Let \( R \) be the region bounded by the following curves. Use the method of your choice to find the volume of the solid generated when \( R \) is revolved about the \( x \)-axis. \( y=4 x^{2} \) and \( y=20-x^{2} \) The volume of the solid is (Type an exact answer.)
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To find the volume of the solid generated by revolving the region \( R \) bounded by the curves \( y = 4x^2 \) and \( y = 20 - x^2 \) about the \( x \)-axis, we first need to find the points of intersection of the two curves. Setting \( 4x^2 = 20 - x^2 \): \[ 4x^2 + x^2 = 20 \\ 5x^2 = 20 \\ x^2 = 4 \\ x = \pm 2 \] The points of intersection are \( x = -2 \) and \( x = 2 \). Next, we determine the volume \( V \) using the washer method: \[ V = \pi \int_{-2}^{2} \left( (20 - x^2)^2 - (4x^2)^2 \right) \, dx \] Calculating \( (20 - x^2)^2 \) and \( (4x^2)^2 \): \[ (20 - x^2) = 400 - 40x^2 + x^4 \] \[ (4x^2)^2 = 16x^4 \] Now substituting these values in the volume integral: \[ V = \pi \int_{-2}^{2} \left( (400 - 40x^2 + x^4) - (16x^4) \right) \, dx \] \[ = \pi \int_{-2}^{2} (400 - 40x^2 - 15x^4) \, dx \] Because the integrand is even, we can simplify the volume calculation: \[ V = 2\pi \int_{0}^{2} (400 - 40x^2 - 15x^4) \, dx \] Calculating the integral: \[ \int (400 - 40x^2 - 15x^4) \, dx = 400x - \frac{40}{3}x^3 - 3x^5/5 \] Evaluating from 0 to 2: \[ = \left[ 400(2) - \frac{40}{3}(2^3) - 3\frac{(2^5)}{5} \right] - \left[ 400(0) - \frac{40}{3}(0^3) - 3\frac{(0^5)}{5} \right] \] \[ = 800 - \frac{40}{3}(8) - \frac{3}{5}(32) \] \[ = 800 - \frac{320}{3} - \frac{96}{5} \] Finding a common denominator (15): \[ = 800 - \frac{1600}{15} - \frac{288}{15} \\ = 800 - \frac{1888}{15} \\ = \frac{12000}{15} - \frac{1888}{15} \\ = \frac{10112}{15} \] Now multiplying by \( 2\pi \): \[ V = 2\pi \cdot \frac{10112}{15} = \frac{20224\pi}{15} \] Thus, the volume of the solid is: \[ \boxed{\frac{20224\pi}{15}} \]