Show all your work to receive full credit. Write your answers as complete sentences. 1. Use the disk method to find the volume of the solid of revolution generated by rotating the region between the graph of \( f(x)=\sqrt[2]{x} \) and the \( x \)-axis over the interval \( [0,8] \) about the \( x \) - axis.

To find the volume of the solid of revolution generated by rotating the region between the graph of \( f(x)=\sqrt[2]{x} \) and the \( x \)-axis over the interval \( [0,8] \) about the \( x \)-axis, we use the disk method. The volume \( V \) can be calculated using the formula: \[ V = \pi \int_{a}^{b} [f(x)]^2 \, dx \] Here, \( f(x) = \sqrt{x} \), so we need to find \( [f(x)]^2 \): \[ [f(x)]^2 = (\sqrt{x})^2 = x \] Next, we set up our integral from \( 0 \) to \( 8 \): \[ V = \pi \int_{0}^{8} x \, dx \] Now, we can compute the integral: \[ \int x \, dx = \frac{x^2}{2} \] Evaluating this from \( 0 \) to \( 8 \): \[ \left[ \frac{x^2}{2} \right]_{0}^{8} = \frac{8^2}{2} - \frac{0^2}{2} = \frac{64}{2} - 0 = 32 \] Therefore, substituting back into our volume formula: \[ V = \pi (32) = 32\pi \] Thus, the volume of the solid of revolution is \( 32\pi \) cubic units. Would you like assistance with anything else?