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 need to express \( x \) in terms of \( y \). Since \( y = \sqrt{x} \), we can rewrite this as: \[ x = y^2. \] Next, we determine the limits for \( y \) by evaluating \( y \) at the bounds of \( x \): - When \( x = 0 \), \( y = \sqrt{0} = 0 \). - When \( x = 4 \), \( y = \sqrt{4} = 2 \). Thus, we are rotating around the y-axis from \( y = 0 \) to \( y = 2 \). The volume \( V \) generated by rotating around the y-axis is given by the formula: \[ V = \pi \int_{a}^{b} [R(y)]^2 - [r(y)]^2 \, dy, \] where \( R(y) \) is the outer radius and \( r(y) \) is the inner radius. In this case, since we are rotating the area between the curve \( y = \sqrt{x} \) and the x-axis, we do not have an inner radius. Therefore: \[ R(y) = y^2 \quad \text{and} \quad r(y) = 0. \] Substituting into the volume formula gives us: \[ V = \pi \int_{0}^{2} (y^2)^2 \, dy = \pi \int_{0}^{2} y^4 \, dy. \] Now we compute the integral: \[ V = \pi \left[ \frac{y^5}{5} \right]_{0}^{2} = \pi \left( \frac{2^5}{5} - \frac{0^5}{5} \right) = \pi \left( \frac{32}{5} \right). \] Therefore, the volume is: \[ V = \frac{32\pi}{5}. \] So the final volume generated by revolving the area is \( \frac{32\pi}{5} \) cubic units.