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 solution: ### 1. **Identify 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 Integral** When revolving around the x-axis, the volume \( V \) can be calculated using the washer method: \[ V = \pi \int_{a}^{b} \left[ (R(x))^2 - (r(x))^2 \right] dx \] Where: - \( R(x) \) is the outer radius (distance from the x-axis to the upper curve). - \( r(x) \) is the inner radius (distance from the x-axis to the lower curve). In this case: - The **outer curve** is \( y = 4 \), so \( R(x) = 4 \). - The **inner curve** is \( y = x^2 \), so \( r(x) = x^2 \). ### 3. **Express the Integral** Substitute \( R(x) \) and \( r(x) \) into the integral: \[ V = \pi \int_{-2}^{2} \left[ 4^2 - (x^2)^2 \right] dx = \pi \int_{-2}^{2} \left[ 16 - x^4 \right] dx \] Since the integrand is even, we can simplify the calculation by integrating from 0 to 2 and then doubling the result: \[ V = 2\pi \int_{0}^{2} \left[ 16 - x^4 \right] dx \] ### 4. **Compute the Integral** Evaluate the integral: \[ \int_{0}^{2} \left[ 16 - x^4 \right] dx = \left[ 16x - \frac{x^5}{5} \right]_0^2 = \left( 16 \times 2 - \frac{32}{5} \right) = 32 - \frac{32}{5} = \frac{160 - 32}{5} = \frac{128}{5} \] ### 5. **Calculate the Volume** Multiply by \( 2\pi \): \[ V = 2\pi \times \frac{128}{5} = \frac{256}{5} \pi \] ### **Final Answer** The volume of the solid is: \[ V = \frac{256}{5} \pi \text{ cubic units} \] Answer: Its volume is 256⁄5 multiplied by π, that is V = (256⁄5) · π cubic units