Question Find the volume of a solid of revolution formed by revolving the region bounded above by the graph of \( f(x)=x+4 \) below by the \( x \)-axis over the interval \( [1,4] \) about the line \( y=-2 \). Enter an exact value in terms of \( \pi \).

To find the volume of the solid of revolution formed by revolving the region bounded by \( f(x) = x + 4 \) and the \( x \)-axis over the interval \([1, 4]\) about the line \( y = -2 \), we can use the washer method. First, we determine the outer radius \( R \) and the inner radius \( r \): 1. The upper curve is \( f(x) = x + 4 \). 2. The distance from \( y = -2 \) to \( y = f(x) \) (outer radius) is given by: \[ R(x) = (x + 4) - (-2) = x + 6 \] 3. The distance from \( y = -2 \) to the \( x \)-axis (inner radius) is: \[ r(x) = 0 - (-2) = 2 \] Next, we set up the volume integral: \[ V = \pi \int_{1}^{4} \left( R(x)^2 - r(x)^2 \right) \, dx = \pi \int_{1}^{4} \left( (x + 6)^2 - (2)^2 \right) \, dx \] Calculating the squares: \[ R(x)^2 = (x + 6)^2 = x^2 + 12x + 36 \] \[ r(x)^2 = 2^2 = 4 \] Now substitute back into the volume integral: \[ V = \pi \int_{1}^{4} (x^2 + 12x + 36 - 4) \, dx = \pi \int_{1}^{4} (x^2 + 12x + 32) \, dx \] Integrating term by term: \[ V = \pi \left[ \frac{x^3}{3} + 6x^2 + 32x \right]_{1}^{4} \] Calculating at the bounds: 1. Upper bound at \( x = 4 \): \[ \frac{4^3}{3} + 6(4^2) + 32(4) = \frac{64}{3} + 6(16) + 128 = \frac{64}{3} + 96 + 128 = \frac{64}{3} + \frac{288}{3} + \frac{384}{3} = \frac{836}{3} \] 2. Lower bound at \( x = 1 \): \[ \frac{1^3}{3} + 6(1^2) + 32(1) = \frac{1}{3} + 6 + 32 = \frac{1}{3} + \frac{18}{3} + \frac{96}{3} = \frac{115}{3} \] Now subtract: \[ V = \pi \left( \frac{836}{3} - \frac{115}{3} \right) = \pi \left( \frac{721}{3} \right) = \frac{721\pi}{3} \] So, the volume of the solid of revolution is: \[ \frac{721\pi}{3} \]