Calculate the volume of the solid generated by revolving the region bounded by the curve \( y = x^2 \) and the line \( y = 4 \) around the x-axis.

To calculate the volume of the solid generated by revolving the region bounded by the curve \( y = x^2 \) and the line \( y = 4 \) around the \( x \)-axis, we'll use the **washer method**. Here's a step-by-step explanation: 1. **Find the Points of Intersection:** First, determine where the curves \( y = x^2 \) and \( y = 4 \) intersect: \[ x^2 = 4 \implies x = \pm 2 \] So, the region of interest is between \( x = -2 \) and \( x = 2 \). 2. **Set Up the Washer Method:** When revolving around the \( x \)-axis, the washer has: - **Outer radius (\( R \))**: The distance from the \( x \)-axis to the line \( y = 4 \), which is constant at \( R = 4 \). - **Inner radius (\( r \))**: The distance from the \( x \)-axis to the curve \( y = x^2 \), which is \( r = x^2 \). The volume \( V \) is given by the integral: \[ V = \pi \int_{-2}^{2} \left( R^2 - r^2 \right) dx = \pi \int_{-2}^{2} \left( 4^2 - (x^2)^2 \right) dx = \pi \int_{-2}^{2} \left( 16 - x^4 \right) dx \] 3. **Simplify the Integral:** Recognize that the integrand \( 16 - x^4 \) is an even function, so we can simplify the integral by doubling the integral from 0 to 2: \[ V = 2\pi \int_{0}^{2} \left( 16 - x^4 \right) dx \] 4. **Compute the Integral:** \[ \int_{0}^{2} \left( 16 - x^4 \right) dx = \left[ 16x - \frac{x^5}{5} \right]_{0}^{2} = \left( 16(2) - \frac{2^5}{5} \right) - \left( 0 - 0 \right) = 32 - \frac{32}{5} = \frac{160}{5} - \frac{32}{5} = \frac{128}{5} \] 5. **Multiply by \( 2\pi \):** \[ V = 2\pi \times \frac{128}{5} = \frac{256}{5} \pi \] **Final Answer:** \[ \boxed{\dfrac{256}{5}\, \pi} \]