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 \( y = \sqrt{x} \). - The limits of integration are 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 \( y = 0 \) to \( y = 2 \). 4. **Set up the volume integral using the washer method**: - The volume \( V \) of the solid of revolution is given by the integral: \[ V = \pi \int_{a}^{b} [R(y)^2 - r(y)^2] \, dy \] - In this case, since we are revolving around the y-axis and there is no inner radius (the area is bounded by the x-axis), we have: - Outer radius \( R(y) = y^2 \) - Inner radius \( r(y) = 0 \) - Thus, the volume integral simplifies to: \[ V = \pi \int_{0}^{2} (y^2)^2 \, dy = \pi \int_{0}^{2} y^4 \, dy \] 5. **Calculate the integral**: - We will now compute the integral: \[ V = \pi \int_{0}^{2} y^4 \, dy \] Now, let's calculate the integral. \[ \int y^4 \, dy = \frac{y^5}{5} \] Evaluating this from \( 0 \) to \( 2 \): \[ \left[ \frac{y^5}{5} \right]_{0}^{2} = \frac{2^5}{5} - \frac{0^5}{5} = \frac{32}{5} - 0 = \frac{32}{5} \] Thus, the volume \( V \) is: \[ V = \pi \cdot \frac{32}{5} = \frac{32\pi}{5} \] 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 \[ \boxed{\frac{32\pi}{5}}. \]