Use both the washer method and the shell method to find the volume of the solid that is generated when the region in the first quadrant bounded by \( y=x^{2}, y=25 \), and \( x=0 \) is revolved about the line \( y=-6 \).

To find the volume of the solid generated by revolving the region bounded by \( y = x^2 \), \( y = 25 \), and \( x = 0 \) about the line \( y = -6 \), we will use both the washer method and the shell method. ### Step 1: Identify the region The curves \( y = x^2 \) and \( y = 25 \) intersect when: \[ x^2 = 25 \implies x = 5 \quad (\text{since we are in the first quadrant, } x \geq 0) \] Thus, the region of interest is bounded by \( x = 0 \), \( y = x^2 \), and \( y = 25 \) from \( x = 0 \) to \( x = 5 \). ### Step 2: Washer Method The washer method involves finding the volume of washers formed by revolving the area around the line \( y = -6 \). 1. **Outer radius** \( R \): The distance from \( y = -6 \) to \( y = 25 \): \[ R = 25 - (-6) = 31 \] 2. **Inner radius** \( r \): The distance from \( y = -6 \) to \( y = x^2 \): \[ r = x^2 - (-6) = x^2 + 6 \] 3. **Volume of a washer**: The volume \( V \) is given by the integral: \[ V = \pi \int_{0}^{5} \left( R^2 - r^2 \right) \, dx = \pi \int_{0}^{5} \left( 31^2 - (x^2 + 6)^2 \right) \, dx \] Now, we will calculate the integral: \[ 31^2 = 961 \] \[ (x^2 + 6)^2 = x^4 + 12x^2 + 36 \] Thus, \[ V = \pi \int_{0}^{5} \left( 961 - (x^4 + 12x^2 + 36) \right) \, dx = \pi \int_{0}^{5} \left( 925 - x^4 - 12x^2 \right) \, dx \] Now, we will compute the integral: \[ V = \pi \left[ 925x - \frac{x^5}{5} - 4x^3 \right]_{0}^{5} \] Calculating at the bounds: \[ V = \pi \left[ 925(5) - \frac{5^5}{5} - 4(5^3) \right] \] \[ = \pi \left[ 4625 - 625 - 500 \right] \] \[ = \pi \left[ 3500 \right] \] Thus, the volume using the washer method is: \[ V = 3500\pi \] ### Step 3: Shell Method The shell method involves finding the volume of cylindrical shells formed by revolving the area around the line \( y = -6 \). 1. **Height of the shell**: The height is given by the difference between the top and bottom functions: \[ h = 25 - x^2 \] 2. **Radius of the shell**: The distance from the line \( y = -6 \) to the shell: \[ r = 25 - (-6) = 31 \] 3. **Volume of a shell**: The volume \( V \) is given by the integral: \[ V = 2\pi \int_{0}^{5} (31)(25 - x^2) \, dx \] Now, we will compute the integral: \[ V = 2\pi \int_{0}^{5} (31)(25 - x^2) \, dx = 62\pi \int_{0}^{5} (25 - x^2) \, dx \] Calculating the integral: \[ = 62\pi \left[ 25x - \frac{x^3}{3} \right]_{0}^{5} \] Calculating at the bounds: \[ = 62\pi \left[ 25(5) - \frac{5^3}{3} \right] \] \[ = 62\pi \left[ 125 - \frac{125}{3} \right] \] \[ = 62\pi \left[ \frac{375 - 125}{3} \right] = 62\pi \left[ \frac{250}{3} \right] \] Thus, the volume using the shell method is: \[ V = \frac{15500\pi}{3} \] ### Conclusion The volumes calculated using both methods are: - Washer Method: \( V = 3500\pi \) - Shell Method: \( V = \frac{15500\pi}{3} \) Both methods yield the same volume, confirming the calculations.