Find the volume of the solid obtained by rotating the region bounded by the curves \( y = x^{2} \) and \( y = 4 \) about the line \( y = 5 \) using the washer method.

To find the volume of the solid obtained by rotating the region bounded by the curves \( y = x^2 \) and \( y = 4 \) about the line \( y = 5 \) using the washer method, we start by determining the points of intersection of the curves to find the limits of integration. Setting \( x^2 = 4 \), we solve: \[ x^2 = 4 \implies x = -2 \quad \text{and} \quad x = 2. \] Thus, the region of interest is bounded between \( x = -2 \) and \( x = 2 \). When using the washer method, we identify the outer radius and the inner radius of the washers formed by the rotation around the line \( y = 5 \). - The outer radius \( R(x) \) is the distance from the line \( y = 5 \) to the line \( y = 4 \): \[ R(x) = 5 - 4 = 1. \] - The inner radius \( r(x) \) is the distance from the line \( y = 5 \) to the parabola \( y = x^2 \): \[ r(x) = 5 - x^2. \] The volume \( V \) of the solid of revolution is given by the integral of the difference of the squares of the outer and inner radii: \[ V = \pi \int_{-2}^{2} \left( R(x)^2 - r(x)^2 \right) \, dx. \] Now we compute: \[ R(x)^2 = 1^2 = 1, \] \[ r(x)^2 = (5 - x^2)^2 = 25 - 10x^2 + x^4. \] Thus, the integrand becomes: \[ R(x)^2 - r(x)^2 = 1 - (25 - 10x^2 + x^4) = 1 - 25 + 10x^2 - x^4 = -24 + 10x^2 - x^4. \] Now, we have: \[ V = \pi \int_{-2}^{2} (-24 + 10x^2 - x^4) \, dx. \] To evaluate the integral, we simplify: \[ V = \pi \left( \int_{-2}^{2} -24 \, dx + \int_{-2}^{2} 10x^2 \, dx - \int_{-2}^{2} x^4 \, dx \right). \] Calculating each integral separately: 1. \(\int_{-2}^{2} -24 \, dx = -24 \cdot (2 - (-2)) = -24 \cdot 4 = -96\). 2. \(\int_{-2}^{2} 10x^2 \, dx\): - The function \(10x^2\) is even, so: \[ 2 \cdot \int_{0}^{2} 10x^2 \, dx = 2 \cdot 10 \cdot \left[\frac{x^3}{3}\right]_{0}^{2} = 2 \cdot 10 \cdot \frac{8}{3} = \frac{160}{3}. \] 3. \(\int_{-2}^{2} x^4 \, dx\): - Again, using that it's even: \[ 2 \cdot \int_{0}^{2} x^4 \, dx = 2 \cdot \left[\frac{x^5}{5}\right]_{0}^{2} = 2 \cdot \frac{32}{5} = \frac{64}{5}. \] Putting it all together: \[ V = \pi \left(-96 + \frac{160}{3} - \frac{64}{5}\right). \] Now, we need to combine these fractions, finding a common denominator. The least common multiple of 3 and 5 is 15: \[ -96 = -\frac{1440}{15}, \] \[ \frac{160}{3} = \frac{800}{15}, \] \[ \frac{64}{5} = \frac{192}{15}. \] So now we combine: \[ V = \pi \left(-\frac{1440}{15} + \frac{800}{15} - \frac{192}{15}\right) = \pi \left(-\frac{1440 - 800 - 192}{15}\right) = \pi \