Section 5.3: Problem (1 point) Use the method of cylindrical shells to find the volume generated by rotating the region bounded by the curves \( x=\sqrt{y}, x=0 \), and \( y=2 \) about the \( x \)-axis. volume \( =\square \)

To find the volume generated by rotating the region bounded by the curves \( x=\sqrt{y} \), \( x=0 \), and \( y=2 \) about the \( x \)-axis using the method of cylindrical shells, we can set up the integral. The height of each shell at a given \( y \) is given by \( x = \sqrt{y} \) and the radius is \( y \). The volume \( V \) of the solid can be calculated using the formula: \[ V = 2\pi \int_{0}^{2} (radius)(height)\, dy = 2\pi \int_{0}^{2} y \sqrt{y}\, dy \] This can be simplified to: \[ V = 2\pi \int_{0}^{2} y^{3/2}\, dy \] Calculating the integral: \[ V = 2\pi \left[ \frac{2}{5} y^{5/2} \right]_{0}^{2} = 2\pi \cdot \frac{2}{5} \cdot (2^{5/2} - 0) = 2\pi \cdot \frac{2}{5} \cdot 4\sqrt{2} \] Simplifying this gives: \[ V = \frac{16\pi\sqrt{2}}{5} \] Thus, the volume is \( \frac{16\pi\sqrt{2}}{5} \).