Question Define \( R \) as the region bounded by the graphs of \( g(y)=3 y^{2}+1, y=-1, y=1 \), and the \( y \)-axis. Using the disk method, what is the volume of the solid of revolution generated by rotating \( R \) about the \( y \)-axis? Enter your answer in terms of \( \pi \).

To find the volume of the solid of revolution generated by rotating the region \( R \) around the \( y \)-axis using the disk method, we will first determine the boundaries of \( R \) and then set up the integral. The function \( g(y) = 3y^2 + 1 \) describes the right boundary of the region, while the vertical lines \( y = -1 \) and \( y = 1 \) define the height of the region. The left boundary of the region is the \( y \)-axis, where \( x = 0 \). When we rotate the region \( R \) around the \( y \)-axis, the radius of the disk at a given \( y \) is given by \( g(y) \), which represents the distance from the \( y \)-axis to the curve \( g(y) \). The volume \( V \) of the solid formed is given by the formula: \[ V = \pi \int_{y=a}^{y=b} (radius)^2 \, dy \] In this case, the integral is set up as follows: \[ V = \pi \int_{y=-1}^{y=1} (3y^2 + 1)^2 \, dy \] Now, we compute the integral: 1. Expand \( (3y^2 + 1)^2 \): \[ (3y^2 + 1)^2 = 9y^4 + 6y^2 + 1 \] 2. Set up the integral: \[ V = \pi \int_{-1}^{1} (9y^4 + 6y^2 + 1) \, dy \] 3. Calculate the integral term by term: - \(\int_{-1}^{1} 9y^4 \, dy = 9 \cdot \left[ \frac{y^5}{5} \right]_{-1}^{1} = 9 \cdot \left(\frac{1}{5} - \frac{(-1)^5}{5}\right) = 9 \cdot \frac{2}{5} = \frac{18}{5}\) - \(\int_{-1}^{1} 6y^2 \, dy = 6 \cdot \left[ \frac{y^3}{3} \right]_{-1}^{1} = 6 \cdot \left(\frac{1}{3} - \frac{(-1)^3}{3}\right) = 6 \cdot \frac{2}{3} = 4\) - \(\int_{-1}^{1} 1 \, dy = [y]_{-1}^{1} = 1 - (-1) = 2\) 4. Now, combine the results: \[ V = \pi \left( \frac{18}{5} + 4 + 2 \right) = \pi \left( \frac{18}{5} + \frac{20}{5} + \frac{10}{5} \right) = \pi \left( \frac{48}{5} \right) \] Thus, the final answer for the volume of the solid of revolution generated by rotating region \( R \) about the \( y \)-axis is: \[ \boxed{\frac{48\pi}{5}} \]