Question Find the volume of a solid of revolution formed by revolving the region bounded above by the graph of \( f(x)=3 x+2 \) below by the \( x \)-axis over the interval \( [2,3] \) about the line \( y=-2 \). Enter an exact value in terms of \( \pi \).

To find the volume of the solid formed by revolving the region bounded by \( f(x) = 3x + 2 \) and the \( x \)-axis over the interval \([2, 3]\) about the line \( y = -2 \), we can use the washer method. The function intersects the x-axis when \( f(x) = 0 \): \[ 3x + 2 = 0 \implies x = -\frac{2}{3} \] However, for our calculation within the interval \([2, 3]\), we don't reach the x-axis. The outer radius \( R(x) \) and inner radius \( r(x) \) are given by the distance from the line \( y = -2 \): - Outer radius \( R(x) = (3x + 2) - (-2) = 3x + 4 \) - Inner radius \( r(x) = 0 - (-2) = 2 \) The volume \( V \) is calculated using the formula for washers: \[ V = \pi \int_{a}^{b} \left( R(x)^2 - r(x)^2 \right) \, dx \] Substituting our limits \( a=2 \), \( b=3 \): \[ V = \pi \int_{2}^{3} \left( (3x + 4)^2 - (2)^2 \right) \, dx \] To solve this, we first expand the equation: \[ (3x + 4)^2 = 9x^2 + 24x + 16 \] \[ (2)^2 = 4 \] So the integral simplifies to: \[ V = \pi \int_{2}^{3} \left( 9x^2 + 24x + 16 - 4 \right) \, dx = \pi \int_{2}^{3} (9x^2 + 24x + 12) \, dx \] Now we compute the integral: \[ \int (9x^2 + 24x + 12) \, dx = 3x^3 + 12x^2 + 12x \] Evaluating it from 2 to 3: \[ \left[ 3(3)^3 + 12(3)^2 + 12(3) \right] - \left[ 3(2)^3 + 12(2)^2 + 12(2) \right] \] Calculating separately: For \( x = 3 \): \[ 3(27) + 12(9) + 36 = 81 + 108 + 36 = 225 \] For \( x = 2 \): \[ 3(8) + 12(4) + 24 = 24 + 48 + 24 = 96 \] Thus, \[ V = \pi \left( 225 - 96 \right) = 129\pi \] Therefore, the volume of the solid of revolution is: \[ \boxed{129\pi} \]