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 \( y = x^2 \), \( y = 0 \), \( x = 0 \), and \( x = 2 \) about the \( x \)-axis using the **disc method**, follow these steps: ### Step 1: Understand the Disc Method The **disc method** is used to find the volume of a solid of revolution by integrating the area of circular discs perpendicular to the axis of rotation. The formula for the volume \( V \) when rotating around the \( x \)-axis is: \[ V = \pi \int_{a}^{b} [f(x)]^2 \, dx \] where: - \( f(x) \) is the function defining the boundary of the region. - \( a \) and \( b \) are the limits of integration along the \( x \)-axis. ### Step 2: Identify the Function and Limits In this problem: - \( f(x) = x^2 \) (the upper boundary) - The lower boundary is \( y = 0 \) (the \( x \)-axis itself). - The region is bounded from \( x = 0 \) to \( x = 2 \). ### Step 3: Set Up the Integral Using the disc method formula: \[ V = \pi \int_{0}^{2} (x^2)^2 \, dx \] Simplify \( (x^2)^2 \) to \( x^4 \): \[ V = \pi \int_{0}^{2} x^4 \, dx \] ### Step 4: Compute the Integral Integrate \( x^4 \) with respect to \( x \): \[ \int x^4 \, dx = \frac{x^5}{5} + C \] Evaluate from \( 0 \) to \( 2 \): \[ \left. \frac{x^5}{5} \right|_{0}^{2} = \frac{2^5}{5} - \frac{0^5}{5} = \frac{32}{5} - 0 = \frac{32}{5} \] ### Step 5: Multiply by \( \pi \) to Find the Volume \[ V = \pi \times \frac{32}{5} = \frac{32\pi}{5} \] ### Final Answer \[ V = \frac{32}{5} \pi \text{ cubic units} \]