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 \( y=5-x^{2} \)

To find the volume of the solid generated by revolving the region \( R \) bounded by the curves \( y = 4x^2 \) and \( y = 5 - x^2 \) about the \( x \)-axis, we'll use the **method of washers**. Here's a step-by-step solution: ### 1. **Find the Points of Intersection** First, determine where the two curves intersect to establish the limits of integration. Set \( 4x^2 = 5 - x^2 \): \[ 4x^2 + x^2 = 5 \\ 5x^2 = 5 \\ x^2 = 1 \\ x = \pm 1 \] So, the curves intersect at \( x = -1 \) and \( x = 1 \). ### 2. **Determine the Outer and Inner Radii** When revolving around the \( x \)-axis, the **outer radius** \( R(x) \) is the distance from the \( x \)-axis to the higher function, and the **inner radius** \( r(x) \) is the distance to the lower function. For \( -1 \leq x \leq 1 \): - The upper function is \( y = 5 - x^2 \). - The lower function is \( y = 4x^2 \). Thus: \[ R(x) = 5 - x^2 \\ r(x) = 4x^2 \] ### 3. **Set Up the Integral Using the Washer Method** The volume \( V \) is given by: \[ V = \pi \int_{a}^{b} \left[ R(x)^2 - r(x)^2 \right] dx \] Plugging in the expressions for \( R(x) \) and \( r(x) \): \[ V = \pi \int_{-1}^{1} \left[ (5 - x^2)^2 - (4x^2)^2 \right] dx \] ### 4. **Simplify the Integrand** Expand and simplify the integrand: \[ (5 - x^2)^2 = 25 - 10x^2 + x^4 \\ (4x^2)^2 = 16x^4 \\ \Rightarrow (5 - x^2)^2 - (4x^2)^2 = 25 - 10x^2 + x^4 - 16x^4 = 25 - 10x^2 - 15x^4 \] So the integral becomes: \[ V = \pi \int_{-1}^{1} (25 - 10x^2 - 15x^4) dx \] ### 5. **Leverage Symmetry to Simplify Integration** Since the integrand is an even function (symmetric about the \( y \)-axis), we can simplify the integral by doubling the integral from 0 to 1: \[ V = 2\pi \int_{0}^{1} (25 - 10x^2 - 15x^4) dx \] ### 6. **Evaluate the Integral** Compute the integral term by term: \[ \int_{0}^{1} 25 \, dx = 25x \Big|_{0}^{1} = 25(1) - 25(0) = 25 \\ \int_{0}^{1} 10x^2 \, dx = \frac{10}{3}x^3 \Big|_{0}^{1} = \frac{10}{3} \\ \int_{0}^{1} 15x^4 \, dx = 3x^5 \Big|_{0}^{1} = 3 \] Combine these results: \[ \int_{0}^{1} (25 - 10x^2 - 15x^4) dx = 25 - \frac{10}{3} - 3 = \frac{56}{3} \] ### 7. **Calculate the Final Volume** Multiply by \( 2\pi \): \[ V = 2\pi \times \frac{56}{3} = \frac{112}{3}\pi \] ### **Final Answer** The volume of the solid is \(\dfrac{112}{3}\pi\).