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 \( y = 4 \) around the line \( y = 5 \), we will use the washer method. First, we need to determine the points of intersection between the curves \( y = x^2 \) and \( y = 4 \). Setting them equal gives: \[ x^2 = 4 \implies x = \pm 2. \] Thus, the region of integration is bounded by \( x = -2 \) and \( x = 2 \). Next, we identify the outer and inner radii of the washers. The outer radius \( R(x) \) is the distance from the line \( y = 5 \) to the curve \( y = 4 \): \[ R(x) = 5 - 4 = 1. \] The inner radius \( r(x) \) is the distance from the line \( y = 5 \) to the curve \( y = x^2 \): \[ r(x) = 5 - x^2. \] Using the washer method, we can express the volume \( V \) as: \[ V = \pi \int_{-2}^{2} \left[ R(x)^2 - r(x)^2 \right] \, dx = \pi \int_{-2}^{2} \left[ 1^2 - (5 - x^2)^2 \right] \, dx. \] Now we simplify the expression: \[ V = \pi \int_{-2}^{2} \left[ 1 - (5 - x^2)^2 \right] \, dx. \] Calculating \( (5 - x^2)^2 \): \[ (5 - x^2)^2 = 25 - 10x^2 + x^4. \] Thus, the integral becomes: \[ V = \pi \int_{-2}^{2} \left[ 1 - (25 - 10x^2 + x^4) \right] \, dx = \pi \int_{-2}^{2} \left[-24 + 10x^2 - x^4\right] \, dx. \] Now we break it down: \[ V = \pi \left[ \int_{-2}^{2} -24 \, dx + \int_{-2}^{2} 10x^2 \, dx - \int_{-2}^{2} x^4 \, dx \right]. \] Calculating each integral: 1. \(\int_{-2}^{2} -24 \, dx = -24 \times (2 - (-2)) = -24 \times 4 = -96\). 2. For \(\int_{-2}^{2} 10x^2 \, dx\): \[ = 10 \times \left[ \frac{x^3}{3} \right]_{-2}^{2} = 10 \times \left[\frac{2^3}{3} - \frac{(-2)^3}{3}\right] = 10 \times \left[\frac{8}{3} + \frac{8}{3}\right] = 10 \times \frac{16}{3} = \frac{160}{3}. \] 3. For \(\int_{-2}^{2} x^4 \, dx\): \[ = \left[ \frac{x^5}{5} \right]_{-2}^{2} = \frac{2^5}{5} - \frac{(-2)^5}{5} = \frac{32}{5} + \frac{32}{5} = \frac{64}{5}. \] Now substituting these back: \[ V = \pi \left( -96 + \frac{160}{3} - \frac{64}{5} \right). \] To combine these, we first find a common denominator (which is 15): \[ -96 = -\frac{1440}{15}, \quad \frac{160}{3} = \frac{800}{15}, \quad \frac{64}{5} = \frac{192}{15}. \] Thus: \[ V = \pi \left( -\frac{1440}{15} + \frac{800}{15} - \frac{192}{15} \right) = \pi \left( \frac{-1440 + 800 - 192}{15} \right) = \pi \left( \frac{-832}{15} \right) = -\frac{832\pi}{15