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 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 using the washer method, we will follow these steps: 1. **Identify the function and the limits of integration**: The function is \( y = \sqrt{x} \) and we are revolving around the y-axis from \( x = 0 \) to \( x = 4 \). 2. **Express \( x \) in terms of \( y \)**: Since we are revolving around the y-axis, we need to express \( x \) as a function of \( y \). From \( y = \sqrt{x} \), we can square both sides to get: \[ x = y^2 \] 3. **Determine the limits for \( y \)**: When \( x = 0 \), \( y = \sqrt{0} = 0 \). When \( x = 4 \), \( y = \sqrt{4} = 2 \). Therefore, the limits for \( y \) are from \( 0 \) to \( 2 \). 4. **Set up the volume integral**: The volume \( V \) of the solid of revolution can be calculated using the formula: \[ V = \pi \int_{a}^{b} [R(y)]^2 \, dy \] where \( R(y) \) is the radius of the washer at height \( y \). In this case, \( R(y) = y^2 \) (the distance from the y-axis to the curve). 5. **Write the integral**: The volume integral becomes: \[ V = \pi \int_{0}^{2} (y^2)^2 \, dy = \pi \int_{0}^{2} y^4 \, dy \] 6. **Calculate the integral**: Now we will compute the integral: \[ \int y^4 \, dy = \frac{y^5}{5} \] Evaluating this from \( 0 \) to \( 2 \): \[ \left[ \frac{y^5}{5} \right]_{0}^{2} = \frac{2^5}{5} - \frac{0^5}{5} = \frac{32}{5} \] 7. **Calculate the volume**: Now substituting back into the volume formula: \[ V = \pi \cdot \frac{32}{5} = \frac{32\pi}{5} \] Thus, 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: \[ \boxed{\frac{32\pi}{5}} \]