Question What is the volume of the solid of revolution obtained by rotating the region bounded by \( f(x)=x^{2}+1 \) and the \( x \) - over the interval \( [1,3] \) about the \( y \)-axis? Enter your answer in terms of \( \pi \).

To find the volume of the solid of revolution obtained by rotating the region bounded by \( f(x)=x^{2}+1 \) and the \( x \)-axis over the interval \( [1,3] \) about the \( y \)-axis, we can use the shell method. The formula for the volume using this method is: \[ V = 2\pi \int_{1}^{3} x(f(x)) \, dx \] Substituting \( f(x) \) into the equation gives us \( f(x) = x^2 + 1 \). Thus, we have: \[ V = 2\pi \int_{1}^{3} x(x^2 + 1) \, dx = 2\pi \int_{1}^{3} (x^3 + x) \, dx \] Calculating the integral: \[ \int (x^3 + x) \, dx = \frac{x^4}{4} + \frac{x^2}{2} \] Evaluating from \( 1 \) to \( 3 \): \[ = \left[\frac{3^4}{4} + \frac{3^2}{2}\right] - \left[\frac{1^4}{4} + \frac{1^2}{2}\right] = \left[\frac{81}{4} + \frac{9}{2}\right] - \left[\frac{1}{4} + \frac{1}{2}\right] \] Converting \( \frac{9}{2} \) to quarters gives \( \frac{18}{4} \): \[ = \left[\frac{81}{4} + \frac{18}{4}\right] - \left[\frac{1}{4} + \frac{2}{4}\right] = \frac{99}{4} - \frac{3}{4} = \frac{96}{4} = 24 \] Therefore, the volume \( V \) is: \[ V = 2\pi \times 24 = 48\pi \] So, the volume of the solid of revolution is \( \boxed{48\pi} \).