Calculate the volume of the solid formed by rotating the region bounded by the curve \( y = x^2 \) and the line \( y = 1 \) around the x-axis using the washer method.

To find the volume of the solid formed by rotating the region bounded by the curve \( y = x^2 \) and the line \( y = 1 \) around the x-axis using the washer method, we first need to determine the points of intersection between the curve and the line. Setting \( x^2 = 1 \), we have: \[ x^2 = 1 \implies x = \pm 1 \] This indicates that the region of interest in the x-axis is from \( x = -1 \) to \( x = 1 \). The formula for the volume \( V \) of a solid of revolution using the washer method is given by: \[ V = \pi \int_a^b \left( R(x)^2 - r(x)^2 \right) dx \] where \( R(x) \) is the outer radius and \( r(x) \) is the inner radius. In this case: - The outer radius \( R(x) \) is the line \( y = 1 \). - The inner radius \( r(x) \) is the curve \( y = x^2 \). Thus, we have: \[ R(x) = 1, \quad r(x) = x^2 \] Now we substitute these into the volume formula: \[ V = \pi \int_{-1}^{1} \left( (1)^2 - (x^2)^2 \right) dx \] This simplifies to: \[ V = \pi \int_{-1}^{1} \left( 1 - x^4 \right) dx \] Now we can compute the integral: \[ V = \pi \left[ \int_{-1}^{1} 1 \, dx - \int_{-1}^{1} x^4 \, dx \right] \] Calculating the first integral: \[ \int_{-1}^{1} 1 \, dx = 2 \] Now calculating the second integral: \[ \int_{-1}^{1} x^4 \, dx = 2 \int_{0}^{1} x^4 \, dx = 2 \left[ \frac{x^5}{5} \right]_{0}^{1} = 2 \cdot \frac{1}{5} = \frac{2}{5} \] Putting everything together: \[ V = \pi \left( 2 - \frac{2}{5} \right) = \pi \left( \frac{10}{5} - \frac{2}{5} \right) = \pi \left( \frac{8}{5} \right) \] Thus, the final volume of the solid is: \[ V = \frac{8\pi}{5} \]