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 can use the method of disks (or washers). First, we need to determine the points where the curve \( y = x^2 \) intersects the line \( y = 4 \). This can be found by setting \( x^2 = 4 \): \[ x = \pm 2 \] This means our region of interest is between \( x = -2 \) and \( x = 2 \). The volume \( V \) of the solid of revolution is given by the integral: \[ V = \pi \int_{a}^{b} (R(x)^2 - r(x)^2) \, dx \] Here, \( R(x) \) is the outer radius and \( r(x) \) is the inner radius. In this case: - The outer radius \( R(x) = 4 \) (the line) - The inner radius \( r(x) = x^2 \) (the curve) Thus, the volume can be expressed as: \[ V = \pi \int_{-2}^{2} \left( 4^2 - (x^2)^2 \right) dx \] Calculating the integrand: \[ 4^2 = 16, \quad (x^2)^2 = x^4 \] So our integral becomes: \[ V = \pi \int_{-2}^{2} (16 - x^4) \, dx \] Now we can calculate the integral. Since the function \( 16 - x^4 \) is even, we can evaluate from 0 to 2 and double the result: \[ V = 2\pi \int_{0}^{2} (16 - x^4) \, dx \] Evaluating the integral: \[ \int_{0}^{2} (16 - x^4) \, dx = \left[ 16x - \frac{x^5}{5} \right]_{0}^{2} \] Calculating it at the bounds: \[ = \left( 16(2) - \frac{(2)^5}{5} \right) - \left( 16(0) - \frac{(0)^5}{5} \right) \] \[ = 32 - \frac{32}{5} = 32 - 6.4 = 25.6 \] Thus, \[ V = 2 \pi (25.6) = 51.2 \pi \] Therefore, the volume of the solid generated by revolving the region bounded by \( y = x^2 \) and \( y = 4 \) around the x-axis is: \[ \boxed{51.2\pi} \]