Using the disc method, find the volume of the solid formed by revolving the region bounded by \( y = x^2 \) and \( y = 0 \) from \( x = 0 \) to \( x = 2 \) about the x-axis.

To find the volume of the solid formed by revolving the region bounded by \( y = x^2 \) and \( y = 0 \) from \( x = 0 \) to \( x = 2 \) about the x-axis using the disk method, we start by determining the radius of the disks. The radius at any point \( x \) is given by the value of the function \( y = x^2 \). The volume \( V \) can be calculated using the following integral formula: \[ V = \pi \int_{a}^{b} [f(x)]^2 \, dx \] In this case, \( f(x) = x^2 \), \( a = 0 \), and \( b = 2 \). Thus, we can set up the integral: \[ V = \pi \int_{0}^{2} (x^2)^2 \, dx \] This simplifies to: \[ V = \pi \int_{0}^{2} x^4 \, dx \] Now we compute the integral: \[ \int x^4 \, dx = \frac{x^5}{5} \] Evaluating this from 0 to 2 gives: \[ = \left[ \frac{x^5}{5} \right]_{0}^{2} = \frac{2^5}{5} - \frac{0^5}{5} = \frac{32}{5} \] Finally, we multiply by \( \pi \): \[ V = \pi \cdot \frac{32}{5} \] Thus, the volume of the solid is: \[ V = \frac{32\pi}{5} \] So, the volume of the solid formed by revolving the region bounded by \( y = x^2 \) and \( y = 0 \) from \( x = 0 \) to \( x = 2 \) about the x-axis is \( \frac{32\pi}{5} \) cubic units.