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 $\pi$.

Step 1 ：Define $R$ as the region bounded by the graphs of $g(y)=3 y^{2}+1, y=-1, y=1$, and the $y$-axis.

Step 2 ：The volume of a solid of revolution generated by rotating a region about the y-axis can be found using the disk method. The formula for the disk method is $\pi \int_{a}^{b} [f(y)]^2 dy$, where $f(y)$ is the function defining the region and $a$ and $b$ are the bounds of the region.

Step 3 ：In this case, $f(y) = g(y) = 3y^2 + 1$, and the bounds are $y = -1$ and $y = 1$.

Step 4 ：Substitute $f(y)$ and the bounds into the formula, we get $\pi \int_{-1}^{1} [3y^2 + 1]^2 dy$.

Step 5 ：Solve the integral, we get the volume of the solid of revolution generated by rotating the region $R$ about the y-axis is $\frac{48\pi}{5}$.

Step 6 ：Final Answer: $\boxed{\frac{48\pi}{5}}$