer radius (the function that describes the outer boundary, which is \( x = y^2 \)) and \( r(y) \) represents the inner radius (which is \( 0 \) since there is no hole in this case). Thus, the volume integral simplifies to: \[ V = \pi \int_{0}^{2} (y^2)^2 \, dy = \pi \int_{0}^{2} y^4 \, dy \] Now, we compute the integral: \[ \int y^4 \, dy = \frac{y^5}{5} \] Evaluating the definite integral: \[ \int_{0}^{2} y^4 \, dy = \left[ \frac{y^5}{5} \right]_{0}^{2} = \frac{2^5}{5} - \frac{0^5}{5} = \frac{32}{5} \] Thus, the volume \( V \) becomes: \[ V = \pi \cdot \frac{32}{5} = \frac{32\pi}{5} \] 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 \[ \boxed{\frac{32\pi}{5}}. \]